9702/42 Feb/Mar 2024 — Physics Solved Past Paper

Q1MediumGravitational FieldsMotion in a Circle

(a)

Explain why the gravitational potential near to a point mass is negative.

DifficultyMedium-Easy

Worked solution

Answer

Gravitational potential is defined to be zero at infinity.

Since gravity is attractive, work must be done to move a unit mass from near the point mass out to infinity, so the potential energy per unit mass at finite $r$ is less than zero. Hence $\phi$ near the mass is negative.

Final answer

Gravitational potential is negative near a point mass because it is defined as zero at infinity and work is required to move unit mass to infinity against the attractive force, so the potential at finite r is less than zero.

Detailed explanation

Background Concept

Gravitational potential $\phi$ at a point is defined as the work done per unit mass by an external agent in bringing a small test mass from infinity to that point (with no change in kinetic energy). In equations,

\phi(r) = \frac{W_{\text{ext}}}{m}

A standard reference choice is $\phi = 0$ at $r \to \infty$ . This is a choice of zero level, like choosing where to measure height from.

For an attractive force such as gravity, the gravitational force on a test mass points towards the mass. If you move the test mass outwards (away from the mass), you must do positive work against this force.

Understanding the Question

The question asks specifically about the sign of gravitational potential near a point mass, i.e. why the values come out negative rather than positive.

You must connect:

the reference point ( $\phi = 0$ at infinity), and
the fact gravity is attractive (so moving away requires energy input).

Approach

Use the definition of potential with the conventional zero at infinity. Consider moving a unit mass from distance $r$ to infinity: if an external agent must do positive work to reach infinity, then the potential at $r$ must be below the zero level at infinity, i.e. negative.

Step-by-Step Reasoning

Take the standard reference:

\phi(\infty) = 0.

Gravity is attractive, so a test mass at distance $r$ experiences a force towards the mass.
To take the mass from $r$ out to infinity slowly (no gain in kinetic energy), an external agent must apply an outward force and do positive work.
If going from $r$ to $\infty$ requires positive external work per unit mass, then going from $\infty$ to $r$ would require negative external work per unit mass (the field does work on the mass). Therefore,

\phi(r) - \phi(\infty) < 0 \quad \Rightarrow \quad \phi(r) < 0.

So gravitational potential near the mass is negative.

Key Takeaways

$\phi$ is defined relative to a chosen zero, usually $\phi=0$ at infinity.
Because gravity is attractive, the potential at finite distance is below this reference and so is negative.

Common Mistakes

Saying “it is negative because gravity is attractive” without mentioning the reference level at infinity.
Confusing gravitational potential $\phi$ (J kg $^{-1}$ ) with gravitational potential energy $U$ (J). The sign argument is the same, but units differ.

Things to Be Careful About

The negativity is not because energy is “less than nothing” in an absolute sense; it is because of the chosen zero at infinity.
Always link sign to work done against the field and the direction of the gravitational force.

Techniques used

use the definition of gravitational potential relative to infinityrelate sign of work done to attractive gravitational forceinterpret potential as potential energy per unit mass

(b)

A planet may be assumed to be a uniform sphere. It has gravitational potential $\phi$ at distance $r$ from the centre of the planet.

The variation with $\frac{1}{r}$ of $\phi$ is shown in Fig. 1.1.

Fig. 1.1

8M

(i)

Show that the mass of the planet is $8.8 \times 10^{25} \text{ kg}$ .

2M Similar

DifficultyMedium

Worked solution

Working

Outside the planet,

\phi = -\frac{GM}{r}

So a graph of $\phi$ against $\frac{1}{r}$ has gradient $-GM$ .

From Fig. 1.1, using the origin and a point on the line (e.g. $\frac{1}{r}/10^{-8} = 4.0$ , $\phi/10^{8} = -2.35$ ):

\text{gradient} = \frac{-2.35\times 10^{8}}{4.0\times 10^{-8}} = -5.88\times 10^{15}

Hence

GM = 5.88\times 10^{15}

M = \frac{5.88\times 10^{15}}{6.67\times 10^{-11}} = 8.8\times 10^{25}\ \text{kg}

Answer

$M = 8.8 \times 10^{25}\ \text{kg}$

Final answer

8.8 × 10^25 kg

Detailed explanation

Background Concept

For a spherically symmetric mass (such as a uniform spherical planet), the gravitational field and potential outside the sphere are the same as if all the mass were concentrated at the centre.

The gravitational potential due to a point (or spherically symmetric) mass $M$ at distance $r$ is

\phi = -\frac{GM}{r}

This is potential energy per unit mass (units: J kg $^{-1}$ ). If you plot $\phi$ against $1/r$ you expect a straight line through the origin with gradient $-GM$ .

Understanding the Question

You are given a straight-line graph of $\phi$ versus $1/r$ (with scaled axes). You must use the gradient of this line to find $GM$ , then divide by $G$ to find $M$ .

Key clues:

straight line through the origin implies $\phi \propto 1/r$ ;
the known theoretical form is $\phi = -GM(1/r)$ .

Approach

Write $\phi = -GM/r$ .
Compare with the straight-line form $y = mx$ where $y = \phi$ and $x = 1/r$ .
Read two points on the line to get the gradient.
Convert from the axis scales to real units.
Use $\text{gradient} = -GM$ to obtain $M$ .

Step-by-Step Reasoning

Start from:

\phi = -\frac{GM}{r} = -GM\left(\frac{1}{r}\right)

So if we let $y=\phi$ and $x=1/r$ , then

y = (-GM)x

meaning the gradient is $-GM$ .

Use two points on the drawn best-fit line. The axes are labelled $\phi/10^{8}$ and $(1/r)/10^{-8}$ , so you must multiply the plotted values by the scale factors to get $\phi$ in J kg $^{-1}$ and $1/r$ in m $^{-1}$ .
Example using a point near the end of the line (any accurate read from the line is acceptable):

plotted: $(1/r)/10^{-8} = 4.0$ so $1/r = 4.0\times 10^{-8}\ \text{m}^{-1}$
plotted: $\phi/10^8 = -2.35$ so $\phi = -2.35\times 10^8\ \text{J kg}^{-1}$

Gradient:

\text{gradient} = \frac{\Delta \phi}{\Delta (1/r)} = \frac{-2.35\times 10^{8}}{4.0\times 10^{-8}} = -5.88\times 10^{15}

Set $-GM = -5.88\times 10^{15}$ so:

GM = 5.88\times 10^{15}

Divide by $G = 6.67\times 10^{-11}$ :

M = \frac{GM}{G} = \frac{5.88\times 10^{15}}{6.67\times 10^{-11}} = 8.8\times 10^{25}\ \text{kg}

Key Takeaways

Outside a spherical planet, $\phi = -GM/r$ .
A $\phi$ vs $1/r$ graph has gradient $-GM$ .
Always apply axis scale factors before computing gradients.

Common Mistakes

Using the plotted numbers directly without multiplying by $10^8$ and $10^{-8}$ .
Taking the gradient as $\Delta x/\Delta y$ instead of $\Delta y/\Delta x$ .
Forgetting the minus sign and concluding $GM$ is negative (instead, the gradient is negative because $\phi$ is negative).

Things to Be Careful About

Choose two well-separated points on the best-fit line (reduces percentage uncertainty in gradient).
Check units: gradient of $\phi$ (J kg $^{-1}$ ) vs $1/r$ (m $^{-1}$ ) has units J kg $^{-1}$ m, which matches $GM$ .
Small differences in read-off can slightly change $M$ ; exam marking usually allows for reading tolerance.

Techniques used

use the relation $\phi = -GM/r$ outside a spherical massinterpret the gradient of a straight-line graphconvert from scaled axes to physical units

(ii)

The period of rotation of the planet is 0.72 Earth days.

A satellite in orbit around the planet remains above the same point on the surface of the planet.

Use the mass of the planet in (b)(i) to determine the radius $R$ of the orbit of the satellite.

$R$ = ______ $\text{m}$

3M Similar

DifficultyMedium

Worked solution

Working

Geostationary orbit: orbital period $T$ equals planet rotation period.

T = 0.72\times 24\times 3600 = 6.22\times 10^{4}\ \text{s}

For a circular orbit,

T^2 = \frac{4\pi^2 R^3}{GM}

so

R = \left(\frac{GM T^2}{4\pi^2}\right)^{1/3}

With $M = 8.8\times 10^{25}\ \text{kg}$ , $G = 6.67\times 10^{-11}$ :

R = \left(\frac{(6.67\times 10^{-11})(8.8\times 10^{25})(6.22\times 10^{4})^2}{4\pi^2}\right)^{1/3} = 8.3\times 10^{7}\ \text{m}

Answer

$R = 8.3 \times 10^{7}\ \text{m}$

Final answer

8.3 × 10^7 m

Detailed explanation

Background Concept

A satellite that remains above the same point on a rotating planet is in a geostationary (more generally, synchronous) orbit. That requires its orbital period to match the planet’s rotation period.

For a circular orbit of radius $R$ around mass $M$ , gravity provides the centripetal force:

\frac{GMm}{R^2} = \frac{mv^2}{R}

This leads to an expression for the orbital period:

T = 2\pi\sqrt{\frac{R^3}{GM}}

or equivalently

T^2 = \frac{4\pi^2 R^3}{GM}.

Understanding the Question

You are told:

the planet’s rotation period is $0.72$ Earth days,
a satellite stays over the same point on the surface (so $T_{\text{orbit}} = T_{\text{rotation}}$ ),
use the planet’s mass from (b)(i).

You must find the orbital radius $R$ (distance from the planet’s centre).

Approach

Convert $0.72$ days into seconds.
Use the circular-orbit period formula and rearrange for $R$ .
Substitute $G$ , $M$ , and $T$ .

Step-by-Step Reasoning

Convert period to SI:

T = 0.72\times 24\times 3600 = 6.22\times 10^{4}\ \text{s}.

Start from

T^2 = \frac{4\pi^2 R^3}{GM}

Rearrange:

R^3 = \frac{GM T^2}{4\pi^2}

so

R = \left(\frac{GM T^2}{4\pi^2}\right)^{1/3}.

Substitute $M = 8.8\times 10^{25}\ \text{kg}$ and $G = 6.67\times 10^{-11}$ :

GM = (6.67\times 10^{-11})(8.8\times 10^{25}) \approx 5.87\times 10^{15}.

Evaluate:

R = \left(\frac{(5.87\times 10^{15})(6.22\times 10^{4})^2}{4\pi^2}\right)^{1/3} \approx 8.3\times 10^{7}\ \text{m}.

Key Takeaways

Geostationary condition: $T_{\text{orbit}} = T_{\text{rotation}}$ .
Circular orbit gives $T^2 \propto R^3$ .
Use SI units throughout before substituting.

Common Mistakes

Forgetting to convert days to seconds.
Using the planet’s radius instead of orbital radius from the centre.
Using $T = \sqrt{R^3/GM}$ but forgetting the factor $2\pi$ .

Things to Be Careful About

$R$ here is measured from the centre of the planet.
Keep enough significant figures through intermediate steps; round at the end.
The relationship assumes a circular orbit and that the satellite’s mass is negligible compared with the planet.

Techniques used

equate gravitational force to centripetal force for circular orbituse $T = 2\pi/\omega$ and $v = r\omega$rearrange the orbital-period relationship to solve for radiusconvert a time period into SI units

(iii)

The speed of the satellite in (b)(ii) is $8400 \text{ m s}^{-1}$ . The mass of the satellite is $1200 \text{ kg}$ .

Determine the additional energy required to move the satellite from its orbit to infinity.

energy required = ______ $\text{J}$

3M Similar

DifficultyMedium

Worked solution

Working

At infinity: $E = 0$ .

For a circular orbit,

E_{\text{total}} = -\frac{GMm}{2R}

so the energy required to reach infinity is

\Delta E = \frac{GMm}{2R} = \frac{1}{2}mv^2

\Delta E = \frac{1}{2}(1200)(8400)^2 = 4.23\times 10^{10}\ \text{J}

Answer

energy required $= 4.2 \times 10^{10}\ \text{J}$

Final answer

4.2 × 10^10 J

Detailed explanation

Background Concept

For a mass $m$ in a circular gravitational orbit of radius $R$ around a planet of mass $M$ :

Gravitational potential energy:

U = -\frac{GMm}{R}

Orbital speed comes from gravity providing centripetal force:

\frac{GMm}{R^2} = \frac{mv^2}{R} \Rightarrow v^2 = \frac{GM}{R}

Kinetic energy:

K = \frac{1}{2}mv^2 = \frac{GMm}{2R}

So the total mechanical energy in the circular orbit is

E = K + U = \frac{GMm}{2R} - \frac{GMm}{R} = -\frac{GMm}{2R}

At infinity (taking the usual reference), $U=0$ . If the satellite just reaches infinity and comes to rest, then $K=0$ as well, so total energy is $0$ .

Therefore energy that must be supplied to go from orbit to infinity is

\Delta E = 0 - \left(-\frac{GMm}{2R}\right) = \frac{GMm}{2R}

and since $\frac{GMm}{2R} = \frac{1}{2}mv^2$ , the required energy equals the satellite’s orbital kinetic energy.

Understanding the Question

You are given the satellite’s orbital speed $v = 8400\ \text{m s}^{-1}$ and mass $m = 1200\ \text{kg}$ . You must find the additional energy needed to move it from that orbit to infinity (where $E=0$ ).

This is essentially an “escape energy from orbit” calculation.

Approach

Use orbital energy:

Either compute $\Delta E = GMm/(2R)$ using $R$ from (ii),
or use the equivalent result $\Delta E = \frac{1}{2}mv^2$ (simplest here because $v$ is given).

Step-by-Step Reasoning

Energy needed is

\Delta E = \frac{1}{2}mv^2

Substitute $m=1200\ \text{kg}$ , $v=8400\ \text{m s}^{-1}$ :

\Delta E = \frac{1}{2}(1200)(8400)^2

(8400)^2 = 7.056\times 10^{7}

\Delta E = 600\times (7.056\times 10^{7}) = 4.2336\times 10^{10}\ \text{J}

Round suitably:

\Delta E \approx 4.2\times 10^{10}\ \text{J}.

Key Takeaways

For a circular orbit, $E_{\text{total}} = -GMm/(2R)$ .
The energy to escape to infinity (ending with zero speed) equals $|E_{\text{total}}|$ .
Conveniently, $|E_{\text{total}}| = K = \frac{1}{2}mv^2$ for a circular orbit.

Common Mistakes

Using only the potential energy change $GMm/R$ (that would ignore the initial kinetic energy).
Using $\Delta E = GMm/R$ instead of $GMm/(2R)$ .
Forgetting that the final total energy at infinity is taken as zero.

Things to Be Careful About

This result applies to a circular orbit. The equality $\Delta E = K$ depends on that.
Ensure the final condition is “to infinity” with zero speed; if non-zero final speed were required, you would add extra kinetic energy.
Keep units consistent: $\text{kg}$ , $\text{m s}^{-1}$ give $\text{J}$ .

Techniques used

use kinetic energy from orbital speeduse the total energy of a circular orbitapply energy conservation between orbit and infinity

Q2MediumThermodynamicsIdeal GasesTemperature

(a)

By referring to both kinetic energy and potential energy, explain what is meant by the internal energy of an ideal gas.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Internal energy is the total (random) molecular kinetic energy plus the molecular potential energy.

For an ideal gas the intermolecular forces are negligible, so the potential energy is constant (taken as zero) and the internal energy is due to the random kinetic energy of the molecules only.

Final answer

Internal energy = total random molecular kinetic + potential energy; for an ideal gas, potential energy is constant/negligible so internal energy depends only on random molecular kinetic energy.

Detailed explanation

Background Concept

The internal energy $U$ of a substance is the energy associated with the microscopic motion and interactions of its particles. In general,

it includes random kinetic energy of particles (translational, and possibly rotational/vibrational), and
it includes potential energy due to intermolecular forces (because molecules attract/repel each other).

For an ideal gas, one key assumption is that the molecules exert no intermolecular forces on each other except during collisions. That means there is no significant stored intermolecular potential energy that changes with separation; effectively the potential energy can be treated as constant.

Understanding the Question

You are asked to explain what “internal energy of an ideal gas” means, specifically mentioning kinetic energy and potential energy. So you must:

state what internal energy is in general, and
state what changes for an ideal gas.

Approach

Start with the general definition: sum of random kinetic + potential energies of molecules.
Then apply the ideal-gas assumption: negligible intermolecular forces $→$ potential energy is constant/negligible.
Conclude: internal energy of an ideal gas is due to random kinetic energy only (and so depends on temperature only).

Step-by-Step Reasoning

General statement: internal energy $U$ $U$ is the total microscopic energy of the molecules:
- random kinetic energy (due to random motion), and
- potential energy (due to intermolecular forces).
Ideal gas condition: molecules are treated as not interacting (except collisions), so their intermolecular potential energy does not change; it is taken as zero or constant.
Therefore, for an ideal gas, $U$ is the total random kinetic energy of the molecules.

Key Takeaways

Internal energy is a microscopic energy store.
For an ideal gas, changing $U$ means changing molecular kinetic energy, so $U$ depends only on $T$ .

Common Mistakes

Saying “internal energy is kinetic energy only” without first mentioning the general kinetic + potential definition.
Confusing internal energy with “heat” (thermal energy transfer) or “temperature”.
Claiming the ideal gas has no potential energy rather than “potential energy is constant/negligible”.

Things to Be Careful About

The question explicitly demands reference to both kinetic and potential energy.
Use the correct ideal-gas justification: negligible intermolecular forces.

Techniques used

state the definition of internal energy in terms of molecular energiesapply the ideal-gas assumption about intermolecular forceslink internal energy of an ideal gas to molecular kinetic energy only

(b)

A fixed mass of an ideal gas at a temperature of $20^\circ\text{C}$ is sealed in a cylinder by a piston, as shown in Fig. 2.1.

Fig. 2.1

The initial volume of the gas is $1.24 \times 10^{-4} \text{ m}^3$ .
Thermal energy is supplied to the gas and its volume increases by $5.20 \times 10^{-5} \text{ m}^3$ .

6M

(i)

The piston is freely moving so that the gas is always at atmospheric pressure.

Atmospheric pressure is $1.01 \times 10^5 \text{ Pa}$ .

Calculate the work done by the gas.

work done by gas = ______ $\text{J}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At constant pressure,

W = p\Delta V

W = (1.01 \times 10^{5})(5.20 \times 10^{-5}) = 5.252\ \text{J}

Answer

W = 5.25\ \text{J}

Final answer

5.25 J

Detailed explanation

Background Concept

When a gas expands against an external pressure, it does work on its surroundings. For a quasistatic expansion at constant pressure $p$ ,

W = p\Delta V

where:

$W$ is the work done by the gas (in joules, J),
$p$ is the pressure (in pascals, Pa),
$\Delta V$ is the increase in volume (in $\text{m}^3$ ).

(For a freely moving piston open to the atmosphere, the gas pressure stays equal to atmospheric pressure, so $p$ is constant.)

Understanding the Question

The piston moves freely, so the gas always remains at atmospheric pressure $p = 1.01 \times 10^{5}\ \text{Pa}$ . The volume increase is $\Delta V = 5.20 \times 10^{-5}\ \text{m}^3$ . You are asked for the work done by the gas during this expansion.

Approach

Because the pressure is constant, use $W = p\Delta V$ and substitute the given values directly (they are already in SI units).

Step-by-Step Reasoning

diagram-b-i

Constant pressure implies the work is

W = p\Delta V.

Substitute:

W = (1.01 \times 10^{5})(5.20 \times 10^{-5}).

Multiply powers of ten and numbers:

W = 5.252\ \text{J} \approx 5.25\ \text{J}.

Key Takeaways

For constant pressure processes, work is the simple area under a $p$ - $V$ graph: $W = p\Delta V$ .
Ensure $p$ is in Pa and $\Delta V$ in $\text{m}^3$ so that $\text{Pa}\cdot\text{m}^3 = \text{J}$ .

Common Mistakes

Using the final volume instead of the change in volume.
Mixing up work done by the gas vs work done on the gas (sign convention).
Converting units unnecessarily and making a power-of-ten error.

Things to Be Careful About

The piston is “freely moving” specifically to signal constant pressure.
Quote to 2–3 significant figures to match given data (here 3 s.f.).

Techniques used

use work done at constant pressure W = p\Delta Vsubstitute values consistently in SI unitsquote the final answer to appropriate significant figures

(ii)

Calculate the final thermodynamic temperature $T$ of the gas.

$T$ = ______ $\text{K}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Initial temperature:

T_1 = 20^{\circ}\text{C} + 273 = 293\ \text{K}

Final volume:

V_2 = 1.24 \times 10^{-4} + 5.20 \times 10^{-5} = 1.76 \times 10^{-4}\ \text{m}^3

At constant $p$ and fixed mass, $\dfrac{V}{T} = \text{constant}$ , so

T_2 = T_1\frac{V_2}{V_1} = 293\times \frac{1.76 \times 10^{-4}}{1.24 \times 10^{-4}} = 4.16 \times 10^{2}\ \text{K}

Answer

T = 416\ \text{K}

Final answer

416 K

Detailed explanation

Background Concept

The ideal gas equation is

pV = nRT,

where $p$ is pressure, $V$ is volume, $n$ is amount of gas (moles), $R$ is the gas constant, and $T$ is thermodynamic temperature in kelvin.

If the mass (so $n$ ) is fixed and the piston moves freely so $p$ stays constant, then

V \propto T \quad \Rightarrow \quad \frac{V}{T} = \text{constant}.

So a volume ratio directly gives a temperature ratio.

Understanding the Question

Given:

initial temperature $20^{\circ}\text{C}$ (must convert to K),
initial volume $V_1 = 1.24 \times 10^{-4}\ \text{m}^3$ ,
volume increase $\Delta V = 5.20 \times 10^{-5}\ \text{m}^3$ ,
piston is freely moving so pressure is constant.

Find the final temperature $T_2$ in kelvin.

Approach

Convert $20^{\circ}\text{C}$ to kelvin.
Find the final volume $V_2 = V_1 + \Delta V$ .
Use $V/T$ constant at constant $p$ and fixed $n$ :

T_2 = T_1 \frac{V_2}{V_1}.

Step-by-Step Reasoning

Convert to kelvin:

T_1 = 20 + 273 = 293\ \text{K}.

Final volume:

V_2 = 1.24 \times 10^{-4} + 5.20 \times 10^{-5} = 1.76 \times 10^{-4}\ \text{m}^3.

Apply $V/T = \text{constant}$ :

\frac{V_1}{T_1} = \frac{V_2}{T_2} \Rightarrow T_2 = T_1\frac{V_2}{V_1}.

Substitute:

T_2 = 293\times\frac{1.76}{1.24} \approx 293\times 1.419 \approx 416\ \text{K}.

Key Takeaways

Always use kelvin in gas-law calculations.
With constant $p$ and fixed mass, temperature changes in the same ratio as volume.

Common Mistakes

Using $T = 20\ \text{K}$ or $T = 20$ in the formula (forgetting to add 273).
Using $\Delta V$ instead of $V_2$ in the ratio.
Assuming $pV$ constant (that would be constant temperature, not constant pressure).

Things to Be Careful About

Keep the powers of ten consistent; here they cancel in the ratio, but only if you form it correctly.
Give the final answer to 3 s.f. to match the data.

Techniques used

convert Celsius temperature to kelvinuse proportionality from pV = nRT at constant pressure and fixed amount of gascompute final volume from initial volume plus volume change

(iii)

The mass of the gas is $16 \text{ g}$ . For this expansion, there is a net transfer of $960 \text{ J}$ of thermal energy to the gas.

Calculate the specific heat capacity $c$ of the gas at this pressure.

$c$ = ______ $\text{J kg}^{-1} \text{ K}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Mass:

m = 16\ \text{g} = 0.016\ \text{kg}

Temperature rise:

\Delta T = 416 - 293 = 123\ \text{K}

Using $Q = mc\Delta T$ at this pressure:

c = \frac{Q}{m\Delta T} = \frac{960}{0.016 \times 123} = 4.88 \times 10^{2}\ \text{J kg}^{-1}\ \text{K}^{-1}

Answer

c = 4.88 \times 10^{2}\ \text{J kg}^{-1}\ \text{K}^{-1}

Final answer

4.88 × 10^2 J kg^-1 K^-1

Detailed explanation

Background Concept

The specific heat capacity $c$ is defined by

Q = mc\Delta T,

where:

$Q$ is the thermal energy transferred (J),
$m$ is mass (kg),
$\Delta T$ is temperature change (K),
$c$ is in $\text{J kg}^{-1}\ \text{K}^{-1}$ .

At constant pressure, the thermal energy supplied to a gas goes into:

increasing internal energy (raising molecular kinetic energy), and
doing work as the gas expands.

That is why, physically, the constant-pressure heat capacity is larger than the constant-volume one (developed further in part (c)). In this part you are simply asked to calculate $c$ for the given constant-pressure process using the provided $Q$ .

Understanding the Question

From part (b), the gas has been heated from $20^{\circ}\text{C}$ ( $293\ \text{K}$ ) to the final temperature found in (b)(ii) ( $416\ \text{K}$ ). During this expansion, the net thermal energy transferred to the gas is $Q = 960\ \text{J}$ and the mass is $16\ \text{g}$ .

You must find $c$ for this process (constant atmospheric pressure).

Approach

Convert mass to kg.
Compute $\Delta T$ using the initial and final temperatures.
Rearrange $Q = mc\Delta T$ to $c = Q/(m\Delta T)$ and substitute.

Step-by-Step Reasoning

Convert mass:

16\ \text{g} = 16 \times 10^{-3}\ \text{kg} = 0.016\ \text{kg}.

Temperature rise:

\Delta T = 416 - 293 = 123\ \text{K}.

(Temperature differences are the same in K and in $^{\circ}\text{C}$ , but you must use kelvin values consistently.)

Compute $c$ :

c = \frac{Q}{m\Delta T} = \frac{960}{0.016 \times 123}.

Evaluate the denominator:

0.016 \times 123 = 1.968

so

c = \frac{960}{1.968} \approx 4.88 \times 10^{2}\ \text{J kg}^{-1}\ \text{K}^{-1}.

Key Takeaways

Use $Q = mc\Delta T$ with $m$ in kg and $\Delta T$ in K.
For constant pressure heating, $Q$ includes energy for both $\Delta U$ and expansion work.

Common Mistakes

Using $m = 16\ \text{kg}$ or $m = 16\ \text{g}$ directly without converting to kg.
Using $\Delta T = 416$ or $\Delta T = 293$ instead of the difference.
Mixing Celsius and kelvin in a way that changes the temperature difference.

Things to Be Careful About

Use the final temperature from part (b)(ii), not a re-calculated or rounded intermediate that changes $\Delta T$ significantly.
Quote units correctly as $\text{J kg}^{-1}\ \text{K}^{-1}$ and give a sensible number of significant figures.

Techniques used

use the definition of specific heat capacity c = Q/(m\Delta T)convert mass from grams to kilogramsuse the temperature change found from earlier parts

(c)

The gas in (b) is allowed to return to its starting temperature. The piston is now fixed in position.

Thermal energy is supplied to increase the temperature to the same final temperature as in (b).

Use the first law of thermodynamics to suggest and explain how the specific heat capacity of the gas for this situation compares with the value in (b)(iii).

3M Similar

DifficultyMedium

Worked solution

Answer

With the piston fixed, $\Delta V = 0$ so

W = p\Delta V = 0.

From the first law, for the same temperature rise $\Delta U$ is the same, but:

in (b) (constant pressure) some energy is used to do work, so $q = \Delta U + W$ ,
here (constant volume) $q = \Delta U$ .

Therefore less thermal energy is needed for the same $\Delta T$ , so the specific heat capacity is smaller than in (b)(iii) (i.e. $c_V < c_P$ ).

Final answer

Specific heat capacity is smaller than in (b)(iii) (constant volume: W = 0 so less energy needed for same ΔT; cV < cP).

Detailed explanation

Background Concept

The first law of thermodynamics (Cambridge convention) is

\Delta U = q + W,

where:

$\Delta U$ is the change in internal energy of the system,
$q$ is the thermal energy transferred to the system,
$W$ is the work done on the system.

If we instead talk about work done by the gas ( $W_{\text{by}}$ ), then $W = -W_{\text{by}}$ and the first law is often written as

q = \Delta U + W_{\text{by}}.

For a constant-volume process, $\Delta V = 0$ so boundary work is zero:

W_{\text{by}} = p\Delta V = 0.

Specific heat capacity depends on the constraint:

At constant pressure: $Q = mc_P\Delta T$ .
At constant volume: $Q = mc_V\Delta T$ .

Understanding the Question

You compare two ways of heating the same gas through the same temperature rise (from the starting temperature back up to the same final temperature found in (b)):

In (b): piston free (constant pressure), so the gas expands and does work.
Now: piston fixed (constant volume), so no expansion work is done.

You must use the first law to state how the specific heat capacity in the fixed-volume case compares with that in (b)(iii), and explain why.

Approach

Identify that fixed piston means $\Delta V = 0 \Rightarrow$ no work done.
Use the first law to argue how the supplied heat $q$ differs between the two processes for the same $\Delta T$ .
Relate $q$ to $mc\Delta T$ to compare the heat capacities.

Step-by-Step Reasoning

Constant volume implies no work:

\Delta V = 0 \Rightarrow W_{\text{by}} = p\Delta V = 0.

For the same gas and the same temperature increase, the change in internal energy $\Delta U$ is the same (for an ideal gas, $U$ depends only on $T$ ).
Apply the first law in the form $q = \Delta U + W_{\text{by}}$ :

In (b) at constant pressure: $W_{\text{by}} > 0$ because the gas expands, so

q_{P} = \Delta U + W_{\text{by}} \quad \text{(larger)}.

With the piston fixed (constant volume): $W_{\text{by}} = 0$ , so

q_{V} = \Delta U \quad \text{(smaller)}.

Since $q = mc\Delta T$ , for the same $m$ and same $\Delta T$ , a smaller required $q$ means a smaller $c$ .

Therefore the specific heat capacity at constant volume is less than the value in (b)(iii) (which is at constant pressure):

c_V < c_P.

Key Takeaways

Fixing the volume removes expansion work.
At constant pressure, some supplied energy goes into doing work, so more heat is needed per kelvin.
Hence $c_P$ is greater than $c_V$ .

Common Mistakes

Saying $\Delta U = 0$ because volume is fixed (wrong: $\Delta U$ depends on temperature change).
Forgetting that the comparison is for the same final temperature (same $\Delta T$ ).
Mixing up sign conventions for $W$ in the first law; the essential physics is: constant pressure needs extra energy for expansion work.

Things to Be Careful About

Explicitly state $\Delta V = 0 \Rightarrow W = 0$ for the fixed piston case to earn the mark.
Make the comparison clear: smaller heat input needed for same $\Delta T$ implies smaller specific heat capacity.

Techniques used

apply the first law of thermodynamics to compare two processesidentify when work done is zero from constant volumededuce relative sizes of specific heat capacities from energy partition

Q3MediumOscillations

A small object of mass $24 \text{ g}$ rests on a platform. The platform is attached to an oscillator, as shown in Fig. 3.1.

Fig. 3.1

The oscillator moves the platform up and down.

(a)

The total energy of the oscillations of the object is $2.2 \times 10^{-4} \text{ J}$ .
In one oscillation the object travels a total distance of $14 \text{ mm}$ .

Calculate the angular frequency $\omega$ of the oscillations.

$\omega$ = ______ $\text{rad s}^{-1}$

3M Similar

DifficultyMedium

Worked solution

Working

Total distance in one oscillation $=4A$ .

A=\frac{14\times 10^{-3}}{4}=3.5\times 10^{-3}\ \text{m}

Total energy for SHM:

E=\frac{1}{2}m\omega^2A^2

\omega=\sqrt{\frac{2E}{mA^2}}=\sqrt{\frac{2(2.2\times 10^{-4})}{0.024(3.5\times 10^{-3})^2}}

\omega=3.87\times 10^{1}\ \text{rad s}^{-1}

Answer

$\omega \approx 39\ \text{rad s}^{-1}$

Final answer

39 rad s^-1

Detailed explanation

Background Concept

For simple harmonic motion (SHM), the total mechanical energy of the oscillating mass is constant (if we neglect damping). It can be written in terms of the amplitude $A$ and angular frequency $\omega$ as

E=\frac{1}{2}m\omega^2A^2

This comes from the maximum kinetic energy (at the equilibrium position) or the maximum potential energy (at the turning points). The key idea is that the energy depends on both how far it oscillates (the amplitude) and how “fast” the oscillation is (via $\omega$ ).

Understanding the Question

You are told:

mass $m = 24\ \text{g} = 0.024\ \text{kg}$
total energy $E = 2.2\times 10^{-4}\ \text{J}$
in one full oscillation, the object travels a total distance of $14\ \text{mm} = 14\times 10^{-3}\ \text{m}$

You must calculate the angular frequency $\omega$ .

The travel distance clue is how to find the amplitude $A$ .

Approach

Convert the stated “total distance travelled in one oscillation” into amplitude $A$ .
Substitute $m$ , $E$ , and $A$ into

E=\frac{1}{2}m\omega^2A^2

and rearrange for $\omega$ .

Step-by-Step Reasoning

1) Convert distance per cycle to amplitude
In one complete cycle, the object moves:

from $+A$ to $-A$ : distance $2A$
then from $-A$ back to $+A$ : distance $2A$

So total distance in one cycle is $4A$ .

4A = 14\times 10^{-3}\ \text{m} \ \Rightarrow\ A = \frac{14\times 10^{-3}}{4} = 3.5\times 10^{-3}\ \text{m}

2) Use the SHM energy equation

E=\frac{1}{2}m\omega^2A^2

Rearrange:

\omega = \sqrt{\frac{2E}{mA^2}}

Substitute values:

\omega=\sqrt{\frac{2(2.2\times 10^{-4})}{0.024(3.5\times 10^{-3})^2}}\approx 3.87\times 10^{1}\ \text{rad s}^{-1}

So $\omega \approx 39\ \text{rad s}^{-1}$ .

Key Takeaways

In SHM, total distance travelled in one full cycle is $4A$ .
Total energy in SHM is $E = \tfrac12 m\omega^2A^2$ .
Always convert mm to m and g to kg before substitution.

Common Mistakes

Using $2A$ instead of $4A$ for the distance in one full oscillation.
Forgetting unit conversions (e.g. using $24\ \text{g}$ directly).
Rearranging incorrectly (e.g. missing the square root for $\omega$ ).

Things to Be Careful About

$A$ must be in metres for SI consistency.
Quote $\omega$ in $\text{rad s}^{-1}$ .
Significant figures: energy is given to 2 s.f., so $\omega$ should be about 2 s.f. as well.

Techniques used

relate total distance travelled in one cycle to the amplitudeuse the SHM total energy expression to link amplitude and angular frequencyrearrange and substitute to solve for angular frequency

(b)

The frequency of the oscillator is fixed, and the amplitude of the oscillations is gradually increased.

4M

(i)

Calculate the maximum amplitude of the oscillations so the object does not lose contact with the platform.

amplitude = ______ $\text{m}$

2M Similar

DifficultyMedium

Worked solution

Working

For the object (upward positive):

N - mg = ma

At the top of the motion, $x=+A$ so

a=-\omega^2A

Minimum contact is when $N=0$ :

0 - mg = m(-\omega^2A) \Rightarrow g=\omega^2A

A_{\max}=\frac{g}{\omega^2}=\frac{9.81}{(38.7)^2}=6.6\times 10^{-3}\ \text{m}

Answer

$A_{\max} \approx 6.5\times 10^{-3}\ \text{m}$

Final answer

6.5 × 10^-3 m

Detailed explanation

Background Concept

An object stays in contact with a moving platform as long as the platform can provide the required normal contact force $N$ .

For vertical motion (take upward as positive), Newton’s second law gives

N - mg = ma

If $N$ would need to become negative to keep the object following the platform, contact is impossible (a surface cannot “pull” the object), so the object loses contact at the point where $N=0$ .

For SHM, acceleration relates to displacement from equilibrium by

a=-\omega^2 x

At the top turning point, $x=+A$ and acceleration is most downward: $a=-\omega^2A$ .

Understanding the Question

The frequency (hence $\omega$ ) is fixed at the value found in part (a). You gradually increase amplitude $A$ .

You must find the largest amplitude such that the object does not lose contact at any point in the cycle. Loss of contact happens when $N$ first reaches zero.

Approach

Identify where in the SHM the normal reaction is smallest.
Use $N - mg = ma$ and set $N=0$ at the limiting case.
Use SHM acceleration at that position to solve for $A_{\max}$ .

Step-by-Step Reasoning

The normal reaction is smallest when the platform is accelerating downward most strongly, because then the object would need the least upward push from the platform.

In SHM, maximum downward acceleration occurs at the top of the oscillation ( $x=+A$ ):

a=-\omega^2A

Apply Newton’s second law:

N - mg = ma

At the threshold of losing contact, $N=0$ :

-mg = m(-\omega^2A)

Cancel $m$ :

g = \omega^2 A

So

A_{\max} = \frac{g}{\omega^2}

Using $\omega \approx 38.7\ \text{rad s}^{-1}$ from (a):

A_{\max} \approx \frac{9.81}{(38.7)^2} \approx 6.6\times 10^{-3}\ \text{m}

So about $6.5\times 10^{-3}\ \text{m}$ (2 s.f.).

diagram-b1

Key Takeaways

Contact is lost when the normal reaction becomes zero.
In vertical SHM, the most “dangerous” point for contact is the top, where acceleration is maximum downward.
Limiting condition: $\omega^2A = g$ .

Common Mistakes

Using the bottom turning point instead of the top.
Setting $a=g$ without considering direction/sign.
Forgetting that loss of contact corresponds to $N=0$ (not $N=mg$ ).

Things to Be Careful About

Use $\omega$ in $\text{rad s}^{-1}$ , not frequency $f$ unless you convert via $\omega=2\pi f$ .
Ensure you use maximum downward acceleration ( $\omega^2A$ ), not maximum speed.
State amplitude in metres, not mm.

Techniques used

apply Newton's second law to vertical forcesuse the SHM acceleration expression to identify maximum downward accelerationset the normal reaction to zero to find the limiting condition for loss of contact

(ii)

The amplitude of the oscillations is increased so it is greater than the value in (b)(i).

State and explain the position in an oscillation where the object first loses contact with the platform.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The object first loses contact at the highest point (maximum upward displacement).

At this point the platform has maximum downward acceleration $a=-\omega^2A$ , so

N = m(g+a)=m(g-\omega^2A)

is minimum; when $A$ is increased beyond $g/\omega^2$ , $N\to 0$ and contact is lost.

Final answer

At the top (maximum upward displacement), where downward acceleration is maximum so the normal reaction first becomes zero.

Detailed explanation

Background Concept

Loss of contact occurs when the normal reaction force becomes zero. Using upward positive:

N - mg = ma \quad \Rightarrow \quad N = m(g+a)

So $N$ is smallest when the acceleration $a$ is most negative (largest downward acceleration).

In SHM,

a=-\omega^2 x

so acceleration is downward (negative) when displacement $x$ is upward (positive), and its magnitude is largest at the turning points where $|x|=A$ .

Understanding the Question

You have increased the amplitude beyond the safe maximum found in (b)(i). The platform still oscillates at the same fixed $\omega$ .

You must state where in the cycle contact is first lost, and explain using SHM ideas and forces.

Approach

Find where $a$ is most downward.
Use $N=m(g+a)$ to argue where $N$ is smallest.
State that when amplitude is too large, $N$ becomes zero first at that position.

Step-by-Step Reasoning

At the top of the oscillation (maximum upward displacement), $x=+A$ .

SHM acceleration:

a=-\omega^2x=-\omega^2A

This is the maximum downward acceleration.

Now relate this to contact force:

N = m(g+a)=m(g-\omega^2A)

As you increase $A$ , the term $\omega^2A$ increases, so $N$ decreases. Therefore the earliest point at which $N$ reaches zero is at the top of the motion.

Physically: at the top, the platform is trying to accelerate downward very rapidly. If this required downward acceleration exceeds $g$ , gravity alone cannot keep the object following the platform, and the platform would need to pull the object downward (impossible), so they separate.

diagram-b2

Key Takeaways

Loss of contact means $N=0$ .
$N$ is smallest where acceleration is most downward.
In SHM that occurs at the top turning point ( $x=+A$ ).

Common Mistakes

Saying it loses contact at the bottom because “it is moving fastest” (speed is irrelevant for contact here; acceleration is what matters).
Confusing maximum speed (at equilibrium) with maximum acceleration (at turning points).
Not stating the physical reason: a surface cannot exert a negative normal force.

Things to Be Careful About

Always distinguish between displacement $x$ and acceleration $a=-\omega^2x$ ; their directions are opposite.
Be clear that it is the top because acceleration is downward maximum there, not because the object is “lightest” without justification.

Techniques used

identify where SHM acceleration has maximum magnitude and directionuse the condition for loss of contact based on normal reaction becoming zerolink force balance to the physical meaning of the oscillation position

Q4MediumCapacitanceAlternating Currents

(a)

Three capacitors are connected as shown in Fig. 4.1.

Fig. 4.1

Determine the total capacitance, in $\mu\text{F}$ , of the network of three capacitors.

capacitance = ______ $\mu\text{F}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Parallel pair:

C_p = 10 + 20 = 30\ \mu\text{F}

Series with $45\ \mu\text{F}$ :

\frac{1}{C} = \frac{1}{45} + \frac{1}{30} = \frac{5}{90}

C = \frac{90}{5} = 18\ \mu\text{F}

Answer

$18\ \mu\text{F}$

Final answer

18 μF

Detailed explanation

Background Concept

Capacitances combine differently depending on connection:

Parallel (same p.d. across each):

C_{\text{eq}} = C_1 + C_2 + \cdots

Series (same charge on each):

\frac{1}{C_{\text{eq}}} = \frac{1}{C_1} + \frac{1}{C_2} + \cdots

Understanding the Question

The diagram shows a $45\ \mu\text{F}$ capacitor in series with a parallel pair of $10\ \mu\text{F}$ and $20\ \mu\text{F}$ . You must find the single equivalent capacitance of the whole network.

Approach

Replace the two parallel capacitors with one equivalent capacitor using addition.
Replace the resulting series pair with one equivalent capacitor using the reciprocal formula.

Step-by-Step Reasoning

For the parallel section:

C_p = 10\ \mu\text{F} + 20\ \mu\text{F} = 30\ \mu\text{F}

Now $30\ \mu\text{F}$ is in series with $45\ \mu\text{F}$ :

\frac{1}{C} = \frac{1}{45} + \frac{1}{30}

Use a common denominator ( $90$ ):

\frac{1}{C} = \frac{2}{90} + \frac{3}{90} = \frac{5}{90}

So

C = \frac{90}{5} = 18\ \mu\text{F}

Key Takeaways

Parallel capacitances add directly.
Series capacitances add as reciprocals.
Reduce networks step-by-step by spotting simple series/parallel groups.

Common Mistakes

Adding all three directly (only valid if all three were in parallel).
Using the series reciprocal formula on the parallel pair.
Forgetting to invert at the end when using the series formula.

Things to Be Careful About

Identify correctly which components share the same two nodes (parallel) and which lie in a single path (series).
Keep units consistent; here everything is in $\mu\text{F}$ so the final result can stay in $\mu\text{F}$ .

Techniques used

identify series and parallel parts of a capacitor networkcombine parallel capacitances by summingcombine series capacitances using reciprocals

(b)

A capacitor of capacitance $45 \mu\text{F}$ is connected to a variable power supply initially set at $8.0 \text{ V}$ .
The output of the power supply increases so that the potential difference (p.d.) across the capacitor increases to $9.6 \text{ V}$ .

Calculate the increase in energy $\Delta E$ stored in the capacitor.

$\Delta E$ = ______ $\text{J}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\Delta E = \frac{1}{2}C\left(V_2^2 - V_1^2\right)

C = 45\ \mu\text{F} = 45 \times 10^{-6}\ \text{F}

\Delta E = \frac{1}{2}(45 \times 10^{-6})\left(9.6^2 - 8.0^2\right)

\Delta E = 6.34 \times 10^{-4}\ \text{J}

Answer

$6.34 \times 10^{-4}\ \text{J}$

Final answer

6.34 × 10^-4 J

Detailed explanation

Background Concept

The energy stored in a capacitor is

E = \frac{1}{2}CV^2

where $C$ is capacitance in farads and $V$ is the potential difference across the capacitor.
If the voltage changes while $C$ stays constant, the change in stored energy is

\Delta E = E_2 - E_1 = \frac{1}{2}C\left(V_2^2 - V_1^2\right).

Understanding the Question

A single capacitor of capacitance $45\ \mu\text{F}$ has its p.d. increased from $8.0\ \text{V}$ to $9.6\ \text{V}$ . The question asks for the increase in stored energy, not the final energy.

Approach

Convert $C$ from $\mu\text{F}$ to $\text{F}$ .
Use $\Delta E = \tfrac{1}{2}C(V_2^2 - V_1^2)$ .
Substitute and evaluate, giving the answer in joules.

Step-by-Step Reasoning

Convert capacitance:

45\ \mu\text{F} = 45 \times 10^{-6}\ \text{F}.

Calculate the difference in squares of voltage:

V_2^2 - V_1^2 = 9.6^2 - 8.0^2 = 92.16 - 64 = 28.16.

Now compute $\Delta E$ :

\Delta E = \frac{1}{2}(45 \times 10^{-6})(28.16) = (22.5 \times 10^{-6})(28.16) = 6.336 \times 10^{-4}\ \text{J}

Rounded suitably:

\Delta E \approx 6.34 \times 10^{-4}\ \text{J}.

Key Takeaways

Stored energy depends on $V^2$ , so small voltage changes can noticeably change energy.
Always convert $\mu\text{F}$ to $\text{F}$ in calculations.

Common Mistakes

Using $\Delta E = \tfrac{1}{2}C\Delta V^2$ with $\Delta V = 1.6\ \text{V}$ (incorrect; must use $V_2^2 - V_1^2$ ).
Leaving capacitance in $\mu\text{F}$ and writing the final energy in joules without converting.
Forgetting the factor of $\tfrac{1}{2}$ .

Things to Be Careful About

Square the voltages before subtracting.
Give the final answer to appropriate significant figures (limited by the voltages, typically 2–3 s.f.).

Techniques used

use the energy stored in a capacitor equationcalculate a change by subtracting initial from finalconvert microfarads to farads

(c)

A sinusoidal a.c. power supply is connected to the input of a bridge rectifier.
The output of the rectifier is connected to a load resistor.

4M

(i)

Complete the circuit in Fig. 4.2 by adding a capacitor to smooth the p.d. across the load resistor.

Fig. 4.2

1M Similar

DifficultyEasy

Worked solution

Answer

Connect a capacitor in parallel with the load resistor across the rectifier output terminals.

diagram-4ci-smoothing

Final answer

Capacitor in parallel with the load resistor across the rectifier output.

Detailed explanation

Background Concept

A bridge rectifier makes the output unidirectional, but it is still pulsating (it falls between peaks). A smoothing capacitor is connected across the output so that:

it charges quickly to (about) the peak voltage when the rectified voltage rises,
then it discharges through the load when the rectified voltage falls,
keeping the p.d. across the load more nearly constant.

Understanding the Question

The circuit already has the rectifier output connected to a load resistor. You are asked to add one component so that the p.d. across the load is smoother (less ripple).

Approach

To smooth the voltage across the load, the capacitor must be connected across the same two points as the load (so it directly sets/holds the load p.d.). Therefore, place it in parallel with the load resistor, at the rectifier output.

Step-by-Step Reasoning

Identify the two terminals labelled “connections from output of bridge rectifier”.
The load resistor is already connected across these terminals.
Add a capacitor connected between the same two terminals (i.e. across the resistor).

diagram-4ci-smoothing

Key Takeaways

A smoothing capacitor is always placed across the DC output (in parallel with the load).
Series placement would not hold the output voltage and would block DC after charging.

Common Mistakes

Putting the capacitor in series with the load resistor.
Putting the capacitor on the AC input side (before the rectifier).

Things to Be Careful About

In real circuits an electrolytic capacitor would need correct polarity, but for circuit symbols in exams, simply placing it in parallel across the load earns the mark.

Techniques used

identify where a smoothing capacitor is connected in a rectifier circuitplace a component in parallel with the load to reduce ripple

(ii)

The variation with time $t$ of the p.d. $V$ of the smoothed output is shown in Fig. 4.3.

Fig. 4.3

Determine the time constant, in ms, of the smoothing circuit.

time constant = ______ $\text{ms}$

3M Similar

DifficultyMedium

Worked solution

Working

From the graph, between peaks the capacitor discharges from about $4.0\ \text{V}$ to about $3.2\ \text{V}$ in $10\ \text{ms}$ .

V = V_0 e^{-t/\tau}

\frac{3.2}{4.0} = e^{-10/\tau}

\tau = \frac{-10}{\ln(0.80)} = 44.8\ \text{ms}

Answer

$\tau \approx 45\ \text{ms}$

Final answer

45 ms

Detailed explanation

Background Concept

When a smoothing capacitor discharges through a load resistor $R$ , the p.d. across it decays exponentially:

V = V_0 e^{-t/\tau}

where the time constant is

\tau = RC.

A larger $\tau$ means a slower decay, so the ripple is smaller.

Understanding the Question

You are given a graph of the smoothed output voltage. After each peak, the capacitor voltage falls (discharge) until the next peak recharges it. The time constant can be found by choosing two points on a discharge section and applying the exponential decay relationship.

Approach

Pick a discharge interval between two consecutive peaks (one “gap” between recharging events).
Read the initial voltage $V_0$ at the peak and the voltage $V$ after a known time $t$ .
Use $V/V_0 = e^{-t/\tau}$ and rearrange for $\tau$ :

\tau = \frac{-t}{\ln(V/V_0)}.

Step-by-Step Reasoning

From the graph:

Peak voltage $V_0 \approx 4.0\ \text{V}$ .
Just before the next recharge, voltage $V \approx 3.2\ \text{V}$ .
Time between peaks for a full-wave rectified supply on a 50 Hz input is $10\ \text{ms}$ , consistent with the peak separation on the graph.

Use the decay equation:

\frac{V}{V_0} = e^{-t/\tau}

Substitute the read values:

\frac{3.2}{4.0} = e^{-10/\tau} Rightarrow 0.80 = e^{-10/\tau}

Take natural logs:

\ln(0.80) = -\frac{10}{\tau}

So

\tau = \frac{-10}{\ln(0.80)} = 44.8\ \text{ms} \approx 45\ \text{ms}.

Key Takeaways

The smoothing section is an exponential discharge, characterised by $\tau = RC$ .
Time constant can be extracted from a graph using a voltage ratio and the time between those readings.

Common Mistakes

Using the total period of the original AC supply ( $20\ \text{ms}$ for 50 Hz) instead of the peak spacing after full-wave rectification ( $10\ \text{ms}$ ).
Using a linear drop instead of exponential decay.
Forgetting the minus sign and getting a negative time constant.

Things to Be Careful About

Choose points on the smooth discharge curve (not on the dashed rectified waveform).
Your read-off values may vary slightly; exam marking usually allows a range (answers close to $45\ \text{ms}$ depending on readings).

Techniques used

read suitable values from a ripple/smoothing graphmodel the capacitor discharge as an exponential decayrearrange the exponential to solve for the time constant

(d)

A sinusoidal a.c. power supply has a maximum power of $16 \text{ W}$ .

State the value of the mean power when the output of the power supply is:

2M

(i)

full-wave rectified

mean power = ______ $\text{W}$

1M Similar

(ii)

half-wave rectified.

mean power = ______ $\text{W}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For sinusoidal power, mean over a full cycle is $\tfrac{1}{2}P_{\max}$ .
Half-wave rectified gives zero power for half the time, so mean halves again:

\overline{P} = \frac{1}{4}P_{\max} = \frac{1}{4}(16) = 4.0\ \text{W}

Answer

$4.0\ \text{W}$

Final answer

4.0 W

Detailed explanation

Background Concept

With a sinusoidal voltage across a resistor, instantaneous power is proportional to $V^2$ , giving a $\sin^2$ shape:

P(t) = P_{\max}\sin^2(\omega t)

and the mean over a full cycle is

\overline{P}_{\text{full cycle}} = \frac{1}{2}P_{\max}.

Half-wave rectification removes one half-cycle completely (voltage is zero there), so the power is also zero for that half of the time.

Understanding the Question

You are told $P_{\max} = 16\ \text{W}$ and asked for the mean power after half-wave rectification.

Approach

Start with the known mean over a full cycle for $\sin^2$ : $\tfrac{1}{2}P_{\max}$ .
Because half-wave rectification gives power only for half the time, halve the mean again.

Step-by-Step Reasoning

Mean over a full cycle if it were present all the time:

\overline{P} = \frac{1}{2}P_{\max} = \frac{1}{2}(16) = 8\ \text{W}.

But for half-wave rectification, this non-zero power occurs only for half the cycle, so average over the whole time is

\overline{P}_{\text{half-wave}} = \frac{1}{2}(8\ \text{W}) = 4\ \text{W}

Equivalently:

\overline{P}_{\text{half-wave}} = \frac{1}{4}P_{\max} = 4\ \text{W}.

Key Takeaways

Sinusoidal (or full-wave rectified) power has mean $\tfrac{1}{2}P_{\max}$ .
Half-wave rectification reduces mean power by an additional factor of 2 because the signal is zero for half the time.

Common Mistakes

Giving $8\ \text{W}$ (forgetting that half the cycle is removed).
Giving $16\ \text{W}$ (confusing maximum with mean).

Things to Be Careful About

The halving for half-wave rectification is a time-duty-factor effect: zero output for half the cycle means half the average power compared with the full-wave case.

Techniques used

use the relationship between maximum and mean power for a sinusoidal outputaccount for zero power over half a cycle in half-wave rectification

Q5Medium-EasyMotion in a CircleMagnetic FieldsElectric Fields

(a)

An object travels in a circle at constant speed.

State the names of two quantities that vary during the motion of the object.

1 ______
2 ______

2M Similar

DifficultyEasy

Worked solution

Answer

1 $\text{velocity}$

2 $\text{acceleration}$

Final answer

Velocity; acceleration

Detailed explanation

Background Concept

In uniform circular motion, the speed (magnitude of velocity) is constant, but the direction of motion continually changes.

Velocity is a vector, so even if its magnitude stays the same, a changing direction means the velocity is changing.

A changing velocity implies an acceleration. For circular motion this acceleration is the centripetal acceleration, directed towards the centre of the circle:

a = \frac{v^2}{r}

Because $F = ma$ , there must also be a resultant (centripetal) force towards the centre:

F = \frac{mv^2}{r}

Understanding the Question

The object moves in a circle at constant speed. The question asks for two quantities that vary during the motion.

The key idea is that vectors can vary even if their magnitudes are constant, because their directions change continuously.

Approach

Pick vector quantities linked to the motion whose direction changes as the object goes around the circle. Common correct choices include velocity, acceleration, momentum, and resultant force.

Step-by-Step Reasoning

The speed is constant, but the object’s direction of motion changes at every point on the circle.
Therefore the velocity (a vector) changes.
Since velocity changes, there is an acceleration (centripetal acceleration) towards the centre.

So two valid answers are velocity and acceleration (others such as momentum or force would also be acceptable).

Key Takeaways

Constant speed does not mean constant velocity.
In circular motion, vector directions change continuously, so vectors like velocity and acceleration vary.

Common Mistakes

Writing speed (does not vary here).
Writing kinetic energy (also constant because speed is constant).
Giving only one quantity when two are asked.

Things to Be Careful About

The question says constant speed, not constant velocity.
Use vector quantities that genuinely change (usually their direction changes).

Techniques used

distinguish between speed and velocity as scalar vs vectoridentify quantities that change direction in uniform circular motionstate two correct varying quantities

(b)

A charged particle of mass $m$ and with charge $q$ enters a region of uniform magnetic field, perpendicular to the field lines. The magnetic flux density is $B$ .

The particle travels in a circle with period $T$ and radius $r$ .

8M

(i)

By considering the magnetic force acting on the particle, show that

B = \frac{2\pi m}{qT} .

3M Similar

DifficultyMedium-Easy

Worked solution

Working

Magnetic force provides centripetal force:

qvB = \frac{mv^2}{r}

B = \frac{mv}{qr}

For circular motion,

v = \frac{2\pi r}{T}

Substitute:

B = \frac{m}{qr}\left(\frac{2\pi r}{T}\right) = \frac{2\pi m}{qT}

Answer

B = \frac{2\pi m}{qT}

Final answer

B = 2πm/(qT)

Detailed explanation

Background Concept

A charged particle moving in a magnetic field experiences a magnetic force given by

F_B = qvB\sin\theta

where $\theta$ is the angle between the velocity $\vec{v}$ and the magnetic field $\vec{B}$ . If the particle enters perpendicular to the field lines, then $\theta = 90^\circ$ and $\sin\theta = 1$ , so

F_B = qvB

When a force is always perpendicular to the velocity, it does no work (so the speed stays constant) but it changes the direction of the velocity, producing circular motion. The required centripetal force is

F_c = \frac{mv^2}{r}

Also, for uniform circular motion, the speed relates to period $T$ by

v = \frac{2\pi r}{T}

Understanding the Question

A particle of mass $m$ and charge $q$ enters a uniform magnetic field $B$ perpendicular to the field lines and moves in a circle of radius $r$ with period $T$ . The task is to show, using the magnetic force, that

B = \frac{2\pi m}{qT}.

So we need to link $B$ to $T$ by combining the force law with circular-motion formulas.

Approach

Use $F_B = qvB$ (because motion is perpendicular to $B$ ).
Set this equal to the centripetal force $mv^2/r$ .
Replace $v$ using $v = 2\pi r/T$ .
Rearrange to isolate $B$ .

Step-by-Step Reasoning

diagram-5bi-1

Because the particle’s velocity is perpendicular to $\vec{B}$ , the magnetic force magnitude is

F_B = qvB.

This magnetic force always points towards the centre of the circular path (it is perpendicular to $\vec{v}$ ), so it acts as the centripetal force:

qvB = \frac{mv^2}{r}.

Cancel one factor of $v$ :

qB = \frac{mv}{r} \quad \Rightarrow \quad B = \frac{mv}{qr}.

Use the relationship between speed and period for circular motion:

v = \frac{2\pi r}{T}.

Substitute for $v$ :

B = \frac{m}{qr}\left(\frac{2\pi r}{T}\right) = \frac{2\pi m}{qT}.

The $r$ cancels, showing the period does not depend on the radius.

Key Takeaways

For perpendicular entry, $F_B = qvB$ .
Magnetic force provides centripetal force: $qvB = mv^2/r$ .
Using $v = 2\pi r/T$ leads to $B = 2\pi m/(qT)$ .

Common Mistakes

Forgetting the $\sin\theta$ factor and not using $\sin 90^\circ = 1$ .
Using the wrong centripetal term (e.g. $mv/r$ instead of $mv^2/r$ ).
Using $v = \omega r$ but not linking $\omega$ correctly to $T$ (must use $\omega = 2\pi/T$ ).

Things to Be Careful About

The magnetic force must be the resultant force and must be perpendicular to $\vec{v}$ for uniform circular motion.
Keep track of which quantities are vectors vs magnitudes; the derivation uses magnitudes.

Techniques used

equate magnetic force to centripetal forceuse the perpendicular condition so sin(theta) = 1substitute v = 2 pi r / T and rearrange algebraically

(ii)

The particle is an alpha particle. The period of the circular motion is $2.5 \mu\text{s}$ .

Calculate $B$ .

$B$ = ______ $\text{T}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using

B = \frac{2\pi m}{qT}

For an $\alpha$ -particle: $m = 6.64 \times 10^{-27}\ \text{kg}$ , $q = 2e = 3.20 \times 10^{-19}\ \text{C}$ , $T = 2.5\ \mu\text{s} = 2.5 \times 10^{-6}\ \text{s}$ .

B = \frac{2\pi (6.64 \times 10^{-27})}{(3.20 \times 10^{-19})(2.5 \times 10^{-6})}

B = 5.2 \times 10^{-2}\ \text{T}

Answer

$5.2 \times 10^{-2}\ \text{T}$

Final answer

5.2 × 10^-2 T

Detailed explanation

Background Concept

From part (i), the magnetic flux density needed for circular motion of period $T$ is

B = \frac{2\pi m}{qT}.

This is sometimes called the cyclotron relationship: for a given particle ( $m$ and $q$ fixed) in a given $B$ , the period is fixed.

Understanding the Question

We are told the particle is an alpha particle and has period $T = 2.5\ \mu\text{s}$ . We must calculate the magnetic flux density $B$ .

So we need:

$m$ and $q$ for an alpha particle,
convert $\mu\text{s}$ to seconds,
substitute into the formula.

Approach

Use $m \approx 6.64 \times 10^{-27}\ \text{kg}$ for an alpha particle.
Use $q = +2e = 3.20 \times 10^{-19}\ \text{C}$ .
Convert $T = 2.5\ \mu\text{s}$ to $2.5 \times 10^{-6}\ \text{s}$ .
Substitute and calculate.

Step-by-Step Reasoning

Start with

B = \frac{2\pi m}{qT}.

Substitute values:

$m = 6.64 \times 10^{-27}\ \text{kg}$
$q = 3.20 \times 10^{-19}\ \text{C}$
$T = 2.5 \times 10^{-6}\ \text{s}$

B = \frac{2\pi (6.64 \times 10^{-27})}{(3.20 \times 10^{-19})(2.5 \times 10^{-6})}.

Compute the denominator:

(3.20 \times 10^{-19})(2.5 \times 10^{-6}) = 8.0 \times 10^{-25}.

Compute the numerator:

2\pi(6.64 \times 10^{-27}) \approx 4.17 \times 10^{-26}.

Then

B \approx \frac{4.17 \times 10^{-26}}{8.0 \times 10^{-25}} \approx 5.2 \times 10^{-2}\ \text{T}.

Key Takeaways

Use $q = 2e$ for an alpha particle.
Always convert microseconds to seconds before substitution.
The final unit for $B$ is tesla (T).

Common Mistakes

Using $q = 1.60 \times 10^{-19}\ \text{C}$ instead of $2e$ .
Forgetting to convert $2.5\ \mu\text{s}$ to seconds.
Using proton mass instead of alpha particle mass.

Things to Be Careful About

Powers of ten: $10^{-19} \times 10^{-6} = 10^{-25}$ .
Quote $B$ to sensible significant figures (usually 2 s.f. here).

Techniques used

substitute particle constants into the derived expressionconvert microseconds to secondscalculate and round to appropriate significant figures

(iii)

A second alpha particle is in the same uniform field. It travels in a circle of radius $2r$ .

State and explain how the periods of the motion of the two particles compare.

1M Similar

DifficultyEasy

Worked solution

Answer

The periods are the same.

Since

T = \frac{2\pi m}{qB}

and both particles are $\alpha$ -particles in the same $B$ , $T$ is independent of $r$ (radius cancels).

Final answer

Same period

Detailed explanation

Background Concept

For a charged particle moving perpendicular to a uniform magnetic field, the magnetic force provides the centripetal force and leads to

B = \frac{2\pi m}{qT} \quad \Rightarrow \quad T = \frac{2\pi m}{qB}.

This shows that in a fixed magnetic field, the period depends only on:

particle mass $m$ ,
particle charge $q$ ,
field strength $B$ ,

and not on the radius of the circular path.

Understanding the Question

A second alpha particle is in the same uniform magnetic field, but it moves in a circle of radius $2r$ instead of $r$ . We must compare the periods.

The key clue is “same uniform field” and “alpha particle” (so $m$ and $q$ are unchanged).

Approach

Use the expression for the period in a magnetic field:

T = \frac{2\pi m}{qB}.

Since $m$ , $q$ , and $B$ are the same for both particles, the periods must be equal.

Step-by-Step Reasoning

From magnetic circular motion,

T = \frac{2\pi m}{qB}.

For both particles: $m$ is the alpha mass, $q=+2e$ , and $B$ is the same field.
None of these depends on radius $r$ .
Therefore doubling the radius to $2r$ does not change the period.

So the two alpha particles have the same period.

Key Takeaways

In a uniform magnetic field, the circular period is independent of speed and radius.
Changing radius means the speed changes proportionally, keeping the period fixed.

Common Mistakes

Assuming a bigger circle automatically means a longer period (true for some other situations, but not here because speed changes too).
Bringing $r$ into the final expression for $T$ even though it cancels.

Things to Be Careful About

This only holds when the motion is perpendicular to a uniform $B$ and the particle remains non-relativistic (mass effectively constant).

Techniques used

use the derived period relationship for magnetic circular motioncompare dependence of period on radiusstate a clear comparative conclusion

(iv)

The speed of the alpha particle in (b)(ii) is $1.1 \times 10^6 \text{ m s}^{-1}$ . An electric field is applied so that this particle now moves with constant velocity.

Use your answer in (b)(ii) to calculate the electric field strength $E$ . Give the unit with your answer.

$E$ = ______ unit ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For constant velocity, resultant force is zero, so

qE = qvB

E = vB

Using $v = 1.1 \times 10^6\ \text{m s}^{-1}$ and $B = 5.2 \times 10^{-2}\ \text{T}$ :

E = (1.1 \times 10^6)(5.2 \times 10^{-2}) = 5.7 \times 10^4

Answer

E = 5.7 \times 10^4\ \text{V m}^{-1}

Final answer

5.7 × 10^4 V m^-1

Detailed explanation

Background Concept

A charged particle in electric and magnetic fields experiences:

Electric force:

F_E = qE

Magnetic force (magnitude):

F_B = qvB\sin\theta.

If the fields are arranged so that the particle moves at constant velocity in a straight line, the resultant force must be zero. This is the principle of a velocity selector: the electric and magnetic forces balance.

For perpendicular geometry ( $\theta = 90^\circ$ ):

F_B = qvB.

Balancing forces gives:

qE = qvB \Rightarrow E = vB.

Understanding the Question

The alpha particle from (b)(ii) has speed $1.1 \times 10^6\ \text{m s}^{-1}$ . An electric field is now applied so that it moves with constant velocity (meaning no acceleration).

We must use the previously found $B$ to calculate the required electric field strength $E$ , and include the unit.

Approach

Constant velocity implies net force $=0$ .
Set electric force equal to magnetic force: $qE = qvB$ .
Cancel $q$ to get $E=vB$ .
Substitute $v$ and $B$ .

Step-by-Step Reasoning

Condition for constant velocity:

F_E = F_B.

With perpendicular arrangement:

qE = qvB.

Cancel $q$ (works for any non-zero charge):

E = vB.

Substitute $v = 1.1 \times 10^6\ \text{m s}^{-1}$ and $B = 5.2 \times 10^{-2}\ \text{T}$ :

E = (1.1 \times 10^6)(5.2 \times 10^{-2}) \approx 5.7 \times 10^4.

Units: $E$ can be written as $\text{N C}^{-1}$ or equivalently $\text{V m}^{-1}$ . So

E = 5.7 \times 10^4\ \text{V m}^{-1}.

Key Takeaways

Constant velocity means zero resultant force.
In crossed fields, balance $qE$ and $qvB$ to get $E=vB$ .
$\text{N C}^{-1}$ and $\text{V m}^{-1}$ are equivalent units for electric field strength.

Common Mistakes

Adding forces instead of balancing them (should be equal and opposite).
Forgetting that magnetic force uses $v$ and depends on angle.
Using $E = B/v$ (inverting incorrectly).

Things to Be Careful About

Use $B$ from (b)(ii) with correct significant figures.
Ensure $v$ is in $\text{m s}^{-1}$ and $B$ in T so that $E$ comes out in $\text{V m}^{-1}$ (or $\text{N C}^{-1}$ ).

Techniques used

apply the condition for constant velocity (zero resultant force)equate electric and magnetic forces for a velocity selectorsubstitute numerical values with correct units

Q6MediumMagnetic Fields

(a)

A small coil C has 64 turns and cross-sectional area $0.71 \text{ cm}^2$ . The coil is placed inside a solenoid as shown in Fig. 6.1.

Fig. 6.1

The centre of coil C is on the central axis of the solenoid.

8M

(i)

There is a constant current in the solenoid.
Coil C is moved through the solenoid from position X to position Y.

On Fig. 6.2, sketch a line to show the variation of the magnetic flux linkage in coil C with position as it moves from X to Y.

Fig. 6.2

1M Similar

DifficultyMedium-Easy

Worked solution

diagram-a-i

Answer

Flux linkage is approximately constant (maximum) while the coil is well inside the solenoid, then it decreases smoothly to approximately zero as the coil moves out of the end and is near zero at $Y$ .

Final answer

Constant high value inside solenoid then falls to ~0 outside (near Y).

Detailed explanation

Background Concept

Magnetic flux through a coil is

\Phi = BA\cos\theta

and the magnetic flux linkage is

N\Phi

where $N$ is the number of turns. For a coil on the axis of a solenoid, the field $B$ inside a long solenoid is approximately uniform and parallel to the axis, but near the end it “fringes” and drops to a much smaller value outside.

Understanding the Question

You move the small coil from $X$ (inside the solenoid) to $Y$ (outside). The current in the solenoid is constant, so the solenoid’s field pattern in space does not change with time; only the coil’s position changes.

You must sketch how the flux linkage $N\Phi$ in the small coil varies with position.

Approach

Decide how $N\Phi$ depends on position: $N\Phi \propto B$ because $N$ and $A$ are constant and the orientation is fixed.
Use the known field pattern of a solenoid: nearly constant $B$ inside, small $B$ outside, with a transition region near the end.
Sketch a graph with a flat section (inside) and a fall to (about) zero (outside).

Step-by-Step Reasoning

Inside the solenoid (near $X$ ), $B$ is approximately constant, so

N\Phi = NBA

is constant. On the graph this is a horizontal line at some maximum value.

As the coil approaches the solenoid end, the field lines spread out and the field strength along the axis decreases. Therefore $B$ decreases, so $N\Phi$ decreases.
Well outside the solenoid (near $Y$ ), the field is very small compared with inside, so $B \approx 0$ and hence $N\Phi \approx 0$ .

So the curve stays roughly constant then falls smoothly to near zero.

diagram-a-i

Key Takeaways

Flux linkage depends on field strength: $N\Phi \propto B$ for fixed coil area and orientation.
A solenoid produces a uniform field inside and a weak field outside, with an end-transition region.

Common Mistakes

Drawing the flux linkage increasing as the coil leaves the solenoid (it should decrease).
Making the drop instantaneous with no transition region (end effects cause a gradual change).
Not showing it approaching (approximately) zero outside.

Things to Be Careful About

The vertical scale is not required; only the correct qualitative shape matters.
The sign of flux linkage is not being tested here; it is the magnitude/variation with position.
If the solenoid is treated as “long”, the inside section should be clearly flat (uniform field).

Techniques used

use flux linkage as the product of turns and magnetic fluxrelate magnetic flux to magnetic flux density and coil areainfer qualitative variation of field strength along the axis of a solenoid

(ii)

Explain the shape of your line in (a)(i).

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Inside the solenoid, $B$ is (approximately) uniform and the coil’s area/orientation are constant, so

N\Phi = NBA

is constant.

Near the end, the field fringes and $B$ decreases, so $N\Phi$ decreases. Outside the solenoid, $B \approx 0$ so the flux linkage is (approximately) zero.

Final answer

Uniform B inside gives constant NΦ; fringing at end makes B fall; outside B≈0 so NΦ≈0.

Detailed explanation

Background Concept

Flux linkage is the total flux through all turns:

\text{flux linkage} = N\Phi

For a coil in a magnetic field,

\Phi = BA\cos\theta

If $N$ , $A$ and $\theta$ are fixed, then changes in flux linkage come only from changes in $B$ .

A long solenoid with steady current produces an approximately uniform magnetic flux density inside, directed along the axis. Outside, the field is much weaker; near the ends the field lines spread out (fringing), so $B$ reduces gradually.

Understanding the Question

Your sketch in (a)(i) should have a flat section and then a drop. This part asks you to justify those features using physics, not by describing the graph.

Approach

Explain the graph by connecting each region of the motion to what happens to $B$ :

inside: $B$ constant
near end: $B$ falls (fringing)
outside: $B$ very small
Then translate $B$ changes into $N\Phi$ changes using $N\Phi = NBA$ .

Step-by-Step Reasoning

While the coil is well inside the solenoid, the solenoid field is approximately uniform and constant in time because the current is constant. The coil’s area and orientation do not change. Therefore

N\Phi = NBA

is constant with position.

As the coil reaches the end region, the solenoid field is no longer uniform because field lines spread out into the surrounding space. This reduces the axial component of $B$ through the coil, so the flux $\Phi$ and hence $N\Phi$ decrease.
Once outside, the solenoid’s field is weak (compared with inside), so $B \approx 0$ and consequently $N\Phi \approx 0$ .

Key Takeaways

A constant solenoid current means a fixed spatial field pattern.
Flux linkage changes if (and only if) the flux through the coil changes.
End effects in solenoids produce gradual transitions rather than sharp steps.

Common Mistakes

Saying “flux linkage is constant because current is constant” without linking to $B$ and $N\Phi = NBA$ .
Claiming the outside field is exactly zero; in reality it is just much smaller.

Things to Be Careful About

Always reference the defining relationship $N\Phi = NBA\cos\theta$ to justify a graph shape.
Make clear which quantity is changing (here: $B$ with position near the end).

Techniques used

link flux linkage to magnetic flux density in the coiluse the uniform-field model for a solenoid interioruse qualitative reasoning about fringing fields near a solenoid end

(iii)

Coil C is now held stationary at X. The current in the solenoid varies so that the magnetic flux density $B$ at X varies from time 0 to time $4t$ as shown in Fig. 6.3.

Fig. 6.3

Calculate the maximum magnetic flux linkage in coil C.

flux linkage = ______ $\text{Wb}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Maximum $B = 0.080\ \text{T}$ .

A = 0.71\ \text{cm}^2 = 0.71 \times 10^{-4}\ \text{m}^2 = 7.1 \times 10^{-5}\ \text{m}^2

\Phi_{\max} = BA = (0.080)(7.1 \times 10^{-5}) = 5.68 \times 10^{-6}\ \text{Wb}

(N\Phi)_{\max} = 64 \times 5.68 \times 10^{-6} = 3.64 \times 10^{-4}\ \text{Wb}

Answer

$3.6 \times 10^{-4}\ \text{Wb}$

Final answer

3.6 × 10^-4 Wb

Detailed explanation

Background Concept

Magnetic flux through a coil is

\Phi = BA\cos\theta

For a coil whose plane is perpendicular to the field (so the field is normal to the coil area), $\theta = 0$ and $\cos\theta = 1$ , giving $\Phi = BA$ .

Flux linkage is the flux through all turns:

\text{flux linkage} = N\Phi

Understanding the Question

The coil has $N = 64$ turns and cross-sectional area $A = 0.71\ \text{cm}^2$ . It is held at $X$ and the graph shows that $B$ reaches a maximum value of $0.080\ \text{T}$ . You are asked for the maximum magnetic flux linkage $N\Phi$ .

Approach

Read $B_{\max}$ from the graph.
Convert the coil area from $\text{cm}^2$ to $\text{m}^2$ .
Compute $\Phi_{\max} = B_{\max}A$ .
Multiply by $N$ to get $(N\Phi)_{\max}$ .

Step-by-Step Reasoning

From the graph, the maximum magnetic flux density is

B_{\max} = 0.080\ \text{T}

Convert area:

1\ \text{cm}^2 = 10^{-4}\ \text{m}^2

So

A = 0.71 \times 10^{-4}\ \text{m}^2 = 7.1 \times 10^{-5}\ \text{m}^2

Find maximum flux through one turn:

\Phi_{\max} = BA = (0.080)(7.1 \times 10^{-5}) = 5.68 \times 10^{-6}\ \text{Wb}

Multiply by number of turns:

(N\Phi)_{\max} = 64\Phi_{\max} = 64(5.68 \times 10^{-6}) = 3.64 \times 10^{-4}\ \text{Wb}

Rounded appropriately (limited by 2 s.f. data),

(N\Phi)_{\max} \approx 3.6 \times 10^{-4}\ \text{Wb}

Key Takeaways

Convert areas carefully: $\text{cm}^2 \to \text{m}^2$ introduces a factor of $10^{-4}$ .
Use $\Phi = BA$ when the coil is oriented for maximum flux.
Flux linkage is $N\Phi$ .

Common Mistakes

Using $0.71\ \text{cm}^2 = 0.71 \times 10^{-2}\ \text{m}^2$ (wrong power of ten; it must be $10^{-4}$ ).
Forgetting to multiply by $N = 64$ .
Giving too many significant figures.

Things to Be Careful About

The unit of flux is the weber (Wb). Flux linkage is sometimes written as Wb-turns, but the question’s answer line indicates Wb, so follow that.
Ensure $B$ is read correctly from the graph (here $0.080\ \text{T}$ , not $0.040\ \text{T}$ ).

Techniques used

convert area to SI unitscalculate magnetic flux using $\Phi = BA$calculate flux linkage using $N\Phi$

(iv)

On Fig. 6.4, sketch a line to show the induced electromotive force (e.m.f.) $E$ in coil C from time 0 to time $4t$ .

Fig. 6.4

3M Similar

DifficultyMedium

Worked solution

Working

E = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t} = -NA\frac{\mathrm{d}B}{\mathrm{d}t}

From $0 \to t$ , $B$ is constant $\Rightarrow \mathrm{d}B/\mathrm{d}t = 0 \Rightarrow E=0$ .

From $t \to 2t$ ,

\frac{\mathrm{d}B}{\mathrm{d}t} = \frac{0.080-0.040}{t} = \frac{0.040}{t}

so $E$ is constant (one sign).

From $2t \to 4t$ ,

\frac{\mathrm{d}B}{\mathrm{d}t} = \frac{0-0.080}{2t} = -\frac{0.040}{t}

so $E$ is constant with opposite sign and the same magnitude.

Answer

diagram-a-iv

Final answer

Sketch: E = 0 from 0 to t; constant value from t to 2t; constant equal-magnitude opposite sign from 2t to 4t.

Detailed explanation

Background Concept

Faraday’s law gives the induced e.m.f. in a coil:

E = -\frac{\mathrm{d}(N\Phi)}{\mathrm{d}t}

If the coil’s area and orientation are fixed, then $\Phi = BA$ and

E = -NA\frac{\mathrm{d}B}{\mathrm{d}t}

So the induced e.m.f. depends on the rate of change of magnetic flux density, i.e. the gradient of the $B$ -against- $t$ graph.

If $B$ is constant, gradient is zero, so $E = 0$ .
If $B$ changes linearly with time, gradient is constant, so $E$ is constant.
The minus sign means the induced e.m.f. changes sign depending on whether $B$ is increasing or decreasing.

Understanding the Question

The coil is held fixed at $X$ . The graph of $B$ vs time shows three sections:

$0$ to $t$ : constant $B = 0.040\ \text{T}$
$t$ to $2t$ : $B$ increases linearly to $0.080\ \text{T}$
$2t$ to $4t$ : $B$ decreases linearly to $0$

You must sketch the induced e.m.f. $E$ from $0$ to $4t$ .

Approach

Use $E \propto -\mathrm{d}B/\mathrm{d}t$ .
Find the gradient of each section of the $B(t)$ graph.
Convert each gradient into the corresponding e.m.f. level: zero gradient gives $E=0$ , constant non-zero gradient gives a horizontal line above or below the axis.
Use opposite signs for increasing vs decreasing $B$ .

Step-by-Step Reasoning

From $0$ to $t$ , the line on the $B$ graph is horizontal, so

\frac{\mathrm{d}B}{\mathrm{d}t} = 0 \quad \Rightarrow \quad E = 0.

So the $E$ graph lies on the zero line for this interval.

From $t$ to $2t$ , $B$ rises from $0.040\ \text{T}$ to $0.080\ \text{T}$ in a time $t$ . The gradient is

\frac{\mathrm{d}B}{\mathrm{d}t} = \frac{0.080-0.040}{t} = \frac{0.040}{t}.

This is constant and positive. Therefore $E = -NA(\mathrm{d}B/\mathrm{d}t)$ is constant and negative (taking the sign convention that an increasing flux gives negative e.m.f.). So draw a horizontal line at a constant value below the axis from $t$ to $2t$ .

From $2t$ to $4t$ , $B$ falls from $0.080\ \text{T}$ to $0$ in time $2t$ . The gradient is

\frac{\mathrm{d}B}{\mathrm{d}t} = \frac{0-0.080}{2t} = -\frac{0.040}{t}.

This is constant and negative, so $E$ is constant and positive. Its magnitude is the same as in the previous section because the gradient magnitude is the same ( $0.040/t$ ). Therefore draw a horizontal line of the same height above the axis from $2t$ to $4t$ .

Putting these pieces together gives a “rectangular pulse” below zero then an equal “rectangular pulse” above zero.

diagram-a-iv

Key Takeaways

Induced e.m.f. is proportional to the gradient of the flux (or $B$ ) vs time graph.
Flat $B(t)$ gives zero e.m.f.
Opposite gradients produce e.m.f.s of opposite sign; equal gradient magnitudes give equal e.m.f. magnitudes.

Common Mistakes

Drawing $E$ with the same shape as $B$ (it should follow the gradient, not the value).
Making the second pulse half the height because the time interval is longer (here the gradient is the same magnitude, so the height is the same).
Forgetting the sign change between increasing and decreasing $B$ .

Things to Be Careful About

The graph requires correct time intervals: $E=0$ only from $0$ to $t$ .
At $t$ and $2t$ , $E$ changes abruptly because the gradient changes abruptly.
Exact numerical values of $E$ are not required because $t$ is not given; only the correct relative levels and signs matter.

Techniques used

apply Faraday's law $E = -\mathrm{d}(N\Phi)/\mathrm{d}t$use the gradient of a $B$-time graph to determine $\mathrm{d}B/\mathrm{d}t$translate piecewise-constant gradients into piecewise-constant e.m.f.

(b)

A metal spring rests on a smooth table. The turns of the spring are equally spaced. The ends of the spring are connected to a d.c. power supply, as shown in Fig. 6.5.

Fig. 6.5

The spring is connected to the d.c. power supply using flexible leads. The spring is not under tension.

With reference to magnetic fields, describe and explain the change in the distance between the turns of the spring when the power supply is first switched on.

3M Similar

DifficultyMedium

Worked solution

Answer

When the supply is switched on, current flows in the spring so each turn produces a magnetic field. Adjacent turns carry current in the same direction, so their magnetic fields interact and the turns attract (force between parallel currents). The turns move closer together, compressing the spring, until the magnetic attraction is balanced by the spring’s elastic restoring forces.

Final answer

Turns move closer together because adjacent turns carry current in same direction and attract; compression continues until balanced by elastic forces.

Detailed explanation

Background Concept

A current produces a magnetic field around the conductor. When two conductors carry currents, each lies in the magnetic field produced by the other, and a magnetic force acts on them. For two parallel current-carrying conductors:

currents in the same direction (\Rightarrow) attraction
currents in opposite directions (\Rightarrow) repulsion

The force can be understood either from the field interaction picture or from

F = BIL

where a conductor of length $L$ carrying current $I$ in a magnetic flux density $B$ experiences a force $F$ .

A spring is effectively many adjacent loops/turns of wire; neighbouring turns behave like nearby conductors carrying current in the same sense.

Understanding the Question

A metal spring lies on a smooth (low-friction) table and is not stretched initially. Its ends are connected to a d.c. supply with flexible leads, so when current starts, the spring is free to change its length. The question asks what happens to the spacing between turns when the power supply is first switched on, and requires an explanation in terms of magnetic fields.

Approach

Identify what switching on does: current begins to flow through all turns.
Describe the magnetic fields due to the currents in each turn.
Use the known interaction: adjacent turns carry current in the same direction, so they attract.
State the mechanical outcome: spacing decreases until forces balance.

Step-by-Step Reasoning

On switching on, a current flows through the helical spring. Each turn is a current loop and produces a magnetic field.
Each turn is close to its neighbours, and the current in neighbouring turns is in the same direction around the helix (so adjacent sections are like parallel conductors with the same current direction).
Parallel currents in the same direction attract. Therefore neighbouring turns experience forces pulling them together.
Because the spring is on a smooth table and connected with flexible leads, it can move and shorten. The spacing between turns decreases (the spring compresses).
The compression continues until the spring’s elastic restoring force (tending to return it to its natural spacing) balances the magnetic attraction. After that, the spacing becomes steady.

Key Takeaways

Currents create magnetic fields; magnetic fields exert forces on other currents.
Same-direction parallel currents attract, causing a current-carrying spring to compress.
Final separation is set by equilibrium between magnetic forces and elastic forces.

Common Mistakes

Saying the turns repel (they attract for same-direction currents).
Explaining using electrostatic forces (this is magnetic interaction, not charge build-up).
Missing the idea of balance: the spring will not keep compressing indefinitely.

Things to Be Careful About

The question says “first switched on”: it is about the initial change in spacing due to current starting to flow.
Mentioning the role of the smooth table/flexible leads helps justify that the spring is free to move.
Keep the explanation explicitly tied to magnetic fields and forces between current-carrying conductors.

Techniques used

model each turn as a current-carrying loop producing a magnetic fieldapply the rule that parallel currents attractlink magnetic force to mechanical compression until equilibrium

Q7MediumQuantum Physics

(a)

A photon has an energy of $3.11 \times 10^{-19} \text{ J}$ .

Calculate the momentum of the photon.

momentum = ______ $\text{Ns}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

p = \frac{E}{c}

p = \frac{3.11 \times 10^{-19}}{3.00 \times 10^{8}} = 1.04 \times 10^{-27}

Answer

1.04 \times 10^{-27}\ \text{N s}

Final answer

1.04 × 10^−27 N s

Detailed explanation

Background Concept

A photon carries energy $E$ and momentum $p$ . For electromagnetic radiation,

E = hf

and

p = \frac{E}{c}

where $h$ is the Planck constant, $f$ is the frequency, and $c$ is the speed of light in vacuum.

Momentum has SI units $\text{kg m s}^{-1}$ , which is equivalent to $\text{N s}$ .

Understanding the Question

You are given the photon energy $E = 3.11 \times 10^{-19}\ \text{J}$ and asked to calculate the photon momentum in $\text{N s}$ . No wavelength or frequency is needed because $p$ is directly related to $E$ .

Approach

Use the relation $p = E/c$ and substitute the given energy and the speed of light.

Step-by-Step Reasoning

Start from

p = \frac{E}{c}.

Substitute $E = 3.11 \times 10^{-19}\ \text{J}$ and $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ :

p = \frac{3.11 \times 10^{-19}}{3.00 \times 10^{8}}.

Compute powers of ten and the numerical factor:

p = \left(\frac{3.11}{3.00}\right) \times 10^{-27} = 1.04 \times 10^{-27}.

The unit is $\text{J} \div (\text{m s}^{-1}) = (\text{N m}) \div (\text{m s}^{-1}) = \text{N s}$ .

Key Takeaways

Photon momentum can be found directly from energy using $p = E/c$ .
Always track powers of ten and units; $\text{N s}$ is a correct unit for momentum.

Common Mistakes

Using $p = mc$ (photons have no rest mass).
Using $p = E \times c$ instead of $p = E/c$ .
Quoting the answer without a unit.

Things to Be Careful About

Use $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ .
Keep the standard form correct: dividing by $10^8$ makes the exponent $-19-8=-27$ .

Techniques used

apply the photon momentum relation p = E/csubstitute numerical values with correct powers of tenexpress momentum in equivalent SI units

(b)

A laser beam has a power of $350 \text{ mW}$ . The light from the laser has a wavelength of $640 \text{ nm}$ .

4M

(i)

Determine the number of photons emitted by the laser in a time of $1.0 \text{ s}$ .

number = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

E = \frac{hc}{\lambda}

E = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{640 \times 10^{-9}} = 3.11 \times 10^{-19}\ \text{J}

Energy in $1.0\ \text{s}$ :

E_{\text{total}} = Pt = 0.350 \times 1.0 = 0.350\ \text{J}

N = \frac{E_{\text{total}}}{E} = \frac{0.350}{3.11 \times 10^{-19}} = 1.13 \times 10^{18}

Answer

1.13 \times 10^{18}

Final answer

1.13 × 10^18

Detailed explanation

Background Concept

Laser power $P$ is the rate of energy transfer:

P = \frac{\Delta E}{\Delta t}.

Each photon has energy

E_{\gamma} = hf = \frac{hc}{\lambda}.

If a laser emits $N$ photons in time $t$ , then the total emitted energy is

E_{\text{total}} = N E_{\gamma}.

Since $E_{\text{total}} = Pt$ , we get

N = \frac{Pt}{E_{\gamma}}.

Understanding the Question

You are told the laser power is $350\ \text{mW}$ and the wavelength is $640\ \text{nm}$ . In $1.0\ \text{s}$ , the laser outputs energy $Pt$ . Dividing that by the energy per photon gives the number of photons emitted in that second.

Approach

Convert $P$ and $\lambda$ into SI units.
Calculate photon energy using $E_{\gamma} = hc/\lambda$ .
Use $N = Pt/E_{\gamma}$ with $t = 1.0\ \text{s}$ .

Step-by-Step Reasoning

Convert units:

$P = 350\ \text{mW} = 0.350\ \text{W}$ .
$\lambda = 640\ \text{nm} = 640 \times 10^{-9}\ \text{m}$ .

Photon energy:

E_{\gamma} = \frac{hc}{\lambda} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{640 \times 10^{-9}} = 3.11 \times 10^{-19}\ \text{J}.

Energy emitted in $1.0\ \text{s}$ :

E_{\text{total}} = Pt = 0.350 \times 1.0 = 0.350\ \text{J}.

Number of photons:

N = \frac{E_{\text{total}}}{E_{\gamma}} = \frac{0.350}{3.11 \times 10^{-19}} = 1.13 \times 10^{18}.

Key Takeaways

Convert to SI units before substituting.
Use $E_{\gamma} = hc/\lambda$ then $N = Pt/E_{\gamma}$ .

Common Mistakes

Forgetting to convert $\text{mW} \to \text{W}$ or $\text{nm} \to \text{m}$ .
Using $E = h\lambda$ (incorrect).
Using $N = P/E$ but forgetting the time factor (here $t=1.0\ \text{s}$ so it is numerically the same, but the method must be clear).

Things to Be Careful About

Use standard values: $h = 6.63 \times 10^{-34}\ \text{J s}$ and $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ unless otherwise stated.
Keep appropriate significant figures (typically 3 s.f. here).

Techniques used

use E = hc/λ to find photon energyconvert given quantities to SI unitsuse N = energy transferred / energy per photon

(ii)

The laser beam is incident normally on a surface that absorbs all of the photons.

Show that the force $F$ exerted on the surface by the laser beam is given by

F = \frac{P}{c}

where $P$ is the power of the laser beam and $c$ is the speed of light.

2M Similar

DifficultyMedium

Worked solution

Working

Momentum per photon:

p = \frac{E}{c}

Photons per second:

\frac{N}{t} = \frac{P}{E}

Force $=$ rate of change of momentum (absorption):

F = \frac{\Delta p}{\Delta t} = \left(\frac{N}{t}\right)p = \left(\frac{P}{E}\right)\left(\frac{E}{c}\right)

F = \frac{P}{c}

Answer

F = \frac{P}{c}

Final answer

F = P/c

Detailed explanation

Background Concept

Force is the rate of change of momentum:

F = \frac{\Delta p}{\Delta t}.

Light carries momentum. A photon has momentum

p = \frac{E}{c}.

If a surface absorbs light, the light’s momentum is transferred to the surface, producing a force due to the continuous momentum change per second.

Power $P$ is energy transferred per second:

P = \frac{\Delta E}{\Delta t}.

Understanding the Question

A laser beam hits a surface normally and is fully absorbed. You must show (derive) that the force on the surface depends only on the beam power $P$ and the speed of light $c$ , and is given by $F = P/c$ .

Approach

Work with “per photon” quantities (energy and momentum).
Convert power into “photons per second”.
Multiply momentum per photon by photons per second to get momentum transferred per second, which is the force.

Step-by-Step Reasoning

Each photon has energy $E$ and momentum

p = \frac{E}{c}.

If the beam power is $P$ , then the energy delivered each second is $P$ joules per second. The number of photons emitted per second is therefore

\frac{N}{t} = \frac{\text{energy per second}}{\text{energy per photon}} = \frac{P}{E}.

For complete absorption, each photon’s momentum is transferred to the surface, so the momentum transferred per second is

\frac{\Delta p}{\Delta t} = \left(\frac{N}{t}\right)p = \left(\frac{P}{E}\right)\left(\frac{E}{c}\right).

The $E$ cancels:

F = \frac{\Delta p}{\Delta t} = \frac{P}{c}.

This result is independent of wavelength because higher-energy photons carry more momentum, but fewer are emitted per second for the same power, and the effects cancel.

Key Takeaways

Force from light is due to momentum transfer.
For absorption at normal incidence: $F = P/c$ .
The key idea is multiplying “momentum per photon” by “photons per second”.

Common Mistakes

Using $F = Pc$ instead of $F = P/c$ .
Forgetting that force is momentum change per unit time.
Confusing absorption with reflection (reflection would double the momentum change, giving $F = 2P/c$ for normal incidence).

Things to Be Careful About

The question specifies absorbs all photons and normal incidence; those conditions are exactly what makes the derivation simple.
Make the cancellation of $E$ explicit to clearly “show that” the formula follows.

Techniques used

use momentum transfer to relate force to rate of change of momentumexpress number of photons per second in terms of power and photon energyeliminate photon energy to obtain a simplified symbolic result

(c)

Light of a single wavelength is incident on the surface of different metals. The work function energy of the metals is given in Table 7.1.

Table 7.1

metal	work function energy / eV
tungsten	4.49
magnesium	3.68
potassium	2.26

3M

(i)

Explain the term threshold wavelength.

1M Similar

DifficultyEasy

Worked solution

Answer

Threshold wavelength is the maximum wavelength of incident light that will just cause photoemission (photon energy equals the work function, so emitted electrons have zero maximum kinetic energy).

Final answer

Maximum wavelength that just causes photoemission (photon energy = work function).

Detailed explanation

Background Concept

In the photoelectric effect, electrons are emitted from a metal surface if photons have enough energy to overcome the metal’s work function $\phi$ .

Photon energy is

E = hf = \frac{hc}{\lambda}.

The threshold condition is when emitted electrons just have zero kinetic energy:

hf_0 = \phi.

This defines the threshold frequency $f_0$ and corresponding threshold wavelength $\lambda_0$ .

Understanding the Question

You are asked to explain what “threshold wavelength” means. This is a definition question tied to the idea of the minimum photon energy needed for emission.

Approach

State that it is the longest wavelength (i.e. lowest energy) that can still eject electrons, corresponding to photon energy equal to the work function.

Step-by-Step Reasoning

Longer wavelength means smaller frequency.
Smaller frequency means smaller photon energy because $E = hf$ .
There is a minimum energy required to release an electron from the metal: $\phi$ .
At the threshold, $E = \phi$ and the emitted electrons have $K_{\max} = 0$ .
Therefore the threshold wavelength is the maximum wavelength that still produces emission.

Key Takeaways

“Threshold” means “just enough energy to cause emission”.
Wavelength and photon energy are inversely related: larger $\lambda$ gives smaller $E$ .

Common Mistakes

Saying it is the minimum wavelength (it is the maximum wavelength).
Defining it without mentioning photoemission/electron emission.

Things to Be Careful About

Use the words “maximum wavelength” or “longest wavelength”. Both are acceptable and precise.
Linking it to $E = \phi$ (or electrons emitted with zero kinetic energy) secures the mark.

Techniques used

state the definition of a threshold condition for photoemissionlink threshold wavelength to photon energy equalling the work function

(ii)

For the metals in Table 7.1, calculate the value of the largest threshold wavelength.

threshold wavelength = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Largest threshold wavelength corresponds to smallest work function: potassium, $\phi = 2.26\ \text{eV}$ .

\phi = 2.26 \times 1.60 \times 10^{-19} = 3.62 \times 10^{-19}\ \text{J}

At threshold $\phi = \frac{hc}{\lambda_0}$ :

\lambda_0 = \frac{hc}{\phi} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{3.62 \times 10^{-19}}

\lambda_0 = 5.49 \times 10^{-7}\ \text{m}

Answer

5.49 \times 10^{-7}\ \text{m}

Final answer

5.49 × 10^−7 m

Detailed explanation

Background Concept

The work function $\phi$ is the minimum energy needed to remove an electron from the surface of a metal.

At the threshold of photoemission, the photon energy just equals the work function:

\phi = hf_0 = \frac{hc}{\lambda_0}.

So the threshold wavelength is

\lambda_0 = \frac{hc}{\phi}.

Because $\lambda_0$ is inversely proportional to $\phi$ , a smaller work function gives a larger threshold wavelength.

Understanding the Question

You are given three metals and their work functions (in eV). You must find which has the largest threshold wavelength and calculate it in metres.

Approach

Choose the smallest work function from the table (this gives the largest $\lambda_0$ ).
Convert that work function from eV to J.
Use $\lambda_0 = hc/\phi$ .

Step-by-Step Reasoning

From the table:

tungsten: $4.49\ \text{eV}$
magnesium: $3.68\ \text{eV}$
potassium: $2.26\ \text{eV}$ (smallest)

Therefore potassium has the largest threshold wavelength.

Convert $2.26\ \text{eV}$ to joules using $1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$ :

\phi = 2.26 \times 1.60 \times 10^{-19} = 3.62 \times 10^{-19}\ \text{J}.

Now apply

\lambda_0 = \frac{hc}{\phi}.

Substitute $h = 6.63 \times 10^{-34}\ \text{J s}$ and $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ :

\lambda_0 = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{3.62 \times 10^{-19}} = 5.49 \times 10^{-7}\ \text{m}.

This is about $549\ \text{nm}$ , which is a sensible wavelength in the visible range.

Key Takeaways

The largest threshold wavelength comes from the smallest work function.
Threshold condition: $\phi = hc/\lambda_0$ .
Always convert eV to J before using SI constants.

Common Mistakes

Picking the largest work function (this would give the smallest threshold wavelength).
Forgetting to convert eV to J.
Using $\lambda_0 = \phi/(hc)$ (inverting the formula).

Things to Be Careful About

Use consistent significant figures (usually 3 s.f. given the data).
Keep track of units: with $h$ in $\text{J s}$ and $c$ in $\text{m s}^{-1}$ , $\lambda_0$ comes out in metres only if $\phi$ is in joules.

Techniques used

identify which metal gives the largest threshold wavelengthconvert work function from eV to Juse λ0 = hc/φ to calculate the threshold wavelength

Q8MediumNuclear Physics

(a)

State what is meant by the binding energy of a nucleus.

2M Similar

DifficultyEasy

Worked solution

Answer

Binding energy is the energy required to separate a nucleus completely into its individual nucleons (equivalently, the energy released when the nucleus is formed from free nucleons).

Final answer

Energy required to completely separate the nucleus into its nucleons (equal to energy released on formation).

Detailed explanation

Background Concept

Nuclear binding energy is a measure of how strongly the nucleons (protons and neutrons) are held together in a nucleus.

If you pull the nucleus apart into separate, free nucleons, you must supply energy. That energy is the binding energy.
If you assemble the nucleus from free nucleons, energy is released (because the final nucleus has lower mass-energy). The amount released is the same magnitude as the binding energy.

Binding energy is linked to mass defect using

E = \Delta m c^2.

Understanding the Question

You are asked to state what is meant by the binding energy of a nucleus. For full marks you should clearly refer to:

separating the nucleus into individual nucleons, and
energy required (or equivalently energy released on formation).

Approach

Give the standard definition: “energy needed to completely separate the nucleus into its constituent nucleons”. Optionally add the equivalent statement about energy released when the nucleus forms.

Step-by-Step Reasoning

Identify what is being separated: a nucleus into all its protons and neutrons.
State that to do this separation, you must do work against the attractive nuclear force, so you must supply energy.
Mention equivalence: the same energy would be released if the nucleons came together to form the nucleus.

Key Takeaways

Binding energy quantifies nuclear stability.
It is defined using complete separation into free nucleons.
It is equal in magnitude to energy released on formation.

Common Mistakes

Defining it as the energy to remove one nucleon (that is separation energy, not total binding energy).
Mixing up binding energy with binding energy per nucleon.
Forgetting to mention separation into individual nucleons.

Things to Be Careful About

Use the word nucleons (protons and neutrons), not “atoms”.
“Completely separate” means to infinity so nuclear forces are negligible.

Techniques used

state the definition of nuclear binding energyrelate binding energy to work done to separate nucleonsrelate binding energy to energy released when nucleons bind

(b)

A nucleus of uranium-235 absorbs a neutron and becomes unstable. It then undergoes a fission reaction. One possible reaction is

^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{142}_{54}\text{Xe} + ^{90}_{38}\text{Sr} + \text{neutrons}.

7M

(i)

Determine the number of neutrons produced in this fission reaction.

number = ______

1M Similar

DifficultyEasy

Worked solution

Working

Mass number:

235 + 1 = 142 + 90 + x

236 = 232 + x \Rightarrow x = 4

Answer

Number of neutrons produced $= 4$ .

Final answer

4

Detailed explanation

Background Concept

In any nuclear reaction, certain quantities are conserved:

Nucleon (mass) number $A$ (total number of protons + neutrons).
Proton (atomic) number $Z$ (total number of protons / total charge number).

To balance a nuclear equation, the sum of $A$ on the left equals the sum of $A$ on the right, and similarly for $Z$ .

Understanding the Question

The reaction given is

^{235}_{92}\text{U} + ^{1}_{0}\text{n} \rightarrow ^{142}_{54}\text{Xe} + ^{90}_{38}\text{Sr} + \text{neutrons}.

You must determine how many free neutrons are produced.

Approach

Use conservation of mass number $A$ (easiest here):

Add $A$ values on the left.
Add known $A$ values on the right.
The difference is the number of neutrons (each neutron has $A=1$ ).

Step-by-Step Reasoning

Left side mass number:

A_{\text{left}} = 235 + 1 = 236.

Right side known mass numbers:

A_{\text{right, known}} = 142 + 90 = 232.

Let the number of neutrons be $x$ . Each has $A=1$ , so total mass number from neutrons is $x$ .

236 = 232 + x \Rightarrow x = 4.

(You can also check charge conservation: $92 = 54 + 38$ so that is consistent.)

Key Takeaways

Balance nuclear equations by conserving $A$ and $Z$ .
Free neutrons contribute $A=1$ , $Z=0$ .

Common Mistakes

Forgetting the absorbed neutron on the left.
Using proton number $Z$ to find neutrons (neutrons have $Z=0$ , so $Z$ does not help directly here).
Arithmetic error: $236-232$ .

Things to Be Careful About

Make sure to include all products (both fission fragments and neutrons).
Each emitted neutron adds 1 to the total mass number.

Techniques used

apply conservation of nucleon numberapply conservation of proton (atomic) numberbalance a nuclear equation

(ii)

Data for the binding energies per nucleon for this fission reaction are given in Table 8.1.

Table 8.1

isotope	binding energy per nucleon / MeV
uranium-235	7.59
xenon-142	8.37
strontium-90	8.72

Calculate the energy released, in MeV, from the fission of one nucleus of uranium-235.

energy = ______ $\text{MeV}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Total binding energy of reactant nucleus:

E_{\text{U}} = 235 \times 7.59 = 1783.65\ \text{MeV}

Total binding energy of products:

E_{\text{Xe}} = 142 \times 8.37 = 1188.54\ \text{MeV}

E_{\text{Sr}} = 90 \times 8.72 = 784.8\ \text{MeV}

E_{\text{products}} = 1188.54 + 784.8 = 1973.34\ \text{MeV}

Energy released:

\Delta E = 1973.34 - 1783.65 = 189.69\ \text{MeV} \approx 1.90 \times 10^2\ \text{MeV}

Answer

Energy released $\approx 190\ \text{MeV}$ .

Final answer

190 MeV

Detailed explanation

Background Concept

The binding energy per nucleon is the average binding energy shared by each nucleon in that nucleus.

If a nucleus has mass number $A$ and binding energy per nucleon $\left(\frac{E_b}{A}\right)$ , then the total binding energy is

E_b = A \left(\frac{E_b}{A}\right).

In a fission (or fusion) reaction, energy released is found from the increase in total binding energy:

\text{energy released} = E_{b,\text{products}} - E_{b,\text{reactants}}.

(Free neutrons are not bound in nuclei, so they contribute no nuclear binding energy.)

Understanding the Question

You are given binding energy per nucleon values for:

uranium-235: $7.59\ \text{MeV per nucleon}$
xenon-142: $8.37\ \text{MeV per nucleon}$
strontium-90: $8.72\ \text{MeV per nucleon}$

You must calculate the energy released when one $^{235}_{92}\text{U}$ nucleus undergoes fission into $^{142}_{54}\text{Xe}$ and $^{90}_{38}\text{Sr}$ (plus neutrons).

Approach

Convert each binding energy per nucleon into total binding energy by multiplying by mass number.
Add total binding energies of product nuclei.
Subtract the reactant nucleus total binding energy.

Step-by-Step Reasoning

Reactant total binding energy (uranium-235):

E_{\text{U}} = 235 \times 7.59 = 1783.65\ \text{MeV}.

Product total binding energies:

E_{\text{Xe}} = 142 \times 8.37 = 1188.54\ \text{MeV}

E_{\text{Sr}} = 90 \times 8.72 = 784.8\ \text{MeV}.

Total binding energy of products:

E_{\text{products}} = 1188.54 + 784.8 = 1973.34\ \text{MeV}.

Energy released:

\Delta E = 1973.34 - 1783.65 = 189.69\ \text{MeV}.

To appropriate significant figures, this is about $190\ \text{MeV}$ .

Key Takeaways

Energy release in fission comes from products having higher binding energy per nucleon (greater total binding energy).
Multiply “per nucleon” values by mass number to get total binding energies.

Common Mistakes

Forgetting to multiply by mass number and just subtracting the per-nucleon values.
Including the emitted neutrons as if they had binding energy (they are free particles).
Using the total nucleon number $236$ instead of $235$ for uranium when only uranium-235 data are supplied.

Things to Be Careful About

Keep units consistent: the table is already in $\text{MeV}$ .
Round sensibly (typically 2–3 s.f. depending on given data).

Techniques used

convert binding energy per nucleon to total binding energyapply energy release as difference in total binding energyperform arithmetic with MeV energies

(iii)

The isotope xenon-142 is unstable. The isotope xenon-132 is stable.

Suggest a reason why xenon-142 is unstable.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

$^{142}\text{Xe}$ has too many neutrons (neutron-to-proton ratio too large) compared with stable xenon, so it is unstable.

Final answer

Neutron-to-proton ratio is too high (too many neutrons), so the nucleus is unstable.

Detailed explanation

Background Concept

Nuclear stability depends strongly on the neutron-to-proton ratio $\left(\frac{N}{Z}\right)$ .

For lighter nuclei, stability is roughly $N \approx Z$ .
For heavier nuclei, stable nuclei require more neutrons than protons, but there is still a limited range (the “band of stability”).

If a nucleus has:

too many neutrons (too large $N/Z$ ), it tends to undergo $\beta^-$ decay (a neutron turns into a proton).
too many protons (too small $N/Z$ ), it tends to undergo $\beta^+$ decay or electron capture.

Understanding the Question

You are told:

$^{142}\text{Xe}$ is unstable.
$^{132}\text{Xe}$ is stable.

You must suggest why $^{142}\text{Xe}$ is unstable, based on nuclear composition.

Approach

Compare how $^{142}\text{Xe}$ differs from $^{132}\text{Xe}$ :

Same element means same number of protons ( $Z=54$ ).
Larger mass number means more neutrons.
Too many neutrons relative to protons pushes it outside the stability band.

Step-by-Step Reasoning

For xenon, $Z=54$ .

For $^{132}\text{Xe}$ :

N = 132 - 54 = 78.

For $^{142}\text{Xe}$ :

N = 142 - 54 = 88.

So $^{142}\text{Xe}$ has 10 extra neutrons compared with the stable isotope, giving a larger $N/Z$ ratio. That makes it unstable (it will tend to reduce $N/Z$ by converting neutrons into protons via $\beta^-$ decay).

Key Takeaways

Same element $\Rightarrow$ same $Z$ .
Larger mass number $\Rightarrow$ more neutrons.
Too high or too low $N/Z$ leads to instability.

Common Mistakes

Saying “it has too many nucleons” (not specific enough).
Claiming it is unstable because it is “too heavy” without mentioning neutron/proton imbalance.
Confusing xenon-142 with a different element (changing $Z$ ).

Things to Be Careful About

You only need a suggestion (1 mark), so one clear statement about neutron excess is sufficient.
Don’t overcomplicate with binding energy calculations; the question is about stability trends.

Techniques used

compare neutron-to-proton ratio with stable isotopesapply the idea of the band of stabilityinfer likely decay mode from neutron excess

(iv)

Xenon-142 decays into the isotope caesium-142.

A sample initially contains only nuclei of xenon-142. After a time equal to $6.0 \text{ s}$ , the ratio

\frac{\text{number of decayed nuclei of xenon-142}}{\text{number of undecayed nuclei of xenon-142}}

is equal to 31.

Calculate the half-life of xenon-142. Show your working.

half-life = ______ $\text{s}$

3M Similar

DifficultyMedium

Worked solution

Working

Let initial number be $N_0$ and remaining (undecayed) number after $6.0\ \text{s}$ be $N$ .

\frac{N_0 - N}{N} = 31 \Rightarrow \frac{N_0}{N} = 32 \Rightarrow \frac{N}{N_0} = \frac{1}{32}

\frac{1}{32} = \left(\frac{1}{2}\right)^5

So $6.0\ \text{s}$ corresponds to $5$ half-lives:

T_{1/2} = \frac{6.0}{5} = 1.2\ \text{s}

Answer

Half-life $= 1.2\ \text{s}$ .

Final answer

1.2 s

Detailed explanation

Background Concept

Radioactive decay follows exponential decay:

N = N_0 e^{-\lambda t}

where:

$N_0$ is the initial number of undecayed nuclei,
$N$ is the number of undecayed nuclei after time $t$ ,
$\lambda$ is the decay constant.

Half-life $T_{1/2}$ is the time for $N$ to fall to $\frac{N_0}{2}$ , and it is related to $\lambda$ by

T_{1/2} = \frac{\ln 2}{\lambda}.

Sometimes it is easier to use the “number of half-lives” idea:

\frac{N}{N_0} = \left(\frac{1}{2}\right)^{t/T_{1/2}}.

Understanding the Question

Initially the sample contains only xenon-142 nuclei.

After $t = 6.0\ \text{s}$ , you are told the ratio

\frac{\text{number decayed}}{\text{number undecayed}}

equals $31$ .

Let:

undecayed = $N$
decayed = $N_0 - N$

You must find $T_{1/2}$ .

Approach

Convert the given ratio into the fraction remaining $\frac{N}{N_0}$ .
Recognise a convenient power of 2 (or use logs).
Relate the elapsed time to the number of half-lives.

Step-by-Step Reasoning

Use the ratio provided:

\frac{N_0 - N}{N} = 31.

Rearrange:

\frac{N_0}{N} - 1 = 31 \Rightarrow \frac{N_0}{N} = 32.

So the remaining fraction is

\frac{N}{N_0} = \frac{1}{32}.

Recognise

\frac{1}{32} = \frac{1}{2^5} = \left(\frac{1}{2}\right)^5.

So $6.0\ \text{s}$ corresponds to $5$ half-lives.

Therefore

T_{1/2} = \frac{6.0\ \text{s}}{5} = 1.2\ \text{s}.

(Alternative log method: set $\frac{N}{N_0}=e^{-\lambda t}=\frac{1}{32}$ , solve for $\lambda$ , then use $T_{1/2}=\ln2/\lambda$ ; it gives the same result.)

Key Takeaways

“Decayed : undecayed” can be turned into a remaining fraction $N/N_0$ .
Recognising powers of 2 can avoid logarithms.
Half-life relates to exponential decay through repeated halving.

Common Mistakes

Using $\frac{N}{N_0}=\frac{31}{32}$ instead of $\frac{1}{32}$ .
Treating the ratio $31$ as the number of half-lives directly.
Mixing up decayed with undecayed in the ratio.

Things to Be Careful About

Define clearly: decayed $= N_0 - N$ , undecayed $= N$ .
Check the rearrangement: a large decayed/undecayed ratio should mean very little remains (here only $1/32$ remains), which is a good sanity check.

Techniques used

express a ratio in terms of remaining nucleiuse exponential decay or half-life relationsolve for half-life using logarithms or powers of two

Q9MediumMedical Physics

(a)

Electrons in a vacuum are accelerated through a potential difference of $84 \text{ kV}$ . The electrons then strike a metal target and X-rays are produced.

4M

(i)

Calculate the minimum wavelength of the X-rays that are produced.

wavelength = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Maximum photon energy:

E = eV = (1.60 \times 10^{-19})(84 \times 10^{3}) = 1.34 \times 10^{-14}\ \text{J}

\lambda_{\min} = \frac{hc}{E} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{1.34 \times 10^{-14}} = 1.48 \times 10^{-11}\ \text{m}

Answer

$\lambda_{\min} = 1.48 \times 10^{-11}\ \text{m}$

Final answer

1.48 × 10^-11 m

Detailed explanation

Background Concept

When electrons are accelerated through a potential difference $V$ , they gain kinetic energy equal to the electrical work done:

E = eV

where $e$ is the elementary charge.

In an X-ray tube, the minimum wavelength (highest photon energy) occurs when one electron gives up all of its kinetic energy to a single photon. For a photon,

E = hf = \frac{hc}{\lambda}

So for the most energetic (shortest wavelength) photon,

eV = \frac{hc}{\lambda_{\min}}

Understanding the Question

Electrons are accelerated through $84\ \text{kV}$ in vacuum and then hit a metal target producing X-rays. You are asked for the minimum wavelength of the X-rays, i.e. the shortest possible wavelength in the spectrum.

Approach

Find the maximum possible photon energy using $E=eV$ .
Convert this energy into a wavelength using $\lambda = hc/E$ .

Step-by-Step Reasoning

Electrical energy gained by one electron:

E = eV = (1.60 \times 10^{-19}\ \text{C})(84 \times 10^3\ \text{V})

Since $1\ \text{V} = 1\ \text{J C}^{-1}$ , this gives energy in joules:

E = 1.34 \times 10^{-14}\ \text{J}

Set this equal to the maximum photon energy:

E = \frac{hc}{\lambda_{\min}}

Rearrange:

\lambda_{\min} = \frac{hc}{E}

Substitute $h=6.63\times 10^{-34}\ \text{J s}$ and $c=3.00\times 10^8\ \text{m s}^{-1}$ :

\lambda_{\min} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{1.34 \times 10^{-14}} = 1.48 \times 10^{-11}\ \text{m}

Key Takeaways

An electron accelerated through $V$ gains energy $eV$ .
The shortest X-ray wavelength corresponds to the maximum photon energy: $eV = hc/\lambda_{\min}$ .

Common Mistakes

Using $E=V$ instead of $E=eV$ .
Forgetting that $84\ \text{kV} = 84 \times 10^3\ \text{V}$ .
Using $c=3.00\times 10^8$ but $h$ in eV·s without consistent units.

Things to Be Careful About

Keep units consistent: convert to joules (or consistently use eV with $hc$ in eV·m).
Quote the wavelength in metres as requested.

Techniques used

equate electrical energy gained to maximum photon energyuse the photon energy relation E = hc / lambdaconvert electronvolts to joules

(ii)

The melting points of two metals are given in Table 9.1.

Table 9.1

metal	melting point / $^\circ\text{C}$
copper	1090
tungsten	3420

Suggest why the metal target is made from tungsten rather than copper.

2M Similar

DifficultyEasy

Worked solution

Answer

Tungsten has a much higher melting point than copper, so it is less likely to melt/overheat when struck by high-energy electrons.

Final answer

Tungsten has a much higher melting point so it is less likely to melt when heated by the electron beam.

Detailed explanation

Background Concept

In an X-ray tube, most of the kinetic energy of the incident electrons is converted into thermal energy in the target (only a small fraction becomes X-ray photons). This makes the target heat up significantly.

A practical target material therefore needs to withstand high temperatures without melting or deforming.

Understanding the Question

You are given melting points for copper ( $1090\ ^\circ\text{C}$ ) and tungsten ( $3420\ ^\circ\text{C}$ ). You must suggest why tungsten is chosen for the target rather than copper, using this data.

Approach

Use the fact that the target is heated strongly by the electron beam. A higher melting point makes the material more suitable.

Step-by-Step Reasoning

Electron impact produces substantial heating of the target.
Tungsten’s melting point ( $3420\ ^\circ\text{C}$ ) is much higher than copper’s ( $1090\ ^\circ\text{C}$ ).
Therefore tungsten is far less likely to melt, allowing the X-ray tube to operate safely at high accelerating voltages/currents.

Key Takeaways

X-ray targets get very hot because most electron energy becomes heat.
High melting point materials are preferred for durability.

Common Mistakes

Stating “tungsten is a better conductor” (not relevant to the provided melting point data).
Discussing attenuation coefficients or ultrasound properties (wrong section of the question).

Things to Be Careful About

The question says “Suggest why … rather than copper” and provides melting points, so the expected argument should explicitly use melting point/overheating.

Techniques used

relate target heating to melting pointuse given data to justify material choice

(b)

An X-ray beam is incident normally on a sample of soft tissue and bone as shown in Fig. 9.1.

Fig. 9.1

Data for the two materials are given in Table 9.2.

Table 9.2

medium	linear attenuation coefficient $\mu / \text{cm}^{-1}$	specific acoustic impedance $Z / 10^6 \text{ kg m}^{-2} \text{ s}^{-1}$
soft tissue	0.22	1.7
bone	3.0	7.8

The total thickness of soft tissue is $x$ . The total thickness of bone is also $x$ .

The incident intensity of the X-ray beam is $I_0$ . The transmitted intensity of the X-ray beam is $13\%$ of the incident intensity.

Determine $x$ , in cm.

$x$ = ______ $\text{cm}$

3M Similar

DifficultyMedium

Worked solution

Working

For successive layers:

\frac{I}{I_0} = e^{-\mu_s x} e^{-\mu_b x} = e^{-(\mu_s+\mu_b)x}

Given $I/I_0 = 0.13$ , $\mu_s = 0.22\ \text{cm}^{-1}$ and $\mu_b = 3.0\ \text{cm}^{-1}$ :

0.13 = e^{-(3.22)x}

x = \frac{-\ln(0.13)}{3.22} = 0.634\ \text{cm}

Answer

$x = 0.63\ \text{cm}$

Final answer

0.63 cm

Detailed explanation

Background Concept

X-ray intensity decreases as it passes through matter due to absorption and scattering. For a narrow, monoenergetic beam this is modelled by:

I = I_0 e^{-\mu x}

where:

$I_0$ is the incident intensity,
$I$ is the transmitted intensity,
$\mu$ is the linear attenuation coefficient (units $\text{length}^{-1}$ ),
$x$ is thickness.

For several layers in sequence, attenuation happens in each layer, so the overall transmission is the product of transmissions, which is equivalent to adding the exponents.

Understanding the Question

An X-ray beam passes normally through a sample made of soft tissue and bone. The total thickness of soft tissue is $x$ and the total thickness of bone is also $x$ . Data are:

soft tissue: $\mu_s = 0.22\ \text{cm}^{-1}$
bone: $\mu_b = 3.0\ \text{cm}^{-1}$

The transmitted intensity is $13\%$ of $I_0$ , i.e. $I/I_0 = 0.13$ . You must determine $x$ .

diagram-b-layers

Approach

Write down the transmission through soft tissue of thickness $x$ .
Write down the transmission through bone of thickness $x$ .
Multiply them to get total transmission, then solve for $x$ using natural logs.

Step-by-Step Reasoning

Transmission through soft tissue thickness $x$ :

\frac{I_1}{I_0} = e^{-\mu_s x}

Transmission through bone thickness $x$ :

\frac{I}{I_1} = e^{-\mu_b x}

Multiply to eliminate the intermediate intensity $I_1$ :

\frac{I}{I_0} = e^{-\mu_s x} e^{-\mu_b x} = e^{-(\mu_s+\mu_b)x}

Substitute values:

0.13 = e^{-(0.22+3.0)x} = e^{-3.22x}

Take natural logs:

\ln(0.13) = -3.22x

So:

x = \frac{-\ln(0.13)}{3.22} \approx 0.634\ \text{cm}

Rounded appropriately:

x \approx 0.63\ \text{cm}

Key Takeaways

Exponential attenuation: $I=I_0 e^{-\mu x}$ .
For multiple layers, add $\mu x$ terms in the exponent.
Use $\ln$ to solve for thickness.

Common Mistakes

Adding transmitted intensities instead of multiplying transmissions.
Using $13$ instead of $0.13$ for the fraction transmitted.
Mixing units (e.g. converting $x$ into metres when $\mu$ is in $\text{cm}^{-1}$ ).

Things to Be Careful About

The question states the total thicknesses are $x$ for each material; you do not need to split $x$ between the two soft tissue layers.
Keep $x$ in cm because $\mu$ is given in $\text{cm}^{-1}$ .

Techniques used

apply exponential attenuation I = I0 e^(-mu x)combine attenuations for successive layers by adding exponentsrearrange using natural logarithms

(c)

6M

(i)

Define the specific acoustic impedance of a medium.

2M Similar

(ii)

Use data from Table 9.2 to calculate the percentage of the intensity of ultrasound that is transmitted at a boundary between soft tissue and bone.

percentage transmitted = ______ $\%$

2M Similar

DifficultyMedium

Worked solution

Working

Intensity reflection coefficient:

R = \left( \frac{Z_2-Z_1}{Z_2+Z_1} \right)^2

With $Z_1=1.7$ and $Z_2=7.8$ :

R = \left( \frac{7.8-1.7}{7.8+1.7} \right)^2 = \left( \frac{6.1}{9.5} \right)^2 = 0.412

Fraction transmitted:

T = 1-R = 0.588

Answer

Percentage transmitted $= 58.8\% \approx 59\%$

Final answer

59 %

Detailed explanation

Background Concept

When an ultrasound wave hits a boundary between two media at normal incidence, part of the intensity is reflected because of the change in acoustic impedance.

The intensity reflection coefficient is:

R = \left( \frac{Z_2-Z_1}{Z_2+Z_1} \right)^2

The transmitted fraction (ignoring absorption in the media) is:

T = 1-R = \frac{4Z_1Z_2}{(Z_1+Z_2)^2}

Understanding the Question

You must use the impedances in Table 9.2:

soft tissue: $Z_1 = 1.7 \times 10^6\ \text{kg m}^{-2}\ \text{s}^{-1}$
bone: $Z_2 = 7.8 \times 10^6\ \text{kg m}^{-2}\ \text{s}^{-1}$

and calculate the percentage intensity transmitted at a single soft tissue–bone boundary.

diagram-c-boundary

Approach

Compute $R$ using the impedance mismatch formula.
Convert to transmitted fraction with $T=1-R$ .
Multiply by $100$ to get a percentage.

Step-by-Step Reasoning

Because both impedances are given with the same $\times 10^6$ factor, you can use the numbers $1.7$ and $7.8$ directly in the ratio (the common factor cancels).

Calculate reflection coefficient:

R = \left( \frac{7.8-1.7}{7.8+1.7} \right)^2 = \left( \frac{6.1}{9.5} \right)^2

Compute:

\frac{6.1}{9.5} \approx 0.642 \quad \Rightarrow \quad R \approx (0.642)^2 \approx 0.412

So about $41.2\%$ of the intensity is reflected.

Transmitted fraction:

T = 1-R = 1-0.412 = 0.588

Percentage transmitted:

0.588 \times 100\% = 58.8\% \approx 59\%

Key Takeaways

Reflection at an ultrasound boundary depends on impedance mismatch.
Use $R=((Z_2-Z_1)/(Z_2+Z_1))^2$ and then $T=1-R$ .

Common Mistakes

Using amplitude (pressure) coefficients instead of intensity coefficients.
Forgetting to square the ratio in $R$ .
Using $T=1+R$ or subtracting the wrong way round.

Things to Be Careful About

The question asks for intensity transmitted, not amplitude.
At normal incidence the formula above applies; for oblique incidence there can be additional complications (not needed here).

Techniques used

apply the intensity reflection coefficient at a boundarycalculate transmitted fraction using impedance valuesconvert a fraction into a percentage

(iii)

The ultrasound is now incident on the sample of soft tissue and bone shown in Fig. 9.1.

Suggest two reasons why the transmitted intensity through the sample is less than the answer in (c)(ii).

1 ______
2 ______

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

There are two boundaries (soft tissue–bone and bone–soft tissue), so further reflections reduce the transmitted intensity.
Ultrasound is attenuated (absorbed/scattered) as it passes through the soft tissue and bone, so intensity decreases with distance in the materials.

Final answer

Further reflections at multiple boundaries; attenuation (absorption/scattering) in the tissue and bone.

Detailed explanation

Background Concept

The calculation in (c)(ii) treats transmission at a single boundary and (usually) assumes no losses other than reflection at that boundary.

In a real sample, ultrasound intensity can be reduced by:

multiple reflections at several interfaces,
attenuation within materials (absorption and scattering), often modelled by an exponential decay with distance,
other effects such as beam divergence or non-normal incidence.

Understanding the Question

The sample in Fig. 9.1 consists of soft tissue, then bone, then soft tissue again. So an ultrasound wave entering the sample encounters more than one boundary and also travels through finite thicknesses of both media.

The question asks for two reasons why the intensity transmitted through the whole sample is less than the single-boundary transmission found in (c)(ii).

Approach

Pick two distinct mechanisms that necessarily reduce transmitted intensity through the entire stack:

additional boundary reflections, and 2) attenuation within the materials.

Step-by-Step Reasoning

More than one boundary:
- In Fig. 9.1, the wave first goes from soft tissue to bone (one boundary) and later from bone back to soft tissue (second boundary).
- At each boundary, some intensity is reflected due to impedance mismatch.
- Therefore the final transmitted intensity after both boundaries is smaller than the transmission of just one boundary.
Attenuation inside soft tissue and bone:
- As ultrasound travels, its intensity reduces due to absorption (conversion to internal energy/heat) and scattering.
- Bone, in particular, tends to attenuate ultrasound strongly.
- So even the portion that is transmitted at the first boundary will lose additional intensity while passing through the thicknesses of bone and soft tissue.

(Other creditable possibilities include beam spreading or non-normal incidence leading to refraction and reduced intensity at the detector.)

Key Takeaways

Total transmission through layered media is reduced by both boundary reflections and propagation losses.
A single-interface transmission calculation is an overestimate for a multi-layer sample.

Common Mistakes

Repeating the same reason twice (e.g. “reflection” and “echoes”) without identifying a second distinct mechanism.
Saying “intensity decreases because it travels further” without naming attenuation mechanisms (absorption/scattering).

Things to Be Careful About

The question compares to (c)(ii), so you must mention effects that were not included in the single-boundary model, especially the second boundary and attenuation through thickness.

Techniques used

identify additional boundaries causing further reflectionsaccount for attenuation within materialslink physical mechanisms to reduced transmitted intensity

Q10Medium-EasyAstronomy and Cosmology

(a)

The Sun has a surface temperature of $5780 \text{ K}$ . The luminosity of the Sun is $3.85 \times 10^{26} \text{ W}$ .

6M

(i)

Calculate the radius of the Sun.

radius = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using Stefan–Boltzmann:

L = 4\pi r^2 \sigma T^4

r = \sqrt{\frac{L}{4\pi\sigma T^4}}

r = \sqrt{\frac{3.85\times 10^{26}}{4\pi(5.67\times 10^{-8})(5780)^4}}

r \approx 6.96\times 10^8\ \text{m}

Answer

$6.96\times 10^8\ \text{m}$

Final answer

6.96 × 10^8 m

Detailed explanation

Background Concept

A star’s luminosity $L$ is the total power it radiates (energy per unit time). If a star’s surface behaves approximately like a black body, the power radiated per unit area is given by the Stefan–Boltzmann law:

\frac{L}{4\pi r^2} = \sigma T^4

where:

$r$ is the star’s radius,
$T$ is its surface temperature,
$\sigma = 5.67\times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$ is the Stefan–Boltzmann constant.

Rearranging gives $L = 4\pi r^2\sigma T^4$ .

Understanding the Question

You are given the Sun’s surface temperature $T = 5780\ \text{K}$ and luminosity $L = 3.85\times 10^{26}\ \text{W}$ . The question asks for the Sun’s radius $r$ .

Approach

Use:

L = 4\pi r^2 \sigma T^4

Rearrange to make $r$ the subject, then substitute the given values (all already in SI units).

Step-by-Step Reasoning

Start with:

L = 4\pi r^2 \sigma T^4

Solve for $r$ :

r^2 = \frac{L}{4\pi\sigma T^4} \qquad\Rightarrow\qquad r = \sqrt{\frac{L}{4\pi\sigma T^4}}

Substitute:

$L = 3.85\times 10^{26}\ \text{W}$
$\sigma = 5.67\times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$
$T = 5780\ \text{K}$

and evaluate to obtain:

r \approx 6.96\times 10^8\ \text{m}

This is consistent with the known solar radius (about $7\times 10^8\ \text{m}$ ), so the magnitude is sensible.

Key Takeaways

Use $L = 4\pi r^2\sigma T^4$ for a (black-body) star.
Rearranging for radius introduces a square root, so powers of ten must be handled carefully.

Common Mistakes

Forgetting the factor $4\pi r^2$ (using $\pi r^2$ instead).
Using $T^2$ instead of $T^4$ .
Using the wrong value or units for $\sigma$ .

Things to Be Careful About

Ensure $T$ is in kelvin.
Keep track of brackets when calculating $T^4$ and the denominator.
Give the final radius with an appropriate number of significant figures and unit $\text{m}$ .

Techniques used

apply the Stefan–Boltzmann lawrearrange an inverse-square area relationshipsubstitute values with correct SI unitstake a square root to obtain a length scale

(ii)

The Earth is a distance of $1.50 \times 10^{11} \text{ m}$ from the Sun.

Calculate the radiant flux intensity $F$ of the radiation from the Sun at a distance of $1.50 \times 10^{11} \text{ m}$ . Give a unit with your answer.

$F$ = ______ unit ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

F = \frac{L}{4\pi d^2}

F = \frac{3.85\times 10^{26}}{4\pi(1.50\times 10^{11})^2}

F \approx 1.36\times 10^3\ \text{W m}^{-2}

Answer

$1.36\times 10^3\ \text{W m}^{-2}$

Final answer

1.36 × 10^3 W m^-2

Detailed explanation

Background Concept

Radiant flux intensity $F$ (sometimes called irradiance) is the power received per unit area at some distance from a source.

If a source emits total power $L$ uniformly in all directions, that power spreads over the surface area of a sphere of radius $d$ :

\text{area} = 4\pi d^2

So:

F = \frac{L}{4\pi d^2}

Units: $L$ is in watts (W), area is in $\text{m}^2$ , so $F$ is in $\text{W m}^{-2}$ .

Understanding the Question

The Sun’s luminosity is $3.85\times 10^{26}\ \text{W}$ . Earth is at distance $d = 1.50\times 10^{11}\ \text{m}$ . The question asks for the radiant flux intensity at that distance.

Approach

Use the inverse square law for flux:

F = \frac{L}{4\pi d^2}

Calculate $4\pi d^2$ , then divide $L$ by this area.

Step-by-Step Reasoning

Square the distance:

d^2 = (1.50\times 10^{11})^2 = 2.25\times 10^{22}\ \text{m}^2

Sphere area:

4\pi d^2 = 4\pi(2.25\times 10^{22}) \approx 2.83\times 10^{23}\ \text{m}^2

Now:

F = \frac{3.85\times 10^{26}}{2.83\times 10^{23}} \approx 1.36\times 10^3\ \text{W m}^{-2}

This is close to the known solar constant ( $\approx 1.36\ \text{kW m}^{-2}$ ), so it is reasonable.

Key Takeaways

Flux decreases with square of distance: doubling $d$ makes $F$ four times smaller.
Always include the unit $\text{W m}^{-2}$ .

Common Mistakes

Using $\pi d^2$ instead of $4\pi d^2$ .
Squaring $1.50\times 10^{11}$ incorrectly (common power-of-ten error).
Giving unit as $\text{W m}^{-1}$ or just W.

Things to Be Careful About

Keep $d$ in metres.
Use consistent significant figures (typically 3 s.f. here).
Don’t confuse flux intensity with luminosity (total power).

Techniques used

apply the inverse square law for radiant fluxsubstitute values in scientific notationinclude correct derived SI units for flux

(iii)

The variation with wavelength of the intensity of radiation emitted from the Sun is shown in Fig. 10.1.

Fig. 10.1

Another star has the same radius as the Sun but has a lower surface temperature.

On Fig. 10.1, sketch a line to show the variation with wavelength of the intensity of the radiation emitted for this star.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

diagram-a-iii

Final answer

Sketch: peak at longer wavelength and lower intensity than the Sun.

Detailed explanation

Background Concept

A star’s spectrum is often approximated by black-body radiation.

Two key results:

Wien’s displacement law:

\lambda_{\max} T = b

so lowering $T$ makes the peak wavelength $\lambda_{\max}$ increase (shift to the right).

Stefan–Boltzmann law (total power per unit area):

\text{power per area} = \sigma T^4

So a cooler star emits less power per unit area. If radius is the same, its luminosity is smaller, so the area under the curve is smaller and the curve is generally lower.

Understanding the Question

You are told the other star has the same radius as the Sun but a lower surface temperature. You must sketch its intensity vs wavelength curve on the same axes.

Approach

Use black-body curve features:

Lower $T$ (\Rightarrow) peak moves to longer wavelength (right).
Lower $T$ (\Rightarrow) overall intensity decreases (curve lower; smaller area).

Step-by-Step Reasoning

On the provided Sun curve, draw a second smooth curve that:

starts at the origin and rises smoothly,
reaches a lower maximum than the Sun’s peak,
has its peak to the right of the Sun’s peak (greater wavelength),
then falls off at long wavelengths.

diagram-a-iii

Key Takeaways

Cooler black bodies peak at longer wavelengths.
Cooler black bodies emit less total radiation (smaller area under curve).

Common Mistakes

Shifting the peak to the left for a cooler star (wrong direction).
Keeping the peak height the same when temperature changes.
Drawing sharp corners instead of a smooth black-body curve.

Things to Be Careful About

Same radius does not mean same luminosity; luminosity still depends strongly on $T^4$ .
The sketch is qualitative: correct relative position (right shift) and lower intensity are the key marking points.

Techniques used

use Wien’s displacement law to predict peak shiftuse Stefan–Boltzmann scaling to compare total emitted powersketch a qualitative black-body spectrum with correct relative changes

(b)

A galaxy in the constellation Corona Borealis is moving away from the Earth.

4M

(i)

The visible emission spectrum for the Sun is shown in Fig. 10.2.

Fig. 10.2

The lines are at wavelengths of $397 \text{ nm}$ , $410 \text{ nm}$ , $434 \text{ nm}$ , $486 \text{ nm}$ and $656 \text{ nm}$ . The compositions of the Sun and a star in the Corona Borealis galaxy are similar.

On Fig. 10.3, sketch the emission spectrum for the star in the Corona Borealis galaxy as observed from the Earth. No calculations are required.

Fig. 10.3

1M Similar

DifficultyEasy

Worked solution

Answer

All emission lines shifted to longer wavelength (to the right), same pattern.

diagram-b-i

Final answer

Emission lines redshifted: all lines shifted to the right (longer wavelength).

Detailed explanation

Background Concept

If a light source is moving relative to an observer, the observed wavelength changes due to the Doppler effect.

If the source is moving away, the observed wavelength increases: redshift.
If the source is moving towards, the observed wavelength decreases: blueshift.

For spectra, this means all spectral lines appear at different wavelengths, but the pattern (relative separations) is preserved.

Understanding the Question

The galaxy is moving away from Earth. The star has similar composition to the Sun, so it produces the same set of emission lines, but due to motion those lines will be observed at longer wavelengths. You must sketch them on the blank wavelength axis.

Approach

Take each solar emission line position and shift it rightwards by the same fractional amount (no calculation needed here). Keep the number of lines and relative spacing pattern.

Step-by-Step Reasoning

Identify that “moving away” implies redshift.
Therefore each of the wavelengths $397\ \text{nm}$ , $410\ \text{nm}$ , $434\ \text{nm}$ , $486\ \text{nm}$ , $656\ \text{nm}$ would be observed at values greater than these.
On the blank spectrum, draw the same five vertical lines, but all displaced to the right compared with the Sun’s spectrum.

diagram-b-i

Key Takeaways

Receding sources produce redshift: wavelength increases.
Spectral “fingerprints” remain the same but shift along the wavelength scale.

Common Mistakes

Shifting lines to shorter wavelengths (blueshift).
Changing the number of lines or the relative pattern.

Things to Be Careful About

The question says “observed from the Earth”: always use Doppler shift direction based on relative motion to Earth.
No calculation is required; a clear rightward shift is sufficient.

Techniques used

apply the idea of Doppler redshift for a receding sourcetransfer a line spectrum pattern by shifting all wavelengths to larger valuessketch a spectrum with correct qualitative displacement

(ii)

The galaxy in Corona Borealis is moving away from the Earth at a speed of $21400 \text{ km s}^{-1}$ .

Use information from (b)(i) to calculate, in nm, the observed wavelength of the lowest visible energy emission for the star in the Corona Borealis galaxy.

wavelength = ______ $\text{nm}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Lowest visible energy corresponds to the largest wavelength: $\lambda_0 = 656\ \text{nm}$ .

\frac{\Delta \lambda}{\lambda_0} \approx \frac{v}{c} \qquad\Rightarrow\qquad \lambda_{\text{obs}} = \lambda_0\left(1+\frac{v}{c}\right)

v = 21400\ \text{km s}^{-1} = 2.14\times 10^7\ \text{m s}^{-1}

\lambda_{\text{obs}} = 656\left(1+\frac{2.14\times 10^7}{3.00\times 10^8}\right) \approx 7.03\times 10^2\ \text{nm}

Answer

$703\ \text{nm}$

Final answer

703 nm

Detailed explanation

Background Concept

Photon energy is related to wavelength by:

E = hf = \frac{hc}{\lambda}

So a larger wavelength means a lower photon energy.

For Doppler shift at speeds not too close to $c$ , the redshift approximation is:

\frac{\Delta\lambda}{\lambda_0} \approx \frac{v}{c}

For a source moving away, $\Delta\lambda > 0$ , so:

\lambda_{\text{obs}} = \lambda_0\left(1+\frac{v}{c}\right)

Understanding the Question

From (b)(i), the star has the same set of emission lines as the Sun, but shifted to longer wavelengths. You are asked for the observed wavelength of the lowest visible energy emission line.

“Lowest energy” in the visible corresponds to the longest wavelength of the listed lines. The longest given is $656\ \text{nm}$ .

The galaxy’s recession speed is $21400\ \text{km s}^{-1}$ , so the line is redshifted.

Approach

Choose the correct rest wavelength: $656\ \text{nm}$ .
Convert $v$ into $\text{m s}^{-1}$ if using $c$ in $\text{m s}^{-1}$ .
Apply $\lambda_{\text{obs}} = \lambda_0(1+v/c)$ .

Step-by-Step Reasoning

Identify the correct line:

Among $397, 410, 434, 486, 656\ \text{nm}$ , the largest is $656\ \text{nm}$ .
Therefore it corresponds to the lowest-energy visible emission.

Convert speed:

v = 21400\ \text{km s}^{-1} = 21400\times 10^3\ \text{m s}^{-1} = 2.14\times 10^7\ \text{m s}^{-1}

Compute fractional speed:

\frac{v}{c} = \frac{2.14\times 10^7}{3.00\times 10^8} \approx 0.0713

Apply the Doppler redshift approximation:

\lambda_{\text{obs}} = 656(1+0.0713) \approx 702.7\ \text{nm} \approx 703\ \text{nm}

Key Takeaways

Lowest photon energy in a set of lines corresponds to the longest wavelength.
Receding sources give redshift: observed wavelength increases.

Common Mistakes

Choosing the shortest wavelength line (that would be highest energy).
Using $\lambda_{\text{obs}} = \lambda_0(1-v/c)$ for a receding source (wrong sign).
Forgetting to convert $\text{km s}^{-1}$ to $\text{m s}^{-1}$ when using $c$ in $\text{m s}^{-1}$ .

Things to Be Careful About

Keep consistent units for $v$ and $c$ .
Quote the final wavelength in nm as requested.
At $v \approx 0.07c$ , the simple approximation is typically accepted at this level unless stated otherwise.

Techniques used

identify the lowest-energy visible photon from wavelength dataapply the Doppler redshift approximation for small v/cconvert speed into consistent units before substitution

(iii)

The wavelength in (b)(ii) is used to calculate a value for the surface temperature of the star in the Corona Borealis galaxy. The calculation does not give an accurate value.

State and explain whether this value of temperature is too high or too low.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Too low: motion away causes redshift so the observed wavelength is larger than the emitted wavelength; using this larger wavelength in Wien’s law gives a smaller temperature.

Final answer

Too low

Detailed explanation

Background Concept

Wien’s displacement law relates the wavelength of maximum intensity to temperature:

\lambda_{\max} T = b

So:

larger $\lambda_{\max}$ implies smaller $T$ .

Also, for a source moving away from the observer, Doppler redshift increases the observed wavelengths:

\lambda_{\text{obs}} > \lambda_{\text{emitted}}

Understanding the Question

In (b)(ii) you found an observed wavelength that is redshifted due to the galaxy’s recession. If someone then uses that wavelength to calculate the star’s surface temperature (as if it were the wavelength corresponding to emission at the surface), the temperature will be wrong. You must say whether it is too high or too low and explain.

Approach

Connect the two effects:

recession makes $\lambda$ appear bigger,
Wien’s law converts a bigger $\lambda$ into a smaller $T$ .

Step-by-Step Reasoning

Because the galaxy is moving away, all wavelengths are redshifted:

\lambda_{\text{obs}} = \lambda_{\text{emitted}}(1+z) \quad\Rightarrow\quad \lambda_{\text{obs}} > \lambda_{\text{emitted}}

If you mistakenly treat $\lambda_{\text{obs}}$ as the wavelength linked to temperature in Wien’s law, you would compute:

T = \frac{b}{\lambda}

Using a larger $\lambda$ gives a smaller $T$ .

Therefore the calculated surface temperature would be too low.

Key Takeaways

Redshift increases observed wavelengths.
Wien’s law has an inverse relationship between $\lambda_{\max}$ and $T$ .

Common Mistakes

Saying “too high” because confusing redshift with higher energy.
Forgetting that the wavelength used is observed at Earth, not emitted at the star.

Things to Be Careful About

Always distinguish between emitted and observed wavelength.
Be explicit about the causal chain: recession (\rightarrow) larger observed $\lambda$ (\rightarrow) smaller calculated $T$ .

Techniques used

link redshift to increased observed wavelengthapply Wien’s law qualitatively to infer temperature directionstate a conclusion with a causal explanation

Physics 9702/42 — February/March 2024

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