A helium atom may be modelled as a nucleus surrounded by two electrons in diametrically opposite circular orbits, each of radius $170\ \text{pm}$, as shown in Fig. 2.1. (a) State Coulomb’s law. (i) State the charge on the nucleus, in terms of the elementary charge $e$. charge = \_\_\_\_\_\_ $e$ (ii) Show that the electric force between the nucleus and one of the electrons is $1.6 \times 10^{-8}\ \text{N}$. (c) Assume that the force in (b)(ii) is the only force on the…

(a) For two point charges Q1 and Q2 separated by r: F = (1/4πϵ0) Q1Q2 /r^2 along the line joining them; repulsive for like charges and attractive for unlike. · (b)(i) +2e · (b)(ii) 1.6 × 10^-8 N · (c)(i) 1.7 × 10^6 m s^-1 · (c)(ii) 6.3 × 10^-16 s · (d)(i) 0.125 · (d)(ii) Repulsion from the other electron opposes the inward nuclear attraction, reducing the resultant centripetal force, so the orbit is less tight (larger radius / lower speed).

9702/43 May/June 2025 — Physics Solved Past Paper

Q1MediumGravitational Fields

(a)

Define gravitational potential at a point.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Gravitational potential at a point is the work done per unit mass by an external agent in bringing a small test mass from infinity to the point (with no change in kinetic energy).

Final answer

Work done per unit mass to bring a test mass from infinity to the point (no change in KE).

Detailed explanation

Background Concept

Gravitational potential, (\phi), is a way of describing a gravitational field using energy ideas.

It is defined as the potential energy per unit mass at a point, taking zero potential at infinity:

\phi = \frac{U}{m}

Equivalently, it is the work done per unit mass by an external agent to move a small test mass from infinity to that point slowly (so that its kinetic energy does not change).

Understanding the Question

You are asked for a definition, not a calculation. For full marks you must:

mention “per unit mass”,
use “from infinity” as the reference,
make clear it is work done (or potential energy) with no change in kinetic energy.

Approach

Write the standard definition used in Cambridge mark schemes: “work done per unit mass to bring a test mass from infinity to the point, without change in kinetic energy”.

Step-by-Step Reasoning

“Potential” is an energy quantity, but it is defined per unit mass.
The reference is chosen as infinity, where (\phi = 0).
To make work done equal to the change in potential energy, the mass must be moved slowly so its kinetic energy stays the same.

Key Takeaways

Gravitational potential is work done per unit mass.
The standard reference is infinity.
State the condition “no change in kinetic energy”.

Common Mistakes

Defining (g) (field strength) instead of (\phi).
Missing “per unit mass”.
Using the surface of the planet as the reference instead of infinity.
Forgetting to state “no change in kinetic energy”.

Things to Be Careful About

Use “external agent does work” (not “work done by the field”, which would have the opposite sign).
The potential is a scalar and is typically negative near a planet (because infinity is zero).

Techniques used

state a definition in terms of work done per unit massspecify the reference point at infinityinclude the condition of no change in kinetic energy

(b)

Mars is a planet that may be considered to be an isolated uniform sphere of radius $3.4 \times 10^{6}\ \text{m}$ .

A satellite of mass $122\ \text{kg}$ is in orbit around Mars at a constant height of $1.7 \times 10^{6}\ \text{m}$ above the surface of the planet.

The height of the orbit is increased to $6.8 \times 10^{6}\ \text{m}$ above the surface. This increases the gravitational potential energy of the satellite by $5.1 \times 10^{8}\ \text{J}$ .

5M

(i)

Show that the mass of Mars is $6.4 \times 10^{23}\ \text{kg}$ .

3M Similar

DifficultyMedium

Worked solution

Working

For a satellite of mass (m),

U = -\frac{GMm}{r}

Increase in GPE when moving from (r_1) to (r_2):

\Delta U = -\frac{GMm}{r_2} - \left(-\frac{GMm}{r_1}\right)=GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)

(r_1 = 3.4\times10^6 + 1.7\times10^6 = 5.1\times10^6\ \text{m})

(r_2 = 3.4\times10^6 + 6.8\times10^6 = 1.02\times10^7\ \text{m})

5.1\times10^8 = (6.67\times10^{-11})\,(122)\,M\left(\frac{1}{5.1\times10^6}-\frac{1}{1.02\times10^7}\right)

M = 6.4\times10^{23}\ \text{kg}

Answer

(M = 6.4\times10^{23}\ \text{kg})

Final answer

6.4 × 10^23 kg

Detailed explanation

Background Concept

For a point (or spherically symmetric) mass (M), the gravitational potential energy (U) of a mass (m) at distance (r) from the centre is

U = -\frac{GMm}{r}

The negative sign occurs because we define (U=0) at infinity; at finite (r) the mass is bound, so (U) is negative.

A change in gravitational potential energy when moving between radii (r_1) and (r_2) is

\Delta U = U_2 - U_1 = -\frac{GMm}{r_2} + \frac{GMm}{r_1} = GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)

Understanding the Question

You are told:

Mars radius (R = 3.4\times10^6\ \text{m})
satellite mass (m = 122\ \text{kg})
initial height (h_1 = 1.7\times10^6\ \text{m})
final height (h_2 = 6.8\times10^6\ \text{m})
increase in GPE (\Delta U = 5.1\times10^8\ \text{J})

You must show that the mass of Mars is (6.4\times10^{23}\ \text{kg}).

Key point: heights are above the surface, but the formula needs (r) from the centre.

Approach

Convert the two heights into orbital radii: (r_1 = R + h_1), (r_2 = R + h_2).
Use (\Delta U = GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)).
Rearrange to solve for (M) and substitute values.

Step-by-Step Reasoning

Find the radii:

r_1 = 3.4\times10^6 + 1.7\times10^6 = 5.1\times10^6\ \text{m}

r_2 = 3.4\times10^6 + 6.8\times10^6 = 1.02\times10^7\ \text{m}

Use the change in potential energy:

\Delta U = GMm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)

Here (r_2 > r_1), so (\frac{1}{r_1}-\frac{1}{r_2}) is positive, giving a positive increase in (U), consistent with the statement.

Substitute and solve for (M):

5.1\times10^8 = (6.67\times10^{-11})(122)M\left(\frac{1}{5.1\times10^6}-\frac{1}{1.02\times10^7}\right)

Evaluate the bracket:

\frac{1}{5.1\times10^6}=1.96\times10^{-7},\quad \frac{1}{1.02\times10^7}=9.80\times10^{-8}

Difference (\approx 9.80\times10^{-8}\ \text{m}^{-1}).

Then

M \approx \frac{5.1\times10^8}{(6.67\times10^{-11})(122)(9.80\times10^{-8})} \approx 6.4\times10^{23}\ \text{kg}

Key Takeaways

Use (U=-GMm/r) and work with (\Delta U) between two radii.
Always convert “height above surface” into (r) from the centre.
The increase in GPE for moving outward comes from becoming less negative.

Common Mistakes

Using heights (h_1, h_2) directly as (r).
Sign errors (e.g. writing (\Delta U = -GMm(1/r_1 - 1/r_2))).
Forgetting that (\Delta U) is (U_2-U_1).

Things to Be Careful About

Check that (r_2 > r_1) gives a positive (\Delta U) as stated.
Keep powers of ten consistent; these calculations are sensitive to exponent errors.
Use (G = 6.67\times10^{-11}\ \text{N m}^2\ \text{kg}^{-2}).

Techniques used

use the expression for gravitational potential energy of a point masscalculate orbital radii from surface radius plus heightequate change in gravitational potential energy to a given value and solve for the planet mass

(ii)

Calculate the gravitational potential $\phi$ at the surface of Mars. Give a unit with your answer.

$\phi$ = ______ unit ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At the surface, (r=R) and

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

\phi = -\frac{(6.67\times10^{-11})(6.4\times10^{23})}{3.4\times10^6} = -1.3\times10^7

Answer

(\phi = -1.3\times10^7\ \text{J kg}^{-1})

Final answer

-1.3 × 10^7 J kg^-1

Detailed explanation

Background Concept

Gravitational potential (\phi) at distance (r) from the centre of a spherical mass (M) is

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

It is related to potential energy by (U = m\phi). The unit is

\text{J kg}^{-1}\quad (\text{equivalently } \text{m}^2\ \text{s}^{-2})

The potential is negative because we define (\phi=0) at infinity.

Understanding the Question

You must calculate (\phi) at the surface of Mars, so (r = R = 3.4\times10^6\ \text{m}). You can use the mass of Mars found in part (b)(i): (M = 6.4\times10^{23}\ \text{kg}). The question also asks you to give a unit.

Approach

Use

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

Substitute (G), (M), and (R), keeping the negative sign, then quote units (\text{J kg}^{-1}).

Step-by-Step Reasoning

Write the expression at the surface:

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

Substitute values:

\phi = -\frac{(6.67\times10^{-11})(6.4\times10^{23})}{3.4\times10^6}

Calculate:

Numerator: (6.67\times10^{-11}\times 6.4\times10^{23} = 4.27\times10^{13})
Divide by (3.4\times10^6): (4.27\times10^{13} / 3.4\times10^6 \approx 1.26\times10^7)

So

\phi \approx -1.3\times10^7\ \text{J kg}^{-1}

Key Takeaways

For a spherical planet, (\phi = -GM/r).
At the surface, (r = R).
The unit of potential is (\text{J kg}^{-1}).

Common Mistakes

Missing the negative sign.
Using height above surface instead of radius from centre.
Giving unit as (\text{J}) instead of (\text{J kg}^{-1}).

Things to Be Careful About

The potential is not the same as field strength: (g = GM/r^2), whereas (\phi = -GM/r).
Quote sensible significant figures (usually 2 s.f. or 3 s.f. consistent with data).

Techniques used

use the definition of gravitational potential for a spherical masssubstitute values into $\phi = -GM/r$quote an appropriate unit for potential

(c)

The satellite in (b) is moved to an orbit in which the satellite remains at the same point above the surface of Mars.

2M

(i)

The orbit has a period of 25 hours.

State what can be deduced from this about the rotation of Mars on its axis.

1M Similar

(ii)

State one other feature of this orbit.

1M Similar

Q2MediumElectric FieldsMotion in a Circle

A helium atom may be modelled as a nucleus surrounded by two electrons in diametrically opposite circular orbits, each of radius $170\ \text{pm}$ , as shown in Fig. 2.1.

Fig. 2.1

(a)

State Coulomb’s law.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

For two point charges $Q_1$ and $Q_2$ separated by distance $r$ , the magnitude of the electrostatic force is

F = \frac{1}{4\pi\varepsilon_0}\frac{|Q_1Q_2|}{r^2}.

The force acts along the line joining the charges; it is repulsive for like charges and attractive for unlike charges.

Final answer

For two point charges Q1 and Q2 separated by r: F = (1/4πϵ0)|Q1Q2|/r^2 along the line joining them; repulsive for like charges and attractive for unlike.

Detailed explanation

Background Concept

Coulomb's law describes the electrostatic force between two point charges. It is an inverse-square law: doubling the separation makes the force four times smaller. The constant

\frac{1}{4\pi\varepsilon_0}

is sometimes written as $k$ .

The force is a vector. Its line of action is the straight line joining the charges, and its direction depends on whether the charges are like (repel) or unlike (attract).

Understanding the Question

You are asked to state Coulomb's law, so you must give the relationship for the force between two point charges, including how it depends on charge, separation, and the direction/attraction-repulsion condition.

Approach

Write the standard formula for the magnitude and then add the key directional information: line of action and whether it attracts or repels.

Step-by-Step Reasoning

Identify the variables: charges $Q_1$ , $Q_2$ and separation $r$ .
State the inverse-square dependence on separation and proportionality to the product of charges:

F \propto \frac{Q_1Q_2}{r^2}.

Include the constant of proportionality in SI units:

F = \frac{1}{4\pi\varepsilon_0}\frac{|Q_1Q_2|}{r^2}.

Add direction: along the line joining the charges; attraction for opposite charges and repulsion for like charges.

Key Takeaways

Coulomb's law: inverse-square in $r$ , proportional to $Q_1Q_2$ .
Direction is essential: along the line joining the charges.
Like charges repel; unlike charges attract.

Common Mistakes

Missing the $r^2$ (writing $1/r$ instead of $1/r^2$ ).
Omitting the statement about attraction/repulsion.
Forgetting that the force acts along the line joining the charges.

Things to Be Careful About

Use point charges (or spherical charges) assumption.
If you include signs in $Q_1Q_2$ , then be clear you are giving a vector direction; otherwise use magnitude bars and state attraction/repulsion separately.

Techniques used

state an inverse-square law with proportionalitiesspecify direction of force along the line joining chargesdistinguish attraction and repulsion based on signs of charge

(b)

2M

(i)

State the charge on the nucleus, in terms of the elementary charge $e$ .

charge = ______ $e$

1M Similar

(ii)

Show that the electric force between the nucleus and one of the electrons is $1.6 \times 10^{-8}\ \text{N}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For nucleus–electron separation $r = 170\ \text{pm} = 170 \times 10^{-12}\ \text{m}$ .

F = \frac{1}{4\pi\varepsilon_0}\frac{|Qq|}{r^2}

with $Q=2e$ and $q=e$ :

F = (8.99\times 10^{9})\frac{2(1.60\times 10^{-19})^2}{(170\times 10^{-12})^2}

F = 1.59\times 10^{-8}\ \text{N} \approx 1.6\times 10^{-8}\ \text{N}.

Answer

$1.6\times 10^{-8}\ \text{N}$

Final answer

1.6 × 10^-8 N

Detailed explanation

Background Concept

The magnitude of the electrostatic force between two point charges is given by Coulomb's law:

F = \frac{1}{4\pi\varepsilon_0}\frac{|Qq|}{r^2} = k\frac{|Qq|}{r^2},

where $k = 8.99\times 10^{9}\ \text{N m}^2\text{C}^{-2}$ .

The separation $r$ must be in metres and charge must be in coulombs.

Understanding the Question

You must show that the force between the helium nucleus ( $+2e$ ) and one electron ( $-e$ ) at radius $170\ \text{pm}$ has magnitude $1.6\times 10^{-8}\ \text{N}$ . This is a direct Coulomb's law substitution.

Approach

Convert $170\ \text{pm}$ to metres.
Substitute $Q = 2e$ , $q = e$ into Coulomb's law (magnitude).
Evaluate and round to the stated value.

Step-by-Step Reasoning

Convert the radius:

170\ \text{pm} = 170\times 10^{-12}\ \text{m}.

Use magnitudes in Coulomb's law:

F = k\frac{2e^2}{r^2}.

Substitute numbers:

F = (8.99\times 10^{9})\frac{2(1.60\times 10^{-19})^2}{(170\times 10^{-12})^2}.

Work through the powers of ten carefully.

$2(1.60\times 10^{-19})^2 = 2(2.56\times 10^{-38}) = 5.12\times 10^{-38}$
$(170\times 10^{-12})^2 = 2.89\times 10^{-20}$

So

F = 8.99\times 10^{9}\times \frac{5.12\times 10^{-38}}{2.89\times 10^{-20}} = 1.59\times 10^{-8}\ \text{N} \approx 1.6\times 10^{-8}\ \text{N}.

Key Takeaways

Always convert pm to m.
For magnitudes, you can use $|Qq|$ and avoid negative signs.
A quick sense-check: very small $r$ gives a relatively large force.

Common Mistakes

Using $r=170\times 10^{-10}\ \text{m}$ (mixing up pm and angstroms).
Forgetting the factor of $2$ from the nuclear charge $+2e$ .
Squaring $170$ but not squaring $10^{-12}$ .

Things to Be Careful About

The question asks for the force between nucleus and one electron, not between the two electrons.
Quote the final value to match the given answer (here $1.6\times 10^{-8}\ \text{N}$ ).

Techniques used

apply Coulomb's law to two point chargesconvert picometres to metressubstitute constants and evaluate to standard form

(c)

Assume that the force in (b)(ii) is the only force on the electrons.

4M

(i)

Calculate the speed of the orbiting electrons.

speed = ______ $\text{m s}^{-1}$

2M Similar

DifficultyMedium

Worked solution

Working

Assuming the only force is the electric attraction $F = 1.6\times 10^{-8}\ \text{N}$ providing centripetal force:

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

v = \sqrt{\frac{Fr}{m}}

with $r=170\times 10^{-12}\ \text{m}$ and $m = 9.11\times 10^{-31}\ \text{kg}$ :

v = \sqrt{\frac{(1.6\times 10^{-8})(170\times 10^{-12})}{9.11\times 10^{-31}}} = 1.7\times 10^{6}\ \text{m s}^{-1}.

Answer

$1.7\times 10^{6}\ \text{m s}^{-1}$

Final answer

1.7 × 10^6 m s^-1

Detailed explanation

Background Concept

For uniform circular motion of radius $r$ and speed $v$ , the required centripetal acceleration is

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

The net inward (radial) force must therefore be

F_{\text{centripetal}} = ma = \frac{mv^2}{r}.

In this model, the only force is the electrostatic attraction between the nucleus and the electron, so that force must be the centripetal force.

Understanding the Question

You are told to assume the force found in (b)(ii), $1.6\times 10^{-8}\ \text{N}$ , is the only force acting on each electron. That force acts towards the nucleus, so it provides the centripetal force needed for circular motion at radius $170\ \text{pm}$ . You must find the electron speed.

Approach

Set

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

and solve for $v$ using the given $F$ and $r$ and the electron mass.

Step-by-Step Reasoning

diagram-c-i

Use the centripetal force equation:

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

Rearrange for $v$ :

v^2 = \frac{Fr}{m} \quad\Rightarrow\quad v = \sqrt{\frac{Fr}{m}}.

Substitute values:

$F = 1.6\times 10^{-8}\ \text{N}$
$r = 170\times 10^{-12}\ \text{m}$
$m = 9.11\times 10^{-31}\ \text{kg}$

\frac{Fr}{m} = \frac{(1.6\times 10^{-8})(170\times 10^{-12})}{9.11\times 10^{-31}} \approx 3.0\times 10^{12}.

Take the square root:

v \approx \sqrt{3.0\times 10^{12}} \approx 1.7\times 10^{6}\ \text{m s}^{-1}.

Key Takeaways

For circular motion, a net inward force is required: $F = mv^2/r$ .
If a single force acts radially inward, it must equal the centripetal force.
Rearranging to $v = \sqrt{Fr/m}$ is a common step.

Common Mistakes

Using $F = mvr$ or other incorrect centripetal expressions.
Using diameter $340\ \text{pm}$ instead of the radius $170\ \text{pm}$ .
Forgetting to use the electron mass (or using proton mass).

Things to Be Careful About

The centripetal force is provided by the resultant radial force, not “an extra force”.
Keep powers of ten consistent: $10^{-8}\times 10^{-12} = 10^{-20}$ , then dividing by $10^{-31}$ gives $10^{11}$ , etc.
Quote the speed to a sensible number of significant figures (typically 2 s.f. here).

Techniques used

equate electrostatic attraction to centripetal forceuse F = m v^2 / r for uniform circular motionrearrange and evaluate a square root expression

(ii)

Calculate the period of the orbit of the electrons.

period = ______ $\text{s}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

T = \frac{2\pi(170\times 10^{-12})}{1.7\times 10^{6}} = 6.3\times 10^{-16}\ \text{s}.

Answer

$6.3\times 10^{-16}\ \text{s}$

Final answer

6.3 × 10^-16 s

Detailed explanation

Background Concept

For uniform circular motion:

the distance travelled in one orbit is the circumference $2\pi r$ ,
the time for one orbit is the period $T$ ,
speed is distance/time.

So

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

Understanding the Question

You have already found the electron's speed (from part (c)(i)) and you know the orbit radius is $170\ \text{pm}$ . The question asks for the period (time for one complete orbit).

Approach

Use

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

with $r$ in metres and $v$ in $\text{m s}^{-1}$ .

Step-by-Step Reasoning

Convert (or use) the radius in metres:

r = 170\times 10^{-12}\ \text{m}.

Substitute into the period formula:

T = \frac{2\pi(170\times 10^{-12})}{1.7\times 10^{6}}.

Evaluate:

$2\pi r \approx 1.07\times 10^{-9}\ \text{m}$
dividing by $1.7\times 10^{6}$ gives

T \approx 6.3\times 10^{-16}\ \text{s}.

Key Takeaways

Period is circumference divided by speed: $T = 2\pi r / v$ .
Very small orbits and high speeds produce extremely small periods.

Common Mistakes

Using $T = v/(2\pi r)$ (inverting the relationship).
Forgetting that pm must be converted to m.
Using $r=340\ \text{pm}$ (diameter) instead of the radius.

Things to Be Careful About

Use the speed from (c)(i) consistently (error carried forward would usually be allowed).
Express the final time in standard form and include the unit $\text{s}$ .

Techniques used

use T = 2 pi r / v for uniform circular motionsubstitute radius and speed with correct unit conversionsexpress a very small time in standard form

(d)

In practice, the orbit of each electron is affected by the presence of the other electron.

3M

(i)

For the position of one of the electrons, determine the ratio

\frac{\text{electric field strength due to the other electron}}{\text{electric field strength due to the nucleus}}

ratio = ______

2M Similar

DifficultyMedium

Worked solution

Working

At one electron:

distance to nucleus: $r = 170\ \text{pm}$ , nucleus charge $= +2e$
distance to other electron: $2r = 340\ \text{pm}$ , electron charge magnitude $= e$

\text{ratio} = \frac{k\,e/(2r)^2}{k\,(2e)/r^2} = \frac{e}{4r^2}\times\frac{r^2}{2e} = \frac{1}{8} = 0.125.

Answer

$0.125$

Final answer

0.125

Detailed explanation

Background Concept

The electric field strength due to a point charge $Q$ at distance $r$ is

E = \frac{1}{4\pi\varepsilon_0}\frac{|Q|}{r^2} = k\frac{|Q|}{r^2}.

When comparing two fields at the same point, taking a ratio is powerful because the constant $k$ cancels.

Understanding the Question

You look at the position of one electron. Two sources produce electric field at that point:

the nucleus at the centre (charge $+2e$ ) at distance $170\ \text{pm}$ ,
the other electron (charge magnitude $e$ ) which is diametrically opposite, so it is a distance of one diameter, $340\ \text{pm}$ , away.

You must find

\frac{E_{\text{other electron}}}{E_{\text{nucleus}}}.

Approach

Write $E = k|Q|/r^2$ for each source at that point and divide, using the correct distances ( $r$ and $2r$ ) and charges ( $e$ and $2e$ ).

Step-by-Step Reasoning

diagram-d-i

Field due to the other electron at distance $2r$ :

E_e = k\frac{e}{(2r)^2} = k\frac{e}{4r^2}.

Field due to nucleus at distance $r$ :

E_N = k\frac{2e}{r^2}.

Form the ratio:

\frac{E_e}{E_N} = \frac{k\,e/(4r^2)}{k\,(2e)/r^2}.

Cancel $k$ , cancel $e$ , and simplify:

\frac{E_e}{E_N} = \frac{1}{4}\times\frac{1}{2} = \frac{1}{8} = 0.125.

Key Takeaways

For point charges, $E \propto Q/r^2$ .
When charges and distances are multiples, ratios simplify quickly.
Here the other electron is twice as far away and has half the charge magnitude of the nucleus (compared with $2e$ ), giving a small ratio.

Common Mistakes

Using $170\ \text{pm}$ for both distances (forgetting the electrons are opposite ends of the diameter).
Using nucleus charge $+e$ instead of $+2e$ .
Squaring the distance incorrectly: $(2r)^2 = 4r^2$ .

Things to Be Careful About

The question asks for electric field strength, not force. (They are related by $F=qE$ , but the ratio of fields is independent of the test charge.)
Use magnitudes for the ratio; direction is not needed until part (d)(ii).

Techniques used

use E = kQ/r^2 for point chargescompare field strengths by forming a ratiouse geometry of the model to identify separations

(ii)

Use your answer in (d)(i) to suggest and explain how the orbit of the electron is affected by the presence of the other electron.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The other electron repels the electron outwards, opposing the inward attraction to the nucleus. Since its field is about $0.125$ of the nucleus field, the resultant inward (centripetal) force is reduced, so the electron is less tightly bound and the orbit would be at a larger radius (or equivalently a smaller speed for the same radius).

Final answer

Repulsion from the other electron opposes the inward nuclear attraction, reducing the resultant centripetal force, so the orbit is less tight (larger radius / lower speed).

Detailed explanation

Background Concept

For an electron in a circular orbit, the net inward (towards the centre) force must provide the centripetal force:

F_{\text{net,in}} = \frac{mv^2}{r}.

Electrostatic forces can be attractive or repulsive:

nucleus ( $+2e$ ) attracts the electron inward,
the other electron ( $-e$ ) repels it, and because it is on the opposite side of the nucleus, that repulsion on the chosen electron points radially outward.

Understanding the Question

Part (d)(i) showed that at the position of one electron, the electric field magnitude from the other electron is $0.125$ times that from the nucleus. You must use this to decide what happens to the orbit when both interactions are present.

Approach

Decide the direction of each force on the electron.
Use the ratio to judge how much the outward effect reduces the inward attraction.
Link reduced net inward force to the requirements for circular motion (it changes the speed and/or radius of the orbit).

Step-by-Step Reasoning

diagram-d-ii

Direction of forces on one electron:

Force due to nucleus: attractive, towards nucleus (inward).
Force due to other electron: repulsive, away from the other electron. Since the other electron is on the opposite side, this is outward from the nucleus.

Magnitude comparison:
From (d)(i), the other electron's field is $0.125$ of the nucleus field. Because $F = qE$ for the same electron charge $q$ , the force magnitudes have the same ratio. So the outward repulsion is about $12.5\%$ of the inward attraction.
Resultant radial force:
Net inward force is reduced compared with the nucleus-only model.
Effect on orbit:
With a smaller inward force available, the electron cannot sustain the same circular motion at the same radius and speed as before. Physically it will be less tightly bound, so the orbit tends to be larger (or the speed would need to be smaller for the same radius).

Key Takeaways

Forces from multiple charges add as vectors; directions matter.
A repulsive force from the other electron acts outward and reduces the net centripetal force.
Reduced net inward force implies a less tight orbit (larger radius / lower speed).

Common Mistakes

Saying the other electron increases the inward force (wrong direction: electrons repel).
Using the ratio to claim the effect is negligible without linking to centripetal force.
Confusing electric field direction with force direction on an electron (force is opposite to field for a negative charge).

Things to Be Careful About

Part (d)(i) was about field strength magnitudes; for (d)(ii) you must consider direction as well.
Because the electron is negative, it is often clearer to think in terms of forces (attraction/repulsion) rather than field direction.

Techniques used

compare directions of forces due to attraction and repulsionuse the ratio of field strengths to infer change in resultant forcepredict qualitative effect on circular orbit from reduced centripetal force

Q3MediumTemperature

(a)

Define specific latent heat.

2M Similar

DifficultyEasy

Worked solution

Answer

Specific latent heat is the thermal energy required to change the state of unit mass of a substance with no change in temperature.

Final answer

Thermal energy required to change the state of unit mass of a substance with no change in temperature.

Detailed explanation

Background Concept

During a phase change (melting, boiling, condensing, freezing), energy can be transferred to or from a substance without changing its temperature. This energy goes into changing the arrangement/separation of molecules rather than increasing their average kinetic energy.

The specific latent heat $L$ of a particular phase change is defined using

Q = mL

where $Q$ is the energy transferred, $m$ is the mass undergoing the change of state, and $L$ is the energy required per unit mass.

Understanding the Question

You are asked to define specific latent heat. For full credit you must include:

it is energy per unit mass,
it is for a change of state,
the temperature does not change during the change.

Approach

Write a single sentence definition including the three points above.

Step-by-Step Reasoning

“Latent heat” refers to energy transferred during a phase change.
“Specific” means per unit mass.
During a phase change for a pure substance at constant pressure, the temperature stays constant even though energy is being transferred.

So the definition must explicitly say “change of state” and “no change in temperature”.

Key Takeaways

Specific latent heat is energy per unit mass for a phase change.
Temperature remains constant during the phase change.
The relation $Q=mL$ links the definition to calculations.

Common Mistakes

Defining it as energy to raise temperature (that is specific heat capacity, not latent heat).
Omitting “per unit mass”.
Omitting the condition that temperature does not change.

Things to Be Careful About

Use “change of state” (or “phase change”) explicitly.
Do not mix up “latent heat” (total) with “specific latent heat” (per unit mass).

Techniques used

state the definition of a thermodynamic quantity preciselyspecify the condition of constant temperature during phase changeinclude unit mass in the definition

(b)

Explain why, for a substance, the specific latent heat of vaporisation is usually greater than the specific latent heat of fusion.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

For fusion (solid to liquid), energy is mainly used to loosen bonds so molecules can move past each other, but they remain close together.

For vaporisation (liquid to gas), molecules must separate widely and overcome intermolecular attractions (and do work against external pressure), so more energy per unit mass is required. Hence $L_v$ is usually greater than $L_f$ .

Final answer

Vaporisation requires separating molecules widely and overcoming intermolecular attraction (and doing work against pressure), whereas fusion mainly loosens bonds with molecules still close together, so more energy per unit mass is needed for vaporisation.

Detailed explanation

Background Concept

Latent heat is the energy associated with changing a substance’s internal energy during a phase change without changing temperature.

At constant temperature, the average kinetic energy of the molecules does not change. Therefore, the energy transferred goes mainly into changing molecular potential energy by overcoming intermolecular attractions and changing molecular separation.

Understanding the Question

You must explain why, for the same substance,

the specific latent heat of vaporisation $L_v$ is usually larger than
the specific latent heat of fusion $L_f$ .

This is asking for a molecular / energy argument, not a numerical one.

Approach

Compare what happens to molecular separation and intermolecular forces in:

melting (solid (\to) liquid)
boiling/evaporating (liquid (\to) gas)

Then relate “more separation / more overcoming forces” to “more energy per unit mass”, i.e. larger specific latent heat.

Step-by-Step Reasoning

Fusion (melting):
- In a solid, molecules are in fixed positions (vibrate about equilibrium positions).
- To melt, the molecules only need enough energy to break/loosen some of the bonds so they can move around.
- The molecules in a liquid are still close together, so not all attractive forces are fully overcome.
Vaporisation (boiling/evaporation):
- In a liquid, molecules are close together and strongly attracted.
- To become a gas, molecules must be separated to much larger distances.
- This requires overcoming intermolecular attractions much more completely, increasing molecular potential energy by a larger amount.
- In addition, at constant pressure the expanding gas does work on the surroundings (work done $W = p\Delta V$ ), which also requires energy.
Therefore the energy required per unit mass is larger for vaporisation, so typically

L_v > L_f.

Key Takeaways

Latent heats are mainly about changing molecular potential energy at constant temperature.
Liquid (\to) gas involves much greater separation than solid (\to) liquid.
Vaporisation often includes work done in expansion, increasing the required energy.

Common Mistakes

Saying “because boiling happens at a higher temperature”: latent heats are defined at the phase change temperature, but the reason is molecular separation/forces, not the numerical temperature.
Talking about kinetic energy increase: temperature is constant during the phase change.
Vague statements like “it needs more energy” without explaining what the energy is used for.

Things to Be Careful About

Make it clear that molecules remain close in a liquid but are far apart in a gas.
Mention overcoming intermolecular forces (and optionally work done against pressure) to fully justify the larger $L_v$ .

Techniques used

compare molecular arrangements in solid, liquid and gasrelate latent heat to work done against intermolecular forcesuse energy considerations to justify a qualitative comparison

(c)

An ice cube of mass $37.0\ \text{g}$ at temperature $0.0\ ^{\circ}\text{C}$ is placed in a beaker containing water of mass $208\ \text{g}$ at temperature $26.4\ ^{\circ}\text{C}$ .

When all the ice has melted, and all the water in the beaker has reached thermal equilibrium, the final temperature of all the water is $10.3\ ^{\circ}\text{C}$ .

The specific heat capacity of water is $4.18\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$ .

The beaker has negligible specific heat capacity and is perfectly insulated from the surroundings.

Determine a value, to three significant figures, for the specific latent heat of fusion of water.

specific latent heat of fusion = ______ $\text{J g}^{-1}$

4M Similar

DifficultyMedium

Worked solution

Working

Heat lost by warm water:

Q_{\text{lost}} = mc\Delta T = (208)(4.18)(26.4-10.3) = 1.40 \times 10^{4}\ \text{J}

Heat gained by ice = heat to melt + heat to warm melted water from $0.0\ ^{\circ}\text{C}$ to $10.3\ ^{\circ}\text{C}$ :

Q_{\text{gain}} = (37.0)L_f + (37.0)(4.18)(10.3)

Energy balance ( $Q_{\text{lost}}=Q_{\text{gain}}$ ):

1.40 \times 10^{4} = 37.0L_f + (37.0)(4.18)(10.3)

L_f = \frac{1.40\times 10^{4} - (37.0)(4.18)(10.3)}{37.0} = 3.35 \times 10^{2}\ \text{J g}^{-1}

Answer

$L_f = 335\ \text{J g}^{-1}$

Final answer

335 J g^-1

Detailed explanation

Background Concept

In a perfectly insulated calorimetry problem, the total energy is conserved:

energy lost by the warmer part of the system
= energy gained by the cooler part of the system.

Two kinds of thermal energy changes can occur:

Sensible heating/cooling (temperature changes):

Q = mc\Delta T

Latent heating (phase change at constant temperature):

Q = mL

Here, ice at $0\ ^{\circ}\text{C}$ melts (latent heat of fusion) and then the meltwater warms up to the final equilibrium temperature.

Understanding the Question

Given:

ice mass $m_i = 37.0\ \text{g}$ at $0.0\ ^{\circ}\text{C}$
water mass $m_w = 208\ \text{g}$ at $26.4\ ^{\circ}\text{C}$
final equilibrium temperature $\theta_f = 10.3\ ^{\circ}\text{C}$
$c_{\text{water}} = 4.18\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$
beaker negligible heat capacity and insulated

Unknown: specific latent heat of fusion of water $L_f$ in $\text{J g}^{-1}$ .

Key idea: the warm water cools down, providing energy to (i) melt the ice and (ii) warm the melted ice-water from $0.0$ to $10.3\ ^{\circ}\text{C}$ .

Approach

Calculate heat lost by the original warm water as it cools from $26.4$ to $10.3\ ^{\circ}\text{C}$ using $Q=mc\Delta T$ .
Write heat gained by the ice as the sum:
- $m_i L_f$ to melt at $0\ ^{\circ}\text{C}$
- $m_i c (10.3 - 0.0)$ to warm the meltwater to the final temperature
Equate heat lost = heat gained and solve for $L_f$ .

Step-by-Step Reasoning

Heat lost by warm water

Temperature drop:

\Delta T_w = 26.4 - 10.3 = 16.1\ ^{\circ}\text{C}

So

Q_{\text{lost}} = m_w c \Delta T_w = (208)(4.18)(16.1)

First multiply $4.18 \times 16.1 = 67.298$ , then

Q_{\text{lost}} = 208 \times 67.298 = 1.3998 \times 10^{4}\ \text{J} \approx 1.40 \times 10^{4}\ \text{J}

Heat gained by the ice

Melting at $0\ ^{\circ}\text{C}$ needs latent heat:

Q_{\text{melt}} = m_i L_f = 37.0L_f

After melting, the meltwater warms from $0.0$ to $10.3\ ^{\circ}\text{C}$ :

Q_{\text{warm}} = m_i c (10.3 - 0.0) = (37.0)(4.18)(10.3)

Compute:

(4.18)(10.3) = 43.054

so

Q_{\text{warm}} = 37.0 \times 43.054 = 1.593 \times 10^{3}\ \text{J}

Energy balance and solve for $L_f$

Because the system is insulated and the beaker’s heat capacity is negligible:

Q_{\text{lost}} = Q_{\text{melt}} + Q_{\text{warm}}

So

1.3998\times 10^{4} = 37.0L_f + 1.593\times 10^{3}

Rearrange:

37.0L_f = 1.3998\times 10^{4} - 1.593\times 10^{3} = 1.2405\times 10^{4}

Divide by $37.0$ :

L_f = \frac{1.2405\times 10^{4}}{37.0} = 3.3527\times 10^{2}\ \text{J g}^{-1}

To three significant figures:

L_f = 335\ \text{J g}^{-1}.

Key Takeaways

In insulated mixing problems, set heat lost = heat gained.
Ice at $0\ ^{\circ}\text{C}$ requires both latent heat to melt and then energy to warm the meltwater.
Use consistent units: here $c$ is in $\text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$ so masses must be in grams.

Common Mistakes

Forgetting to include the heating of the melted ice-water from $0$ to the final temperature.
Using the total final mass $(208+37)\ \text{g}$ in the cooling calculation instead of just the warm water mass.
Using $\Delta T = 26.4$ instead of $26.4-10.3$ for the warm water.
Mixing units (using kg with $c$ in $\text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$ ).

Things to Be Careful About

The ice starts at $0\ ^{\circ}\text{C}$ , so there is no warming of ice before melting.
The final temperature is above $0\ ^{\circ}\text{C}$ , so all ice melts (consistent with the wording) and the meltwater must be warmed.
Quote $L_f$ to three significant figures as requested, and include unit $\text{J g}^{-1}$ .

Techniques used

apply conservation of energy in a thermal equilibrium problemcalculate thermal energy change using Q = mcΔTset up and solve an equation including latent heat and sensible heating

Q4MediumThermodynamicsIdeal Gases

(a)

4M

(i)

State what is meant by the internal energy of a system.

2M Similar

DifficultyEasy

Worked solution

Answer

Internal energy is the total microscopic energy of a system, i.e. the sum of the random kinetic energies of its particles and the potential energies due to intermolecular forces (and separations).

Final answer

Sum of random molecular kinetic energy and intermolecular potential energy.

Detailed explanation

Background Concept

Internal energy $U$ is an energy store associated with the microscopic motion and arrangement of particles in a substance.

For a gas, the particles have:

random kinetic energy (translational, and possibly rotational/vibrational depending on the model), and
potential energy due to intermolecular forces (important in real gases; negligible for an ideal gas).

Internal energy is not the macroscopic kinetic energy of the whole object moving, and not the macroscopic gravitational potential energy of the object in an external field.

Understanding the Question

You are asked to state what “internal energy of a system” means. For full credit you must mention that it is a microscopic total and include both kinetic and potential contributions.

Approach

Give the standard definition used at A Level:

random kinetic energy of the particles
plus potential energy associated with intermolecular forces.

Step-by-Step Reasoning

A “system” contains many particles.
Each particle has kinetic energy because it is moving randomly.
Particles may also have potential energy because of their positions/separations in the intermolecular force field.
Internal energy is the sum of these microscopic energies over all particles.

Key Takeaways

Internal energy is a microscopic energy store.
It is the sum of random kinetic energies plus intermolecular potential energies.

Common Mistakes

Saying only “energy due to temperature” without stating what energies are included.
Confusing internal energy with heat (thermal energy transfer) or with macroscopic kinetic energy of the whole gas.

Things to Be Careful About

Use “random” or “microscopic” to make clear it is not bulk motion.
Include both kinetic and potential terms to secure the 2 marks.

Techniques used

define internal energy in terms of microscopic energy storesdistinguish internal energy from macroscopic kinetic and potential energy

(ii)

Explain why the internal energy of an ideal gas is directly proportional to the thermodynamic temperature of the gas.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

For an ideal gas, intermolecular forces are negligible, so the internal energy is the total random kinetic energy of the molecules.

From kinetic theory, the mean translational kinetic energy per molecule is

\frac{3}{2}kT.

Hence total internal energy $U \propto T$ (e.g. $U = \frac{3}{2}NkT$ ), so $U$ is directly proportional to thermodynamic temperature.

Final answer

In an ideal gas, internal energy is total random kinetic energy; mean kinetic energy per molecule is (3/2)kT, so total U ∝ T.

Detailed explanation

Background Concept

An ideal gas is modelled as:

molecules that occupy negligible volume,
molecules that exert no intermolecular forces except during collisions,
perfectly elastic collisions.

Because intermolecular forces are neglected, there is (effectively) no intermolecular potential energy stored in the gas. Therefore, for an ideal gas:

internal energy $U$ depends only on the random kinetic energy of the molecules.

Kinetic theory gives the key result:

\text{mean translational kinetic energy per molecule} = \frac{3}{2}kT,

where $k$ is the Boltzmann constant and $T$ is the thermodynamic temperature in kelvin.

Understanding the Question

You must explain (not just state) why the internal energy of an ideal gas is directly proportional to $T$ . The word “ideal” is the clue: it tells you potential energy due to intermolecular forces is negligible, so $U$ is purely kinetic.

Approach

State that for an ideal gas, internal energy is the total random kinetic energy.
Use kinetic theory to connect kinetic energy to temperature via $\frac{3}{2}kT$ .
Multiply by the number of molecules to show a direct proportionality.

Step-by-Step Reasoning

In general, internal energy is random kinetic + intermolecular potential.
For an ideal gas, intermolecular potential energy can be taken as zero/constant because forces are negligible.
So $U$ equals the total random kinetic energy of all molecules.
Mean kinetic energy per molecule is $\frac{3}{2}kT$ .
For $N$ molecules, total kinetic energy is

U = N\left(\frac{3}{2}kT\right) = \frac{3}{2}NkT.

Since $N$ and $k$ are constant for a fixed sample, $U \propto T$ .

Key Takeaways

For an ideal gas, $U$ depends only on $T$ .
The proportionality comes from the kinetic theory relation between temperature and mean molecular kinetic energy.

Common Mistakes

Referring to $pV = nRT$ only and trying to argue $U \propto pV$ without linking to kinetic energy.
Forgetting the “ideal” condition and talking about changes in potential energy due to intermolecular forces.

Things to Be Careful About

Temperature must be thermodynamic temperature (kelvin).
Use “mean kinetic energy per molecule is proportional to $T$ ” as the key link; that is what makes $U$ proportional to $T$ for an ideal gas.

Techniques used

use the kinetic theory result for mean translational kinetic energylink internal energy to total random kinetic energy for an ideal gasdeduce proportionality between internal energy and thermodynamic temperature

(b)

A sample of an ideal gas at thermodynamic temperature $T$ has internal energy $U$ .

The gas is compressed so that its temperature increases to $3T$ .
During this compression, work $W$ is done on the gas.

The gas is then cooled at constant volume so that its temperature decreases to $2T$ .

Complete Table 4.1 to show, in terms of some or all of $W$ , $T$ and $U$ , the work done on the gas, the thermal energy supplied to the gas and the increase in internal energy of the gas for each of the two processes.

Table 4.1

	work done on gas	thermal energy supplied to gas	increase in internal energy of gas
compression	$+W$
cooling

4M Similar

DifficultyMedium

Worked solution

Working

For an ideal gas, $U \propto T$ .

Initial: at $T$ , internal energy $= U$ .
After compression to $3T$ : internal energy $= 3U$ so

\Delta U_{\text{comp}} = 3U - U = 2U.

First law with $W$ = work done on gas:

\Delta U = q + W \Rightarrow 2U = q + W \Rightarrow q_{\text{comp}} = 2U - W.

Cooling at constant volume from $3T$ to $2T$ :
internal energy changes $3U \to 2U$ so

\Delta U_{\text{cool}} = 2U - 3U = -U.

Constant volume $\Rightarrow W_{\text{cool}} = 0$ .
So

- U = q_{\text{cool}} + 0 \Rightarrow q_{\text{cool}} = -U.

Answer

process	work done on gas	thermal energy supplied to gas $q$	increase in internal energy $\Delta U$
compression	$+W$	$2U - W$	$+2U$
cooling	$0$	$-U$	$-U$

Final answer

Compression: q = 2U − W, ΔU = +2U. Cooling: work = 0, q = −U, ΔU = −U.

Detailed explanation

Background Concept

Two ideas are being tested.

1) Internal energy of an ideal gas

For an ideal gas, internal energy depends only on temperature:

U \propto T.

So if temperature is multiplied by a factor, internal energy is multiplied by the same factor.

2) First law of thermodynamics

Using the Cambridge convention stated in the syllabus:

\Delta U = q + W,

where:

$\Delta U$ is the change in internal energy of the gas,
$q$ is the thermal energy supplied to the gas (negative if energy leaves the gas),
$W$ is the work done on the gas (positive for compression, negative for expansion).

At constant volume, $\Delta V = 0$ , so no $p\Delta V$ work is done:

W = 0.

Understanding the Question

You are told:

Initially: temperature $T$ , internal energy $U$ .
Process 1 (compression): temperature rises to $3T$ , and work $+W$ is done on the gas.
Process 2 (cooling at constant volume): temperature falls from $3T$ to $2T$ .

You must fill the table entries (work, thermal energy supplied, internal energy change) for each process in terms of $W$ , $T$ , and $U$ .

Approach

Use $U \propto T$ to find the internal energy at $3T$ and at $2T$ .
For each process, compute $\Delta U$ from final minus initial internal energy.
Apply $\Delta U = q + W$ to find $q$ .
Use the process conditions (compression, constant volume) to set $W$ correctly.

Step-by-Step Reasoning

Step 1: Internal energies at the three temperatures

Because $U \propto T$ :

At $T$ : internal energy is given as $U$ .
At $3T$ : internal energy is $3U$ .
At $2T$ : internal energy is $2U$ .

Step 2: Compression process

Initial internal energy: $U$ .
Final internal energy: $3U$ .
So:

\Delta U_{\text{comp}} = 3U - U = 2U.

You are told work done on the gas is $+W$ .
Use first law:

\Delta U = q + W.

Substitute:

2U = q + W \Rightarrow q = 2U - W.

So the thermal energy supplied could be positive or negative depending on the size of $W$ ; the expression $2U - W$ correctly captures this.

Step 3: Cooling at constant volume

This starts at $3T$ (so internal energy $3U$ ) and ends at $2T$ (internal energy $2U$ ).
So:

\Delta U_{\text{cool}} = 2U - 3U = -U.

Because the volume is constant, $W=0$ .
First law gives:

- U = q + 0 \Rightarrow q = -U.

Negative $q$ means thermal energy leaves the gas (it is cooled).

Key Takeaways

For an ideal gas, $U$ scales linearly with $T$ .
Use the first law with a consistent sign convention: $\Delta U = q + W$ (work done on the gas).
Constant volume implies zero work done.

Common Mistakes

Using $\Delta U = q - W$ (the alternative sign convention) without adjusting the meaning of $W$ .
Saying the cooling has “no heat transfer” just because volume is constant; constant volume only means no work, not no heating/cooling.
Getting $\Delta U$ sign wrong for cooling (it must be negative because temperature decreases).

Things to Be Careful About

Always compute $\Delta U$ as final minus initial.
“Thermal energy supplied to the gas” can be negative; that represents thermal energy removed from the gas.
Keep track of which temperature each process starts from: the cooling starts from $3T$ , not from $T$ .

Techniques used

use proportionality of internal energy with thermodynamic temperature for an ideal gasapply the first law of thermodynamics with a consistent sign conventionuse constant-volume condition to set work done equal to zerocalculate internal energy changes between temperature states

Q5MediumOscillations

A cuboidal block floats in a liquid with its base horizontal, as shown in Fig. 5.1.

Fig. 5.1

The base of the block is at a depth $h$ below the surface of the liquid.

The block is displaced downwards by a small distance and then released so that it oscillates.

Fig. 5.2 shows the variation with $h$ of the acceleration $a$ of the block.

Fig. 5.2

Fig. 5.3 shows the variation with $h$ of the kinetic energy $E_K$ of the block.

Fig. 5.3

(a)

2M

(i)

Determine the amplitude of the oscillations.

amplitude = ______ $\text{m}$

1M Similar

DifficultyEasy

Worked solution

Turning points where $E_K = 0$ are at $h = 0.8\,\text{m}$ and $h = 2.0\,\text{m}$ .

Equilibrium at maximum $E_K$ is $h = 1.4\,\text{m}$ .

Amplitude

A = 2.0 - 1.4 = 0.60\,\text{m}

Answer

$0.60\,\text{m}$

Final answer

0.60 m

Detailed explanation

Background Concept

For oscillations, the amplitude $A$ is the maximum displacement from the equilibrium position. In simple harmonic motion (SHM), the object oscillates between two turning points where the speed is zero, so the kinetic energy $E_K$ is zero.

Understanding the Question

You are given two graphs against depth $h$ :

Fig. 5.2: acceleration $a$ vs $h$ (a straight line crossing $a=0$ at one value of $h$ ).
Fig. 5.3: kinetic energy $E_K$ vs $h$ (a curve with a maximum and zeros at two values of $h$ ).

The block oscillates about an equilibrium depth. From the graphs we can identify:

equilibrium depth (where $a=0$ and where $E_K$ is maximum)
turning points (where $E_K=0$ )

Then amplitude is the distance from equilibrium to either turning point.

Approach

Read the turning points from Fig. 5.3 where $E_K=0$ .
Read the equilibrium position from Fig. 5.3 where $E_K$ is maximum (and it should match where $a=0$ on Fig. 5.2).
Compute $A = |h_{\text{turn}} - h_{\text{eq}}|$ .

Step-by-Step Reasoning

From Fig. 5.3, $E_K = 0$ at $h = 0.8\,\text{m}$ and $h = 2.0\,\text{m}$ . These are the turning points.
The peak of the $E_K$ curve is at $h = 1.4\,\text{m}$ , so this is the equilibrium depth $h_{\text{eq}}$ .
Amplitude is the maximum displacement from equilibrium:

A = 2.0 - 1.4 = 0.60\,\text{m}

(You could equally use $1.4 - 0.8 = 0.60\,\text{m}$ .)

Key Takeaways

Turning points occur where $v=0$ so $E_K=0$ .
The equilibrium position is where speed (and hence $E_K$ ) is maximum.
Amplitude is the distance from equilibrium to a turning point.

Common Mistakes

Taking amplitude as the full peak-to-peak distance ( $2A$ ): here $2.0-0.8=1.2\,\text{m}$ is not the amplitude.
Using the graph edges rather than the actual intercepts where $E_K=0$ .

Things to Be Careful About

Ensure you identify equilibrium correctly: it should be consistent across both graphs (here $a=0$ at the same $h$ as maximum $E_K$ ).
Quote the amplitude with a sensible number of significant figures from the graph (here $0.60\,\text{m}$ ).

Techniques used

identify equilibrium position from zero acceleration or maximum kinetic energyread turning points from where kinetic energy is zerocalculate amplitude as the maximum displacement from equilibrium

(ii)

State what the line in Fig. 5.2 shows about the nature of the oscillations.

1M Similar

DifficultyMedium-Easy

Worked solution

The straight line with negative gradient shows $a \propto -(h- h_0)$ (acceleration proportional to displacement from equilibrium and opposite in direction), so the motion is simple harmonic.

Answer

Acceleration is proportional to displacement from equilibrium and directed towards equilibrium (SHM).

Final answer

Acceleration is proportional to displacement from equilibrium and directed towards equilibrium (SHM).

Detailed explanation

Background Concept

A defining condition for simple harmonic motion is that the acceleration is proportional to the displacement from equilibrium and is always directed towards the equilibrium position:

a = -\omega^2 x

where:

$x$ is displacement from equilibrium (positive in a chosen direction),
$\omega$ is angular frequency,
the negative sign means the acceleration is a restoring acceleration (opposite to $x$ ).

So, if a graph of $a$ against displacement is a straight line through the equilibrium point with a negative gradient, the motion is SHM.

Understanding the Question

Fig. 5.2 is a graph of acceleration $a$ against depth $h$ . The block oscillates about some equilibrium depth $h_0$ (where $a=0$ ). You are asked what the line shows about the nature of the oscillations.

Approach

Identify what variable represents displacement from equilibrium: here it is $(h-h_0)$ .
Use the shape of the $a$ – $h$ graph: a straight line implies proportionality; negative gradient implies restoring direction.

Step-by-Step Reasoning

The line is straight: this indicates

a \propto h

relative to some offset (equilibrium).

The line crosses $a=0$ at $h=1.4\,\text{m}$ , so equilibrium depth is $h_0=1.4\,\text{m}$ .
Rewriting in terms of displacement from equilibrium $x = (h-h_0)$ gives a relation of the form

a \propto -(h-h_0) = -x

because the gradient is negative.

This matches the SHM condition $a=-\omega^2 x$ .

Therefore, the oscillations are simple harmonic.

Key Takeaways

SHM is identified by a linear acceleration–displacement relationship with negative gradient.
The equilibrium position corresponds to where $a=0$ .

Common Mistakes

Saying “acceleration is constant”: a straight line does not mean constant acceleration; it means acceleration changes linearly with $h$ .
Missing the significance of the negative gradient (restoring direction).

Things to Be Careful About

Displacement is not $h$ itself; it is $h-h_0$ measured from the equilibrium depth.
The negative sign is essential: it distinguishes SHM from unstable motion (positive feedback).

Techniques used

interpret the sign and proportionality of a straight-line acceleration–displacement graphrelate a linear a–x relation to the SHM condition a = -omega^2 x

(b)

State three other quantitative conclusions that can be drawn from Fig. 5.2 and Fig. 5.3 about the block and its oscillations. Use the space for any working.

3M Similar

DifficultyMedium

Worked solution

From Fig. 5.2, equilibrium is at $h_0 = 1.4\,\text{m}$ .

Gradient of Fig. 5.2:

\frac{\Delta a}{\Delta h} = \frac{-0.6 - (+0.6)}{2.0 - 0.8} = \frac{-1.2}{1.2} = -1.0\,\text{s}^{-2}

For SHM, $a=-\omega^2(h-h_0)$ so gradient $= -\omega^2$ :

\omega^2 = 1.0\,\text{s}^{-2} \Rightarrow \omega = 1.0\,\text{rad s}^{-1}

Hence period

T = \frac{2\pi}{\omega} = 2\pi = 6.3\,\text{s}

Amplitude from (a)(i): $A=0.60\,\text{m}$ , so

v_{\max} = \omega A = 1.0 \times 0.60 = 0.60\,\text{m s}^{-1}

From Fig. 5.3, $E_{K\max} = 8.0\,\text{J}$ , and $E_{K\max}=\frac12 mv_{\max}^2$ :

8.0 = \frac12 m(0.60)^2 \Rightarrow m = \frac{16}{0.36} = 44\,\text{kg}

Answer

Examples of three quantitative conclusions:

equilibrium depth $h_0 = 1.4\,\text{m}$
$\omega = 1.0\,\text{rad s}^{-1}$ so $T = 6.3\,\text{s}$
mass of block $m \approx 44\,\text{kg}$

Final answer

h0 = 1.4 m; T = 6.3 s (ω = 1.0 rad s^-1); m ≈ 44 kg

Detailed explanation

Background Concept

For SHM:

a = -\omega^2 x

So on a graph of $a$ against displacement $x$ , the gradient is $-\omega^2$ .

The period is related to angular frequency by:

\omega = \frac{2\pi}{T}

The speed is maximum at equilibrium and given by:

v_{\max} = \omega A

Energy in SHM is conserved (if no damping). At equilibrium, potential energy is minimum and kinetic energy is maximum. So the total energy $E$ equals $E_{K\max}$ :

E = E_{K\max}

Also:

E_{K\max} = \frac{1}{2} m v_{\max}^2

Understanding the Question

You must extract three other numerical facts from Fig. 5.2 ( $a$ vs $h$ ) and Fig. 5.3 ( $E_K$ vs $h$ ) about the oscillation and/or the block.

The graphs show key SHM signatures:

$a$ varies linearly with $h$ and changes sign at an equilibrium depth.
$E_K$ is zero at two turning points and maximum at the equilibrium depth.

Approach

A good way to get quantitative conclusions is:

Use Fig. 5.2 to find $\omega$ from the gradient (since gradient $=-\omega^2$ ).
Convert $\omega$ to period $T$ .
Use amplitude $A$ (from turning points) to find $v_{\max}=\omega A$ .
Use Fig. 5.3 to read $E_{K\max}$ and then use $E_{K\max}=\tfrac12 m v_{\max}^2$ to find $m$ .

Any three correct numerical conclusions earn the marks; the ones above are tightly linked and make full use of both graphs.

Step-by-Step Reasoning

1) Equilibrium depth

In Fig. 5.2, $a=0$ at $h=1.4\,\text{m}$ .
In Fig. 5.3, $E_K$ is maximum at $h=1.4\,\text{m}$ .

So the equilibrium depth is:

h_0 = 1.4\,\text{m}

2) Find $\omega$ from the gradient
Pick two clear points on the straight line in Fig. 5.2, e.g.

$(h,a)=(0.8, +0.6)$
$(h,a)=(2.0, -0.6)$

Gradient:

\frac{\Delta a}{\Delta h} = \frac{-0.6 - (+0.6)}{2.0-0.8} = -1.0\,\text{s}^{-2}

For SHM, with $x = h-h_0$ :

a = -\omega^2 (h-h_0)

So the gradient of $a$ vs $h$ is $-\omega^2$ . Hence:

\omega^2 = 1.0\,\text{s}^{-2} \Rightarrow \omega = 1.0\,\text{rad s}^{-1}

3) Period

T = \frac{2\pi}{\omega} = 2\pi \approx 6.28\,\text{s} \approx 6.3\,\text{s}

4) Use $E_K$ graph to get total energy and find mass

From Fig. 5.3, maximum kinetic energy is $E_{K\max}=8.0\,\text{J}$ .
The amplitude from the turning points is $A=0.60\,\text{m}$ .
Maximum speed:

v_{\max} = \omega A = 1.0 \times 0.60 = 0.60\,\text{m s}^{-1}

Use $E_{K\max}=\tfrac12 m v_{\max}^2$ :

8.0 = \frac12 m(0.60)^2 \Rightarrow m = \frac{16}{0.36} = 44.4\,\text{kg} \approx 44\,\text{kg}

That gives a numerical property of the block (its mass) obtained by combining both graphs.

Key Takeaways

The gradient of an $a$ –displacement graph gives $-\omega^2$ .
$T$ follows from $\omega$ via $T=2\pi/\omega$ .
The maximum kinetic energy equals the total energy in SHM (no damping).
Combining $E_{K\max}$ with $v_{\max}=\omega A$ can determine the mass.

Common Mistakes

Using the gradient as $-\omega$ instead of $-\omega^2$ .
Forgetting that displacement is $(h-h_0)$ ; using $h$ measured from zero depth gives the wrong interpretation.
Using $E_{K\max}=\tfrac12 m (\omega^2 A^2)$ (incorrect); it must be $E_{K\max}=\tfrac12 m (\omega A)^2$ .

Things to Be Careful About

Units of gradient: $a$ is in $\text{m s}^{-2}$ and $h$ is in $\text{m}$ , so gradient has units $\text{s}^{-2}$ , consistent with $\omega^2$ .
Read values accurately from the graph (intercepts and the labelled peak).
Keep $\omega$ in $\text{rad s}^{-1}$ ; then $T$ comes out in seconds without extra conversion.

Techniques used

determine angular frequency from the gradient of an acceleration–displacement graphuse omega = 2 pi / T to find the perioduse v_max = omega x0 to find maximum speeduse maximum kinetic energy to determine mass

(c)

On Fig. 5.4, sketch the variation with $h$ of the potential energy $E_P$ of the oscillations.

Fig. 5.4

3M Similar

DifficultyMedium

Worked solution

Total energy $E = E_{K\max} = 8.0\,\text{J}$ (from Fig. 5.3).

So

E_P = E - E_K

Hence $E_P = 0$ at $h=1.4\,\text{m}$ and $E_P = 8.0\,\text{J}$ at $h=0.8\,\text{m}$ and $h=2.0\,\text{m}$ .

Sketch: upward-opening parabola, symmetric about $h=1.4\,\text{m}$ , passing through $(1.4,0)$ and $(0.8,8.0)$ and $(2.0,8.0)$ .

Answer

See sketch: minimum at $h=1.4\,\text{m}$ with $E_P=0$ , maxima $E_P=8.0\,\text{J}$ at $h=0.8\,\text{m}$ and $2.0\,\text{m}$ .

Final answer

Upward parabola: Ep = 0 at h = 1.4 m; Ep = 8.0 J at h = 0.8 m and 2.0 m.

Detailed explanation

Background Concept

In undamped SHM, the total mechanical energy is constant:

E = E_K + E_P

At the equilibrium position:

speed is maximum, so $E_K$ is maximum
potential energy is minimum (often taken as zero for oscillation energy), so $E_P$ is minimum

At the turning points:

speed is zero, so $E_K=0$
displacement is maximum, so $E_P$ is maximum and equals the total energy $E$

Therefore, if you are given $E_K$ as a function of position, you can get:

E_P = E - E_K

Understanding the Question

You are given Fig. 5.3 (a downward-opening curve for $E_K$ vs depth $h$ ) and asked to sketch $E_P$ vs $h$ on the blank axes (Fig. 5.4).

Key read-offs from Fig. 5.3:

$E_{K\max}=8.0\,\text{J}$ at $h=1.4\,\text{m}$
$E_K=0$ at $h=0.8\,\text{m}$ and $h=2.0\,\text{m}$

Approach

Identify the constant total energy $E$ from the maximum kinetic energy.
Use $E_P = E - E_K$ .
Mark the three crucial points for the sketch:
- equilibrium: where $E_K$ is maximum, $E_P$ must be minimum
- turning points: where $E_K=0$ , $E_P$ must equal $E$
Draw the correct shape: since $E_K$ is a downward parabola in $h$ , $E_P$ is an upward parabola in $h$ , symmetric about the equilibrium depth.

Step-by-Step Reasoning

From Fig. 5.3, the highest point of the $E_K$ curve is $8.0\,\text{J}$ . With no damping, this equals the total energy:

E = 8.0\,\text{J}

Use the energy relation:

E_P = E - E_K

At equilibrium $h=1.4\,\text{m}$ , $E_K=8.0\,\text{J}$ , so

E_P = 8.0 - 8.0 = 0\,\text{J}

At turning points $h=0.8\,\text{m}$ and $h=2.0\,\text{m}$ , $E_K=0$ , so

E_P = 8.0 - 0 = 8.0\,\text{J}

So the $E_P$ curve passes through $(0.8, 8.0)$ , has a minimum at $(1.4, 0)$ , and passes through $(2.0, 8.0)$ , with a smooth upward-opening parabolic shape.

diagram-ep-graph

Key Takeaways

In SHM (no damping), $E_K + E_P$ is constant.
$E_K$ is maximum at equilibrium; $E_P$ is maximum at turning points.
The $E_P$ graph is the “mirror” of the $E_K$ graph about the horizontal line $E = \text{constant}$ .

Common Mistakes

Drawing $E_P$ as the same shape as $E_K$ (it must be inverted relative to the constant total energy).
Putting $E_P=0$ at the turning points (it is $E_K$ that is zero there).
Choosing the wrong total energy (it is the maximum of $E_K$ , not some other y-value).

Things to Be Careful About

Make sure your sketch uses the same $h$ values as the kinetic energy graph: $0.8\,\text{m}$ , $1.4\,\text{m}$ , $2.0\,\text{m}$ .
The minimum should touch $E_P=0$ exactly at $h=1.4\,\text{m}$ .
The curve should be smooth and symmetric about $h=1.4\,\text{m}$ (same as the $E_K$ curve).

Techniques used

use conservation of mechanical energy to relate kinetic and potential energyconstruct a required graph by subtracting one curve from a constant total energyidentify key points (turning points and equilibrium) to shape a sketch

Q6Medium-HardAlternating CurrentsCapacitance

Fig. 6.1 shows a circuit that rectifies an alternating input voltage $V_{IN}$ and produces an output voltage $V_{OUT}$ across a resistor $R$ .

Fig. 6.1

The four terminals of the rectification circuit are labelled W, X, Y and Z.
A capacitor $C$ is connected in parallel with resistor $R$ .

(a)

2M

(i)

State what is meant by rectification.

1M Similar

DifficultyEasy

Worked solution

Answer

Rectification is the conversion of an alternating p.d./current into a p.d./current that is in one direction only (a unidirectional / d.c. output).

Final answer

Conversion of a.c. to a unidirectional (d.c.) output.

Detailed explanation

Background Concept

Rectification is a process carried out using diodes (or diode bridges) where current is allowed to pass more easily in one direction than the other. An alternating supply reverses polarity every half-cycle, so the current would normally reverse direction as well.

A rectifier changes the output so that it does not reverse direction: the output becomes unidirectional (it may still vary in size with time).

Understanding the Question

You are shown a circuit with an a.c. input voltage $V_{IN}$ and an output voltage $V_{OUT}$ across a resistor. The question asks for the meaning of “rectification”, i.e. what a rectifier does to the a.c. signal.

Approach

Give a clear definition in terms of converting a.c. to d.c. / unidirectional current or p.d. Mention “one direction only” to distinguish it from simply “changing the size”.

Step-by-Step Reasoning

In a.c., the p.d. alternates between positive and negative values, so current would reverse.
A rectifier uses diodes to make the output p.d. have only one polarity at the load.
Therefore the output current through the load is in one direction only (unidirectional).

Key Takeaways

Rectification means making an a.c. signal into a unidirectional output.
The output may be pulsating; smoothing is a separate process.

Common Mistakes

Saying “rectification makes the voltage constant”: that describes smoothing/regulation, not rectification.
Saying only “changes a.c. to d.c.” without indicating “one direction only” can be too vague.

Things to Be Careful About

“d.c.” in this context can mean pulsating d.c. (always same sign), not necessarily perfectly steady.

Techniques used

state the meaning of rectification in terms of current directionrelate a.c. and d.c. waveforms to sign of potential difference

(ii)

State the purpose of capacitor $C$ .

1M Similar

DifficultyEasy

Worked solution

Answer

Capacitor $C$ smooths the rectified output by reducing the ripple in $V_{OUT}$ (it charges near peaks and discharges between them).

Final answer

To smooth $V_{OUT}$ by reducing ripple.

Detailed explanation

Background Concept

After rectification, the output is often a series of peaks (pulsating d.c.). A capacitor connected in parallel with the load acts as an energy store:

it charges up when the rectified voltage rises,
it discharges through the load when the rectified voltage falls.

This makes the load voltage vary less, so the “ripple” (peak-to-trough variation) is reduced.

Understanding the Question

In Fig. 6.1, capacitor $C$ is connected in parallel with the resistor $R$ , and $V_{OUT}$ is measured across $R$ . The question asks what the capacitor is doing in that arrangement.

Approach

State that it smooths the output, and (to be precise) mention charging on peaks and discharging between peaks to keep the p.d. from falling too much.

Step-by-Step Reasoning

When the rectified output rises, the capacitor charges quickly up to (approximately) the peak voltage.
When the rectified output begins to fall, the capacitor cannot instantly lose charge; it discharges through $R$ .
This discharge provides current to the resistor, so $V_{OUT}$ stays closer to the peak value, reducing ripple.

Key Takeaways

A smoothing capacitor reduces ripple by providing current between peaks.
The larger the $RC$ time constant, the slower the discharge and the smoother the output.

Common Mistakes

Saying “it increases the voltage”: it does not increase the peak; it reduces the fall between peaks.
Confusing “smoothing” with “rectification”: rectification makes the output one polarity; smoothing reduces ripple.

Things to Be Careful About

The capacitor is in parallel with the load, not in series; in parallel it directly holds up the load p.d.

Techniques used

link capacitor charge/discharge to reduction of ripple voltagestate the function of a component in a rectifier circuit

(b)

Fig. 6.2 shows the variations with time $t$ of the potential differences (p.d.s) $V_{IN}$ and $V_{OUT}$ .

Fig. 6.2

8M

(i)

The variation of $V_{IN}$ with $t$ can be represented by

V_{IN} = A \cos Bt

where $A$ and $B$ are constants.

Determine the values of $A$ and $B$ . Give a unit with your answer for $A$ .

$A$ = ______ unit ______
$B$ = ______ $\text{rad s}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 6.2, peak value of $V_{IN}$ is $12\ \text{V}$ , so

A = 12\ \text{V}

Period $T = 20\ \text{ms} = 2.0 \times 10^{-2}\ \text{s}$ , so

B = \omega = \frac{2\pi}{T} = \frac{2\pi}{2.0 \times 10^{-2}} = 3.14 \times 10^{2}\ \text{rad s}^{-1}

Answer

$A = 12\ \text{V}$

$B = 3.14 \times 10^{2}\ \text{rad s}^{-1}$

Final answer

A = 12 V, B = 3.14 × 10^2 rad s^-1

Detailed explanation

Background Concept

A sinusoidal voltage can be written in the form

V = V_0 \cos(\omega t)

where:

$V_0$ is the peak (maximum) voltage (amplitude),
$\omega$ is the angular frequency in $\text{rad s}^{-1}$ ,
the period $T$ is the time for one complete cycle.

The link between $\omega$ and $T$ is:

\omega = 2\pi f = \frac{2\pi}{T}

Understanding the Question

You are given a graph of $V_{IN}$ against time.

You must identify $A$ in $V_{IN} = A\cos Bt$ as the peak voltage.
You must identify $B$ as the angular frequency by reading the period from the graph.

Approach

Read the peak value (highest point) of the $V_{IN}$ curve to get $A$ .
Read the period $T$ (time between successive peaks) and convert ms to s.
Calculate $B = 2\pi/T$ .

Step-by-Step Reasoning

From the vertical axis, $V_{IN}$ reaches $+12\ \text{V}$ at its peaks, so $A=12\ \text{V}$ .
From the horizontal axis, one full cycle takes $20\ \text{ms}$ .

Convert:

20\ \text{ms} = 20 \times 10^{-3}\ \text{s} = 2.0 \times 10^{-2}\ \text{s}

Then:

B = \frac{2\pi}{2.0 \times 10^{-2}} = 3.14 \times 10^{2}\ \text{rad s}^{-1}

Key Takeaways

Amplitude comes from the peak value on the graph.
Angular frequency is found from the period using $\omega=2\pi/T$ .

Common Mistakes

Using $f=1/T$ but forgetting to convert to $\omega = 2\pi f$ .
Leaving $T$ in ms (gives $\omega$ too small by a factor of $1000$ ).
Quoting $B$ in $\text{s}^{-1}$ instead of $\text{rad s}^{-1}$ .

Things to Be Careful About

For $\cos$ , at $t=0$ the waveform should start at a maximum if there is no phase constant; the graph shown is consistent with this.
Ensure $A$ is given with a unit (volts).

Techniques used

read amplitude and period from a sinusoidal graphuse angular frequency $\omega = 2\pi/T$match a waveform to $V = A\cos Bt$

(ii)

Determine the type of rectification produced by the circuit in Fig. 6.1.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Full-wave rectification.

Final answer

Full-wave rectification.

Detailed explanation

Background Concept

Half-wave rectification: only one half-cycle of the a.c. appears at the output; the output pulses once per input cycle (same frequency as input).
Full-wave rectification: both half-cycles are inverted so the output is always the same polarity; the output pulses twice per input cycle (double the input frequency).

With a smoothing capacitor, you do not see sharp pulses; instead you see a voltage that repeatedly charges near the peaks and then decays between peaks. The key clue is how often the capacitor is “topped up”.

Understanding the Question

From Fig. 6.2, $V_{IN}$ has period $20\ \text{ms}$ . The $V_{OUT}$ trace stays positive and shows repeated peaks (with ripple). The question asks which rectification (half-wave or full-wave) produced that.

Approach

Check whether the output is refreshed every half-cycle (every $10\ \text{ms}$ ) or every full cycle (every $20\ \text{ms}$ ). A circuit with four terminals and a typical rectifier block strongly suggests a bridge rectifier, which gives full-wave rectification.

Step-by-Step Reasoning

The output does not go negative (it is unidirectional).
The ripple peaks occur more frequently than once per $20\ \text{ms}$ cycle; the capacitor is recharged every half-cycle, indicating full-wave rectification.

Key Takeaways

Full-wave rectification gives a unidirectional output with ripple at twice the input frequency.
A bridge rectifier uses four diodes and has four terminals (two input, two output).

Common Mistakes

Saying “d.c.” without specifying half-wave or full-wave.
Confusing smoothing (capacitor effect) with the type of rectification.

Things to Be Careful About

With smoothing, the output is not a set of neat ‘humps’; you must infer the underlying pulse frequency from the recharge intervals.

Techniques used

infer rectification type from the output waveformuse the frequency of ripple to distinguish half-wave from full-wave

(iii)

On Fig. 6.3, draw the circuit diagram for the components inside the rectification circuit.

Fig. 6.3

2M Similar

DifficultyMedium

Worked solution

diagram-6biii

Answer

A four-diode bridge rectifier between $W,X$ (a.c. input) and $Y,Z$ (d.c. output), with $Y$ as the positive output and $Z$ as the negative output.

Final answer

Bridge rectifier (4 diodes) with W,X as a.c. input and Y positive, Z negative output.

Detailed explanation

Background Concept

A bridge rectifier uses four diodes arranged so that, regardless of the polarity of the a.c. input, current through the load flows in the same direction.

In each half-cycle, a different pair of diodes conducts:

On one half-cycle, two diodes conduct to send current through the load one way.
On the next half-cycle, the other two diodes conduct, but the load current direction is unchanged.

Understanding the Question

Fig. 6.1 shows an unknown “rectification circuit” with terminals $W,X$ on the input side and $Y,Z$ on the output side. You are asked to draw what is inside that box. From the context (four terminals and full-wave output), the intended circuit is a bridge rectifier (four diodes).

Approach

Draw a standard diode bridge:

Connect the a.c. input to two opposite corners of the bridge.
Take the d.c. output from the other two corners.
Orient diodes so that the output terminal $Y$ is always at higher potential than $Z$ .

Step-by-Step Reasoning

diagram-6biii

To check it works:

If $W$ is positive relative to $X$ , current flows from $W$ through a diode into $Y$ , through the load (external $R$ ) from $Y$ to $Z$ , then through another diode back to $X$ .
If $X$ is positive relative to $W$ , current flows from $X$ through a different diode into $Y$ , through the load $Y \to Z$ , then through a different diode back to $W$ .

In both cases, the load current is in the same direction, so the output is full-wave rectified.

Key Takeaways

A bridge rectifier uses four diodes to make the output unidirectional for both half-cycles.
The diode orientation is the critical detail: $Y$ must always be fed via a forward-biased diode from whichever input is positive.

Common Mistakes

Drawing only one diode (half-wave rectifier) despite four terminals and full-wave behaviour.
Reversing a diode so that one half-cycle gives no output or makes $Y$ negative.
Connecting output across the same pair of nodes as the input.

Things to Be Careful About

Diode symbols must have correct direction (anode to cathode). In circuit diagrams, the diode’s bar is the cathode; current (conventional) flows from anode to cathode when forward biased.
Keep the terminal labels $W, X, Y, Z$ exactly as given in the question when placing the bridge.

Techniques used

construct a bridge rectifier using four diodeslabel input and output terminals with correct polaritiesensure diode orientations give unidirectional output for both half-cycles

(iv)

Determine a value for the time constant for the discharge of the capacitor $C$ through the resistor $R$ in Fig. 6.1.

time constant = ______ $\text{s}$

3M Similar

DifficultyMedium-Hard

Worked solution

Working

From Fig. 6.2, $V_{OUT}$ falls from about $12\ \text{V}$ to about $8\ \text{V}$ between charging peaks.
For full-wave rectification, peaks are separated by $10\ \text{ms} = 1.0\times 10^{-2}\ \text{s}$ .

Use capacitor discharge:

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

8 = 12 e^{- (1.0\times 10^{-2})/\tau}

\frac{8}{12} = e^{- (1.0\times 10^{-2})/\tau}

\tau = \frac{1.0\times 10^{-2}}{\ln(12/8)} = 2.47\times 10^{-2}\ \text{s}

Answer

Time constant $\tau \approx 2.5\times 10^{-2}\ \text{s}$ .

Final answer

2.5 × 10^-2 s

Detailed explanation

Background Concept

When a capacitor discharges through a resistor, the p.d. across it decreases exponentially:

V = V_0 e^{-t/RC}

The product $RC$ is called the time constant:

\tau = RC

A larger $\tau$ means the voltage falls more slowly (better smoothing).

Understanding the Question

$V_{OUT}$ is the p.d. across $R$ and also across the capacitor $C$ (they are in parallel). From the graph, the capacitor charges to near the peak and then discharges through $R$ between peaks, causing the ripple.

You are asked to find a value of the time constant $\tau$ from the way $V_{OUT}$ falls between two successive recharging peaks.

Approach

Read the maximum and minimum values of the ripple (approximate is acceptable from a graph).
Decide the time interval between peaks (depends on half-wave vs full-wave). For full-wave, the capacitor is topped up every half-cycle.
Apply $V = V_0 e^{-t/\tau}$ and solve for $\tau$ using logarithms.

Step-by-Step Reasoning

From the plotted $V_{OUT}$ , take $V_0 \approx 12\ \text{V}$ (just after charging) and $V \approx 8\ \text{V}$ (just before the next charge).
The input period is $20\ \text{ms}$ , so for full-wave rectification the ripple period is half of this:

\Delta t \approx 10\ \text{ms} = 1.0\times 10^{-2}\ \text{s}

Now use the discharge law:

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

Substitute:

\frac{8}{12} = e^{- (1.0\times 10^{-2})/\tau}

Take natural logs:

\ln\left(\frac{8}{12}\right) = -\frac{1.0\times 10^{-2}}{\tau}

So:

\tau = \frac{1.0\times 10^{-2}}{\ln(12/8)} \approx 2.47\times 10^{-2}\ \text{s}

Rounded suitably from graph readings, $\tau \approx 2.5\times 10^{-2}\ \text{s}$ .

Key Takeaways

Ripple is caused by exponential discharge of the capacitor between peaks.
Use the time between recharging events (set by rectification type) in the exponential equation.

Common Mistakes

Using $20\ \text{ms}$ instead of $10\ \text{ms}$ for full-wave ripple spacing.
Using a linear drop model instead of exponential decay.
Swapping $V$ and $V_0$ in the logarithm (sign error).

Things to Be Careful About

Read values from the graph sensibly (do not claim unrealistic precision).
Convert ms to s before substituting.
Use $\ln$ (natural log) because the equation uses $e^{...}$ .

Techniques used

read peak and trough values from a ripple waveformuse exponential discharge $V = V_0 e^{-t/\tau}$estimate discharge time from the ripple periodrearrange logarithms to solve for the time constant

(c)

The capacitor $C$ has a capacitance of $570\ \mu\text{F}$ .

Use your answer in (b)(iv) to determine the resistance of resistor $R$ .

resistance = ______ $\Omega$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\tau = RC \Rightarrow R = \frac{\tau}{C}

With $\tau = 2.5\times 10^{-2}\ \text{s}$ and $C = 570\ \mu\text{F} = 570\times 10^{-6}\ \text{F}$ ,

R = \frac{2.5\times 10^{-2}}{570\times 10^{-6}} = 4.4\times 10^{1}\ \Omega

Answer

$R \approx 44\ \Omega$ .

Final answer

44 Ω

Detailed explanation

Background Concept

The time constant for an $RC$ circuit is

\tau = RC

This applies to both charging and discharging exponentials. If you know $\tau$ and $C$ , you can find $R$ by rearranging:

R = \frac{\tau}{C}

Understanding the Question

You are told $C = 570\ \mu\text{F}$ and you have already estimated the discharge time constant $\tau$ from part (b)(iv). The question asks for the resistance $R$ .

Approach

Convert $\mu\text{F}$ to $\text{F}$ .
Substitute into $R=\tau/C$ .

Step-by-Step Reasoning

Convert capacitance:

570\ \mu\text{F} = 570\times 10^{-6}\ \text{F}

Substitute:

R = \frac{2.5\times 10^{-2}}{570\times 10^{-6}} \approx 4.4\times 10^{1}\ \Omega

So $R \approx 44\ \Omega$ (to 2 s.f., appropriate given the graph-based $\tau$ ).

Key Takeaways

$\tau$ sets the smoothing behaviour and is directly proportional to $R$ and $C$ .
Unit conversion for capacitance is essential.

Common Mistakes

Using $570$ as if it were in farads instead of microfarads.
Rearranging incorrectly (e.g. $R = C/\tau$ ).

Things to Be Careful About

Keep powers of ten explicit: $\mu = 10^{-6}$ .
Quote the final answer to a sensible number of significant figures matching the uncertainty in $\tau$ .

Techniques used

use $\tau = RC$ to relate time constant, resistance and capacitanceconvert microfarads to faradsrearrange to find resistance

Q7MediumMagnetic FieldsElectric Fields

(a)

Define magnetic flux density.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Magnetic flux density $B$ is the force per unit current per unit length on a straight conductor placed perpendicular to the magnetic field.

B = \frac{F}{IL}

(where $F$ is the force on length $L$ carrying current $I$ at $90^\circ$ to the field).

Final answer

Magnetic flux density is the force per unit current per unit length on a straight conductor placed perpendicular to the field (B = F/IL).

Detailed explanation

Background Concept

Magnetic flux density $B$ describes the strength of a magnetic field. It is defined operationally (i.e. in terms of a measurable effect) using the force a magnetic field exerts on a current-carrying conductor.

For a straight conductor of length $L$ carrying current $I$ in a magnetic field,

F = BIL\sin\theta

where $\theta$ is the angle between the conductor (current direction) and the magnetic field. The definition uses the special case $\theta = 90^\circ$ so that $\sin\theta = 1$ .

Understanding the Question

You are asked to define magnetic flux density. For full credit, the definition must be precise and include the key condition that the conductor is at right angles to the magnetic field.

Approach

Give the standard definition in words and/or the corresponding defining equation for the perpendicular case:

B = \frac{F}{IL}

Step-by-Step Reasoning

Start from the force on a current-carrying conductor:

F = BIL\sin\theta

For the definition, take $\theta = 90^\circ$ :

F = BIL

Rearrange to make $B$ the subject:

B = \frac{F}{IL}

State in words: “force per unit current per unit length” for a conductor perpendicular to the field.

Key Takeaways

$B$ is defined via the magnetic force on a current-carrying conductor.
The perpendicular condition ( $90^\circ$ ) is essential in the definition.

Common Mistakes

Missing the condition “perpendicular to the field”.
Defining $B$ using $F = BQv$ (this is a valid relationship, but the usual definition at A Level is via a current-carrying conductor).
Writing $B = F/I$ (forgetting the length $L$ ).

Things to Be Careful About

Use the phrase “per unit current per unit length” (not just “force per unit current”).
If you quote an equation, it should be the perpendicular-case equation $B = F/(IL)$ .

Techniques used

state a standard physical definition preciselyexpress a definition as an equation relating measured quantities

(b)

A particle of mass $m$ and charge $+Q$ moves at speed $v$ into a region where there is a uniform magnetic field, as shown in Fig. 7.1.

Fig. 7.1

The uniform magnetic field is into the page and has flux density $B$ . The particle enters the region of the field at point Y.

3M

(i)

State an expression, in terms of some or all of $m$ , $Q$ , $B$ and $v$ , for the magnetic force $F$ that acts on the particle when it is at point Y.

$F$ = ______

1M Similar

(ii)

On Fig. 7.1, draw an arrow at point Y to indicate the direction of the force in (b)(i).

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

b-ii-force-direction

Force on the positive charge is upwards at $Y$ .

Final answer

Upwards

Detailed explanation

Background Concept

The direction of the magnetic force is given by the vector product

\vec{F} = Q\,\vec{v} \times \vec{B}.

For a positive charge, the force direction is the same as $\vec{v} \times \vec{B}$ . For a negative charge it would be opposite.

You can find the direction using:

the right-hand rule for the cross product, or
Fleming’s left-hand rule (conventional current direction is the direction of motion of a positive charge).

Understanding the Question

At $Y$ :

$\vec{v}$ is to the right.
$\vec{B}$ is into the page.
You must draw the force direction arrow on the diagram.

Approach

Use $\vec{v} \times \vec{B}$ :

point fingers along $\vec{v}$ (right)
curl towards $\vec{B}$ (into page)
thumb gives $\vec{F}$ (for positive charge).

Step-by-Step Reasoning

Take right as $+x$ and out of the page as $+z$ . Then:

$\vec{v}$ is in $+x$ .
Into the page is $-z$ .

So

\vec{F} \propto \vec{v} \times \vec{B} = (+x) \times (-z) = +y,

which is upwards on the page.

b-ii-force-direction

Key Takeaways

Force is perpendicular to both $\vec{v}$ and $\vec{B}$ .
For $Q>0$ , direction follows $\vec{v} \times \vec{B}$ .

Common Mistakes

Reversing the direction because of confusing “into the page” with “out of the page”.
Forgetting that the particle is positively charged (sign matters).
Drawing force along the direction of motion (magnetic force is perpendicular to motion).

Things to Be Careful About

Use the correct symbol convention: crosses $\times$ mean into the page.
If you use Fleming’s left-hand rule, be consistent: first finger = $\vec{B}$ , second finger = conventional current (direction of positive charge), thumb = force.

Techniques used

apply Fleming’s left-hand rule or the right-hand rule for v × Buse the sign of the charge to determine force direction

(iii)

On Fig. 7.1, draw a line to show a possible path for the particle through the region of the magnetic field.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

b-iii-path

A circular arc through the field region, curving upwards from $Y$ .

Final answer

Circular arc curving upwards

Detailed explanation

Background Concept

In a uniform magnetic field, if a charged particle’s velocity is perpendicular to $\vec{B}$ , the magnetic force is always perpendicular to the velocity. That means:

the force does no work (speed stays constant),
the force acts as a centripetal force,
so the motion is circular in the plane perpendicular to $\vec{B}$ .

Magnitude:

F = BQv

and for circular motion,

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

so the radius is $r = \frac{mv}{BQ}$ (not required to be found here).

Understanding the Question

You must sketch a possible path of the particle while it is in the shaded region where $\vec{B}$ is uniform and into the page. From part (ii), the force at entry is upward, so the path must initially curve upward.

Approach

Use the fact that a constant-magnitude force always perpendicular to velocity gives circular motion.
Make the curve consistent with the initial force direction (upwards), so the centre of the circle is above the entry point.

Step-by-Step Reasoning

At $Y$ , velocity is to the right.
Force is upwards, so acceleration is upwards.
Acceleration points towards the centre of the circle, so the centre lies above the particle.
Therefore the trajectory inside the field is an arc of a circle that bends upwards.

b-iii-path

Key Takeaways

Uniform $\vec{B}$ + velocity perpendicular to $\vec{B}$ $\Rightarrow$ circular motion.
Curvature direction follows the magnetic force direction.

Common Mistakes

Drawing a straight line (would require zero net force).
Drawing the arc curving downward (wrong cross-product direction).
Drawing a spiral (speed does not change because magnetic force does no work).

Things to Be Careful About

The path is only circular while the particle is in the field region; once it leaves, it continues in a straight line tangential to the arc (if there are no other forces).

Techniques used

recognise uniform magnetic field produces circular motionuse force direction as centripetal acceleration direction to sketch trajectory

(c)

5M

(i)

Explain how an electric field can be used with the magnetic field to ensure that the particle in (b) now passes through point Z.

3M Similar

DifficultyMedium

Worked solution

Answer

Apply a uniform electric field across the magnetic-field region (e.g. using parallel plates) so that the electric force on the positive particle is opposite to the magnetic force.

Here the magnetic force is upwards, so make $\vec{E}$ downwards so that the electric force $QE$ is downwards.

Adjust $E$ so that

QE = BQv

so the resultant force is zero and the particle is undeflected, passing through $Z$ .

c-i-velocity-selector

Final answer

Use a uniform electric field (downwards) so that QE opposes the upward magnetic force; adjust E so QE = BQv and the particle goes straight through Z.

Detailed explanation

Background Concept

A charged particle in simultaneous electric and magnetic fields experiences two forces:

Electric force:

\vec{F}_E = Q\vec{E}

Magnetic force:

\vec{F}_B = Q\,\vec{v} \times \vec{B}

If these forces are equal in magnitude and opposite in direction, the net force is zero and the particle travels in a straight line at constant velocity. This arrangement is called a velocity selector.

Understanding the Question

In part (b), the particle curves because the magnetic force deflects it. Now you want it to pass through point $Z$ (i.e. go straight through the region without being deflected). You must explain how to use an electric field together with the magnetic field to achieve this.

Approach

Decide the direction of magnetic force from part (b): it is upward.
Create an electric field that produces an electric force downward (since the charge is positive).
Choose the electric field strength so that the forces balance:

F_E = F_B

so net force is zero and the particle is not deflected.

Step-by-Step Reasoning

From (b)(ii), magnetic force at entry is upwards.
To cancel it, we need a downward force.
For a positive charge, the electric force is in the direction of $\vec{E}$ , so $\vec{E}$ must be downward.

To produce a uniform $\vec{E}$ , place two parallel plates across the field region:

top plate positive, bottom plate negative, giving $\vec{E}$ downward.

Then adjust the potential difference so that

QE = BQv

and the resultant force is zero, so the particle continues straight and reaches $Z$ .

c-i-velocity-selector

Key Takeaways

Straight-line motion through crossed fields requires zero resultant force.
Balance condition (for perpendicular $\vec{v}$ and $\vec{B}$ ): $QE = BQv$ .
Direction matters: choose $\vec{E}$ to oppose the magnetic deflection.

Common Mistakes

Choosing $\vec{E}$ in the wrong direction (which would increase the deflection).
Saying “electric field cancels magnetic field” (fields do not cancel; forces on the particle cancel).
Forgetting that the magnetic force depends on speed $v$ .

Things to Be Careful About

Because the charge is $+Q$ , the electric force is in the same direction as $\vec{E}$ (opposite for electrons).
The balance works only for one particular speed; particles with other speeds will be deflected.

Techniques used

superpose electric and magnetic forces on a charged particlechoose the electric field direction to oppose magnetic deflectionset resultant force to zero for straight-line motion

(ii)

Derive an expression for $v$ in terms of $B$ and the electric field strength $E$ .

$v$ = ______

2M Similar

Q8MediumQuantum PhysicsNuclear Physics

(a)

State what is meant by the de Broglie wavelength.

1M Similar

DifficultyEasy

Worked solution

Answer

The de Broglie wavelength is the wavelength associated with a moving particle, given by

\lambda = \frac{h}{p}

Final answer

The wavelength associated with a moving particle (matter wave), (\lambda = h/p).

Detailed explanation

Background Concept

Louis de Broglie proposed that particles can exhibit wave-like behaviour. The quantitative link between a particle and its associated matter wave is

\lambda = \frac{h}{p}

where:

$\lambda$ is the de Broglie wavelength,
$h$ is the Planck constant,
$p$ is the momentum of the particle.

This relation is general: it applies to electrons, protons, atoms, etc. For slow (non-relativistic) motion, $p = mv$ .

Understanding the Question

You are being asked to state what “de Broglie wavelength” means. For full credit, you should mention that it is a wavelength associated with a moving particle and ideally give the defining equation $\lambda = h/p$ .

Approach

Give a clear definition (wavelength associated with a moving particle) and include the key formula connecting it to momentum.

Step-by-Step Reasoning

A particle moving with momentum $p$ behaves as if it has an associated wave.
The wavelength of this wave is defined by

\lambda = \frac{h}{p}.

This statement is exactly what the term “de Broglie wavelength” means.

Key Takeaways

De Broglie wavelength links particle momentum to wave behaviour.
The defining relation is $\lambda = h/p$ .

Common Mistakes

Writing $\lambda = h/mv$ as the definition (that is only after assuming $p=mv$ ).
Confusing de Broglie wavelength with photon wavelength or with the Compton wavelength.

Things to Be Careful About

The definition must involve momentum $p$ (most general form).
“Associated with a moving particle” is important: at $p=0$ , the relation would imply an infinite wavelength.

Techniques used

state a definition preciselylink wavelength to momentum using the de Broglie relation

(b)

Calculate the de Broglie wavelength of an electron moving at a speed of $4.9 \times 10^{7}\ \text{m s}^{-1}$ .

wavelength = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

p = mv = (9.11 \times 10^{-31})(4.9 \times 10^{7}) = 4.46 \times 10^{-23}\ \text{kg m s}^{-1}

\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{4.46 \times 10^{-23}} = 1.49 \times 10^{-11}\ \text{m}

Answer

$1.49 \times 10^{-11}\ \text{m}$

Final answer

1.49 × 10^-11 m

Detailed explanation

Background Concept

For a particle, the de Broglie wavelength is

\lambda = \frac{h}{p}.

If the speed is not extremely close to $c$ , we usually take the momentum as

p = mv.

(At higher speeds you would need relativistic momentum, but here the question expects the non-relativistic expression.)

Understanding the Question

An electron has speed $v = 4.9 \times 10^{7}\ \text{m s}^{-1}$ . You must calculate its de Broglie wavelength, so you need its momentum first.

Given:

$m_e = 9.11 \times 10^{-31}\ \text{kg}$
$h = 6.63 \times 10^{-34}\ \text{J s}$
$v = 4.9 \times 10^{7}\ \text{m s}^{-1}$

Unknown: $\lambda$ .

Approach

Find momentum $p = mv$ .
Use $\lambda = h/p$ .
Quote the result in metres with sensible significant figures.

Step-by-Step Reasoning

Momentum:

p = mv = (9.11 \times 10^{-31})(4.9 \times 10^{7})

Multiply numbers and combine powers of ten:

9.11 \times 4.9 = 44.6 \approx 4.46 \times 10^{1}

and

10^{-31} \times 10^{7} = 10^{-24}.

So

p \approx 4.46 \times 10^{1} \times 10^{-24} = 4.46 \times 10^{-23}\ \text{kg m s}^{-1}.

De Broglie wavelength:

\lambda = \frac{h}{p} = \frac{6.63 \times 10^{-34}}{4.46 \times 10^{-23}}

Divide the decimal parts and subtract indices:

\lambda = (6.63/4.46) \times 10^{-11} = 1.49 \times 10^{-11}\ \text{m}.

Key Takeaways

Always go via momentum: $\lambda$ is inversely proportional to $p$ .
For typical exam electron speeds well below $c$ , $p=mv$ is used.

Common Mistakes

Forgetting to calculate momentum first and trying $\lambda = h/(mv^2)$ (wrong).
Using $m = 9.11 \times 10^{-28}$ (confusing kg with g).
Power-of-ten error when dividing $10^{-34}$ by $10^{-23}$ .

Things to Be Careful About

Units: $h$ in $\text{J s}$ is consistent with momentum in $\text{kg m s}^{-1}$ , giving $\lambda$ in metres.
Significant figures: the speed is given to 2 s.f., so 2–3 s.f. is appropriate.

Techniques used

use $\lambda = h/p$calculate momentum using $p = mv$substitute values with correct powers of ten

(c)

State one similarity and one difference between an electron and a positron.

similarity: ______

difference: ______

2M Similar

(d)

An electron moving at a speed of $4.9 \times 10^{7}\ \text{m s}^{-1}$ collides with a positron that is travelling at the same speed in the opposite direction. As a result of the collision, two gamma-ray photons are produced.

8M

(i)

State the name of this type of reaction.

1M Similar

(ii)

State what happens to the electron and to the positron.

2M Similar

(iii)

Explain why two gamma-ray photons are produced, rather than just one.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

Momentum must be conserved. A single photon would have non-zero momentum, but the total initial momentum is zero, so two photons are emitted in opposite directions so their momenta cancel.

Final answer

To conserve momentum: one photon cannot have zero momentum, so two are emitted oppositely to give zero net momentum.

Detailed explanation

Background Concept

Momentum is always conserved in an isolated interaction. Photons have momentum even though they have no rest mass:

p = \frac{E}{c}.

So any photon with energy $E$ has non-zero momentum.

Understanding the Question

The electron and positron have equal speeds in opposite directions, so the total momentum of the system before the collision is zero. The question asks why the products must be two photons rather than one.

Approach

Use conservation of momentum:

initial total momentum $=0$ ,
one photon cannot have zero momentum,
two photons can be emitted in opposite directions so net momentum is zero.

Step-by-Step Reasoning

Electron momentum and positron momentum are equal in magnitude and opposite in direction, so they cancel.
If only one photon were produced, it would have momentum $p = E/c \neq 0$ , meaning the final total momentum would not be zero.
With two photons emitted in opposite directions, their momenta can be equal and opposite, giving total final momentum $=0$ , matching the initial momentum.

Key Takeaways

A single photon cannot represent a zero-momentum final state.
Two-photon emission is the standard way to conserve momentum in $e^-e^+$ annihilation at zero net momentum.

Common Mistakes

Explaining using energy conservation only (energy conservation does not force two photons).
Saying “two photons because there are two particles” (not the correct physics reason).

Things to Be Careful About

The crucial point is momentum: the initial total momentum is zero due to equal and opposite momenta.

Techniques used

apply conservation of momentumuse the fact that a photon must carry momentum

(iv)

Show that the kinetic energy of the electron before the collision is $1.1 \times 10^{-15}\ \text{J}$ .

1M Similar

DifficultyMedium-Easy

Worked solution

Working

E_k = \frac{1}{2}mv^2 = \frac{1}{2}(9.11 \times 10^{-31})(4.9 \times 10^{7})^2

E_k = \frac{1}{2}(9.11 \times 10^{-31})(2.401 \times 10^{15}) = 1.09 \times 10^{-15}\ \text{J} \approx 1.1 \times 10^{-15}\ \text{J}

Answer

$1.1 \times 10^{-15}\ \text{J}$

Final answer

1.1 × 10^-15 J

Detailed explanation

Background Concept

For speeds that are not extremely close to $c$ , the kinetic energy of a particle of mass $m$ and speed $v$ is

E_k = \frac{1}{2}mv^2.

(At relativistic speeds you would use $E_k = (\gamma - 1)mc^2$ , but the question clearly expects the classical expression.)

Understanding the Question

You are asked to show that an electron moving at $4.9 \times 10^{7}\ \text{m s}^{-1}$ has kinetic energy $1.1 \times 10^{-15}\ \text{J}$ .

Given:

$m_e = 9.11 \times 10^{-31}\ \text{kg}$
$v = 4.9 \times 10^{7}\ \text{m s}^{-1}$

Approach

Substitute into $E_k = \tfrac{1}{2}mv^2$ and simplify carefully.

Step-by-Step Reasoning

Square the speed:

(4.9 \times 10^{7})^2 = 4.9^2 \times 10^{14} = 24.01 \times 10^{14} = 2.401 \times 10^{15}.

Multiply by the mass:

(9.11 \times 10^{-31})(2.401 \times 10^{15}) = (9.11 \times 2.401) \times 10^{-16} \approx 21.9 \times 10^{-16} = 2.19 \times 10^{-15}.

Take half:

E_k = \frac{1}{2} \times 2.19 \times 10^{-15} = 1.09 \times 10^{-15}\ \text{J} \approx 1.1 \times 10^{-15}\ \text{J}.

Key Takeaways

Use $E_k = \tfrac{1}{2}mv^2$ and track indices carefully.
“Show that” allows rounding to the stated value.

Common Mistakes

Forgetting the factor $\tfrac{1}{2}$ .
Squaring the power of ten incorrectly (e.g. using $10^{7}$ squared as $10^{7}$ instead of $10^{14}$ ).

Things to Be Careful About

Keep the intermediate step $v^2$ in standard form to avoid index errors.
The final answer should be consistent with the given speed (2 s.f. is fine).

Techniques used

use non-relativistic kinetic energy formulasubstitute values with correct powers of tenround to the stated value

(v)

Use the information in (d)(iv) to determine, to three significant figures, the wavelength associated with the gamma radiation emitted in the collision.

wavelength = ______ $\text{m}$

3M Similar

DifficultyMedium

Worked solution

Working

Energy per photon (equal and opposite momenta) is

E_\gamma = mc^2 + E_k

mc^2 = (9.11 \times 10^{-31})(3.00 \times 10^{8})^2 = 8.20 \times 10^{-14}\ \text{J}

Using $E_k = 1.1 \times 10^{-15}\ \text{J}$ ,

E_\gamma = 8.20 \times 10^{-14} + 1.1 \times 10^{-15} = 8.31 \times 10^{-14}\ \text{J}

E_\gamma = \frac{hc}{\lambda} \Rightarrow \lambda = \frac{hc}{E_\gamma} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{8.31 \times 10^{-14}} = 2.39 \times 10^{-12}\ \text{m}

Answer

$2.39 \times 10^{-12}\ \text{m}$

Final answer

2.39 × 10^-12 m

Detailed explanation

Background Concept

In electron–positron annihilation, energy is conserved. The initial total energy is the sum of rest energies and kinetic energies:

E_\text{initial} = 2mc^2 + 2E_k

Two photons are produced. A photon’s energy is related to its wavelength by

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

If the electron and positron have equal and opposite momenta, the total momentum is zero. For the final momentum to be zero, the two photons must have equal and opposite momenta, which means they also have equal energies.

Understanding the Question

You are told (from part (iv)) that the kinetic energy of the electron is $1.1 \times 10^{-15}\ \text{J}$ (and the positron has the same kinetic energy). You must find the wavelength of the gamma radiation emitted.

Because there are two photons and the initial net momentum is zero, each photon has the same energy. The question says to use the information in (d)(iv), so the kinetic energy must be included (not just $mc^2$ ).

Approach

Find the rest energy of one electron/positron: $mc^2$ .
Energy per photon equals the energy per particle: $E_\gamma = mc^2 + E_k$ .
Convert photon energy to wavelength using $\lambda = hc/E_\gamma$ .

Step-by-Step Reasoning

Rest energy of electron:

mc^2 = (9.11 \times 10^{-31})(3.00 \times 10^{8})^2.

Since $(3.00 \times 10^{8})^2 = 9.00 \times 10^{16}$ ,

mc^2 = 9.11 \times 9.00 \times 10^{-31+16} = 81.99 \times 10^{-15} = 8.20 \times 10^{-14}\ \text{J}.

Energy accounting and energy per photon:

Total initial energy:

E_\text{initial} = 2mc^2 + 2E_k.

Two photons share this total energy equally (equal energies), so each photon has

E_\gamma = \frac{E_\text{initial}}{2} = mc^2 + E_k.

Using $E_k = 1.1 \times 10^{-15}\ \text{J}$ :

E_\gamma = 8.20 \times 10^{-14} + 1.1 \times 10^{-15} = 8.31 \times 10^{-14}\ \text{J}.

Convert to wavelength:

\lambda = \frac{hc}{E_\gamma} = \frac{(6.63 \times 10^{-34})(3.00 \times 10^{8})}{8.31 \times 10^{-14}}.

First evaluate $hc$ :

hc = 1.989 \times 10^{-25}\ \text{J m}.

Then

\lambda = \frac{1.989 \times 10^{-25}}{8.31 \times 10^{-14}} = 2.39 \times 10^{-12}\ \text{m}

(to three significant figures).

Key Takeaways

In $e^-e^+$ annihilation with zero net momentum, the two photons have equal energies.
Each photon carries energy $mc^2 + E_k$ (not just $mc^2$ when particles are moving).
Use $\lambda = hc/E$ for photon wavelength.

Common Mistakes

Using total energy $2mc^2 + 2E_k$ as the energy of one photon (forgetting there are two photons).
Ignoring the kinetic energy even though the question explicitly references (iv).
Using $E = hc\lambda$ instead of $E = hc/\lambda$ .

Things to Be Careful About

The “two photons” condition implies equal sharing here because the initial momenta cancel.
Quote the final wavelength to three significant figures as requested.
Keep consistent constants: use $c = 3.00 \times 10^{8}\ \text{m s}^{-1}$ and $h = 6.63 \times 10^{-34}\ \text{J s}$ (or consistent values).

Techniques used

apply conservation of energy in annihilationcalculate photon energy from rest energy plus kinetic energyuse $E = hc/\lambda$ to determine wavelength

Q9MediumNuclear Physics

(a)

Define activity of a radioactive sample.

1M Similar

(b)

Explain why the variation with time of the activity of a radioactive sample is exponential in nature.

3M Similar

DifficultyMedium

Worked solution

Answer

Radioactive decay is random so each nucleus has a constant probability per unit time of decaying.
Hence the activity (rate of decay) is proportional to the number of undecayed nuclei $N$ :

A = \lambda N\quad \text{and}\quad \frac{\mathrm{d}N}{\mathrm{d}t} = -\lambda N.

As $N$ decreases, the rate decreases in the same proportion, giving an exponential decrease of $N$ (and therefore $A$ ) with time.

Final answer

Because each nucleus has a constant probability per unit time of decay, the rate of decay is proportional to the number remaining (A = λN), so dN/dt = −λN and hence N and A fall exponentially with time.

Detailed explanation

Background Concept

Radioactive decay is a random and spontaneous process. For a given isotope, each nucleus has the same constant probability per unit time of decaying. This constant is the decay constant $\lambda$ .

Two key relationships are:

A = \lambda N

where $A$ is the activity (decays per unit time) and $N$ is the number of undecayed nuclei, and the decay-rate equation

\frac{\mathrm{d}N}{\mathrm{d}t} = -\lambda N.

The solution to this differential equation is exponential:

N = N_0 e^{-\lambda t},\quad A = A_0 e^{-\lambda t}.

Understanding the Question

The question asks for a reason why activity decreases exponentially with time. So you must explain the chain:
random decay with constant probability $\Rightarrow$ rate proportional to number remaining $\Rightarrow$ exponential law.

Approach

State the physical assumption: constant probability of decay per nucleus per unit time.
Convert that into a proportionality: total decays per second $\propto N$ .
State that because the rate is proportional to the amount remaining, the decrease is exponential.

Step-by-Step Reasoning

Consider a short time interval $\Delta t$ . If the probability of decay for one nucleus in that time is proportional to $\Delta t$ , then the expected fraction that decays in $\Delta t$ is constant for that isotope.
If there are $N$ nuclei present, the expected number that decay in $\Delta t$ is proportional to $N$ .
That means the rate of decrease of $N$ is proportional to $N$ itself:

\frac{\mathrm{d}N}{\mathrm{d}t} \propto -N.

Introducing the constant of proportionality $\lambda$ gives

\frac{\mathrm{d}N}{\mathrm{d}t} = -\lambda N.

A quantity whose rate of decrease is proportional to its current value must decrease exponentially. Therefore

N = N_0 e^{-\lambda t}.

Since activity is proportional to $N$ via $A = \lambda N$ , activity also obeys

A = A_0 e^{-\lambda t}.

Key Takeaways

Constant decay probability per nucleus $\Rightarrow$ constant fractional decrease per unit time.
Rate law $\mathrm{d}N/\mathrm{d}t=-\lambda N$ produces exponential decay.
Activity follows the same exponential form because $A\propto N$ .

Common Mistakes

Saying “activity decreases exponentially because half-life is constant” (this reverses cause and effect; constant half-life is a consequence of exponential decay).
Not linking activity to number of nuclei (missing the crucial $A=\lambda N$ idea).

Things to Be Careful About

Exponential does not mean “decreases quickly at first then slowly” unless you justify it via proportional rate.
Make clear it is the probability per unit time that is constant, not the number decaying each second.

Techniques used

state the constant probability of decay per nucleus per unit timelink activity to number of undecayed nucleiuse a proportionality argument leading to an exponential law

(c)

A sample contains a single radioactive isotope that decays to form a stable isotope.

The sample has an activity of $180\ \text{Bq}$ at time $t = 0$ .
At a time $8.4$ minutes later, the activity is $120\ \text{Bq}$ .

6M

(i)

Determine the decay constant, in $\text{min}^{-1}$ , of the radioactive isotope.

decay constant = ______ $\text{min}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

A = A_0 e^{-\lambda t}

120 = 180 e^{-\lambda(8.4)}

\frac{2}{3} = e^{-8.4\lambda}

\ln\left(\frac{2}{3}\right) = -8.4\lambda

\lambda = \frac{\ln(3/2)}{8.4} = 4.83 \times 10^{-2}\ \text{min}^{-1}

Answer

$\lambda = 4.83 \times 10^{-2}\ \text{min}^{-1}$

Final answer

4.83 × 10^−2 min^−1

Detailed explanation

Background Concept

For a single radioactive isotope,

A = A_0 e^{-\lambda t}

where:

$A$ is activity at time $t$ ,
$A_0$ is activity at $t=0$ ,
$\lambda$ is the decay constant (probability per unit time that a nucleus decays).

Because $A$ is proportional to the number of undecayed nuclei, it follows the same exponential form as $N$ .

Understanding the Question

You are told:

$A_0 = 180\ \text{Bq}$ at $t=0$ ,
$A = 120\ \text{Bq}$ at $t = 8.4\ \text{min}$ ,

and asked to find $\lambda$ in $\text{min}^{-1}$ .

Approach

Use the exponential decay law for activity and rearrange:

\frac{A}{A_0} = e^{-\lambda t} \Rightarrow \ln\left(\frac{A}{A_0}\right) = -\lambda t.

Then substitute numbers and solve for $\lambda$ .

Step-by-Step Reasoning

Start with

A = A_0 e^{-\lambda t}.

Substitute the two activity values and the time:

120 = 180 e^{-\lambda(8.4)}.

Divide both sides by $180$ to isolate the exponential:

\frac{120}{180} = e^{-8.4\lambda} \Rightarrow \frac{2}{3} = e^{-8.4\lambda}.

Take natural logs:

\ln\left(\frac{2}{3}\right) = -8.4\lambda.

So

\lambda = -\frac{1}{8.4}\ln\left(\frac{2}{3}\right) = \frac{1}{8.4}\ln\left(\frac{3}{2}\right).

Numerically, $\ln(3/2)=0.405$ so

\lambda = \frac{0.405}{8.4} = 0.0483\ \text{min}^{-1} = 4.83\times 10^{-2}\ \text{min}^{-1}.

Key Takeaways

Activity decays exponentially: $A=A_0 e^{-\lambda t}$ .
Solving for $\lambda$ usually involves taking $\ln$ of an activity ratio.

Common Mistakes

Using $\log_{10}$ instead of $\ln$ without adjusting (Cambridge expects natural log for exponentials).
Forgetting the negative sign: $\ln(A/A_0)$ is negative because $A<A_0$ .
Leaving $\lambda$ in $\text{s}^{-1}$ or mixing seconds and minutes.

Things to Be Careful About

Time must match the unit of $\lambda$ (here minutes $\Rightarrow\ \text{min}^{-1}$ ).
Keep enough significant figures in intermediate steps (e.g. use $0.405465...$ ) to avoid rounding errors.

Techniques used

use the exponential decay equation for activityrearrange to isolate the decay constanttake natural logarithms to solve for the constant

(ii)

Use your answer in (c)(i) to determine the half-life, in min, of the radioactive isotope.

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

1M Similar

DifficultyEasy

Worked solution

Working

t_{1/2} = \frac{\ln 2}{\lambda}

t_{1/2} = \frac{0.693}{4.83 \times 10^{-2}} = 14.4\ \text{min}

Answer

$14.4\ \text{min}$

Final answer

14.4 min

Detailed explanation

Background Concept

The half-life $t_{1/2}$ is the time for the activity (or number of undecayed nuclei) to fall to half its initial value.

From

A = A_0 e^{-\lambda t},

define half-life by $A = A_0/2$ :

\frac{1}{2} = e^{-\lambda t_{1/2}} \Rightarrow t_{1/2} = \frac{\ln 2}{\lambda}.

Understanding the Question

You have already found $\lambda$ in (c)(i). This part asks you to use it to calculate the half-life in minutes.

Approach

Apply the standard relationship

t_{1/2} = \frac{\ln 2}{\lambda}

and ensure the unit comes out in minutes because $\lambda$ was in $\text{min}^{-1}$ .

Step-by-Step Reasoning

Substitute $\lambda = 4.83\times 10^{-2}\ \text{min}^{-1}$ :

t_{1/2} = \frac{0.693}{4.83\times 10^{-2}}.

Calculate:

t_{1/2} = 14.4\ \text{min}.

(Units check: $1/(\text{min}^{-1}) = \text{min}$ .)

Key Takeaways

Half-life is linked to decay constant by $t_{1/2}=\ln 2/\lambda$ .
Units follow directly from the units of $\lambda$ .

Common Mistakes

Using $t_{1/2}=1/\lambda$ (missing the $\ln 2$ factor).
Mixing minutes and seconds between (c)(i) and (c)(ii).

Things to Be Careful About

Use consistent significant figures (typically 3 s.f. for this data gives $14.4\ \text{min}$ ).
Ensure you quote the unit “min”.

Techniques used

use the half-life relation t_{1/2} = ln2 / lambdasubstitute the decay constant and compute

(iii)

On Fig. 9.1, sketch the variation of the activity $A$ of the sample with $t$ for values of $t$ between $t = 0$ and $t = 24\ \text{min}$ .

Fig. 9.1

3M Similar

DifficultyMedium-Easy

Worked solution

activity-decay

Answer

Sketch a smooth exponential decay curve starting at $A=180\ \text{Bq}$ at $t=0$ , passing through $A=120\ \text{Bq}$ at $t=8.4\ \text{min}$ , and approaching $A=0$ asymptotically (e.g. $A\approx 57\ \text{Bq}$ at $t=24\ \text{min}$ ).

Final answer

Exponential decay curve from 180 Bq at t=0 through 120 Bq at t=8.4 min, tending to 0 (A ≈ 57 Bq at 24 min).

Detailed explanation

Background Concept

For radioactive decay,

A = A_0 e^{-\lambda t}.

The key graphical features of an exponential decay are:

starts at $A_0$ when $t=0$ ,
decreases rapidly at first then more slowly,
never becomes negative and approaches $A=0$ as a horizontal asymptote.

Half-life gives helpful checkpoints: after each half-life, activity halves.

Understanding the Question

You must sketch $A$ against $t$ from $0$ to $24\ \text{min}$ .
You are given two anchor points:

$(0,180)$
$(8.4,120)$

Your curve must pass through these and have the correct exponential shape.

Approach

Plot the two given points accurately.
Draw a smooth curve (not a straight line) that decreases and flattens out.
Use your $\lambda$ (or half-life) to estimate the value near $t=24\ \text{min}$ to guide the tail of the curve.

Step-by-Step Reasoning

Mark $A_0 = 180\ \text{Bq}$ at $t=0$ on the vertical axis.
Mark $A=120\ \text{Bq}$ at $t=8.4\ \text{min}$ .
Use the decay law with $\lambda \approx 4.83\times 10^{-2}\ \text{min}^{-1}$ to estimate the end value:

A(24) = 180 e^{-\lambda\,24} \approx 180 e^{-(0.0483)(24)}.

(0.0483)(24) \approx 1.16 \Rightarrow e^{-1.16}\approx 0.314.

A(24) \approx 180\times 0.314 \approx 57\ \text{Bq}.

So by $t=24\ \text{min}$ the curve should be around $\sim 60\ \text{Bq}$ and flattening.
Draw a smooth exponential curve through the two given points, passing near $(24,57)$ , approaching but not touching $A=0$ .

activity-decay

Key Takeaways

Exponential decay curves are smooth, decreasing, and asymptotic to zero.
Use given points and one extra estimated point to get the correct shape over the full range.

Common Mistakes

Drawing a straight line between the points.
Making the curve reach $A=0$ at $t=24\ \text{min}$ (exponential decay does not hit zero).
Sketching a curve that goes below zero.

Things to Be Careful About

Ensure the curve passes through the provided data point $(8.4,120)$ .
Keep the curve concave upwards (slope magnitude decreases with time).
Use the axes scales correctly; the $y$ -axis is in Bq and the $x$ -axis is in minutes.

Techniques used

use key points to anchor an exponential-decay sketchdraw a smooth exponential curve with a horizontal asymptoteestimate additional points using half-life or the decay law

Q10Medium-EasyAstronomy and Cosmology

(a)

State Hubble’s law.

2M Similar

(b)

A star in a distant galaxy emits radiation that has a maximum intensity of emission at a wavelength of $4.62 \times 10^{-7}\ \text{m}$ .

Observations of the galaxy made on the Earth detect the maximum intensity of emission from the star at a wavelength of $4.91 \times 10^{-7}\ \text{m}$ .

6M

(i)

Explain why the observed wavelength and the emitted wavelength have different values.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The detected wavelength is larger than the emitted wavelength, so the radiation is redshifted.

This is because the star/galaxy is moving away from the Earth (recession due to the expansion of the Universe), so the waves are Doppler shifted to longer wavelength.

Final answer

Because the galaxy is receding from Earth, the light is Doppler redshifted so the observed wavelength is longer.

Detailed explanation

Background Concept

Light (and all electromagnetic radiation) can exhibit a Doppler shift. If a source moves away from an observer, successive wavefronts are emitted from progressively further positions, so the spacing between arriving wavefronts increases. That means the observed wavelength $\lambda_{\text{obs}}$ is greater than the emitted wavelength $\lambda_{\text{emit}}$ (a redshift).

In cosmology, distant galaxies generally show redshift because space itself is expanding; this produces the same observed effect: wavelengths are stretched.

Understanding the Question

You are told the star emits maximum intensity at $4.62 \times 10^{-7}\ \text{m}$ but Earth detects the maximum at $4.91 \times 10^{-7}\ \text{m}$ . Since $4.91 \times 10^{-7} > 4.62 \times 10^{-7}$ , the wavelength has increased.

The question asks you to explain why the values differ, so you must mention redshift and the physical reason (recession/expansion).

Approach

Compare the two wavelengths to decide redshift vs blueshift.
Connect redshift to the source moving away from the observer (in this context because of the expansion of the Universe).

Step-by-Step Reasoning

Observed wavelength is longer:

\lambda_{\text{obs}} > \lambda_{\text{emit}}

Longer wavelength corresponds to redshift.
Redshift happens when the source is moving away, so the star/galaxy is receding from Earth (consistent with Hubble expansion).

Key Takeaways

Increase in observed wavelength $\Rightarrow$ redshift.
Redshift indicates recession (moving away), often due to expansion of the Universe for distant galaxies.

Common Mistakes

Saying “the frequency increases” (it decreases when wavelength increases).
Mentioning only “Doppler effect” but not stating that the source is moving away.

Things to Be Careful About

Be explicit: moving away and longer wavelength are the key marking points.
Don’t confuse cosmological redshift with gravitational redshift; here the context is a distant galaxy and Hubble expansion.

Techniques used

identify redshift from an increase in observed wavelengthlink redshift to recessional motion due to expansion of the Universe

(ii)

Calculate the speed of the star relative to the Earth.

speed = ______ $\text{m s}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

\frac{\Delta \lambda}{\lambda} = \frac{v}{c}

\Delta \lambda = (4.91 - 4.62) \times 10^{-7} = 2.9 \times 10^{-8}\ \text{m}

v = c\frac{\Delta \lambda}{\lambda} = (3.00 \times 10^{8}) \times \frac{2.9 \times 10^{-8}}{4.62 \times 10^{-7}} = 1.88 \times 10^{7}\ \text{m s}^{-1}

Answer

$1.9 \times 10^{7}\ \text{m s}^{-1}$

Final answer

1.9 × 10^7 m s^-1

Detailed explanation

Background Concept

Redshift $z$ is defined by

z = \frac{\lambda_{\text{obs}} - \lambda_{\text{emit}}}{\lambda_{\text{emit}}} = \frac{\Delta \lambda}{\lambda_{\text{emit}}}

For speeds small compared with $c$ , the Doppler/redshift relation can be approximated by

\frac{v}{c} \approx z = \frac{\Delta \lambda}{\lambda_{\text{emit}}}

where:

$v$ is the speed of recession (positive when moving away),
$c$ is the speed of light.

Understanding the Question

You are given an emitted wavelength (from the star) and an observed wavelength (measured on Earth) for the same peak in the spectrum. Because the observed wavelength is larger, the star is receding, and you can use the fractional change in wavelength to find the speed.

Given:

$\lambda_{\text{emit}} = 4.62 \times 10^{-7}\ \text{m}$
$\lambda_{\text{obs}} = 4.91 \times 10^{-7}\ \text{m}$

Find: $v$ .

Approach

Calculate the wavelength shift $\Delta \lambda = \lambda_{\text{obs}} - \lambda_{\text{emit}}$ .
Compute the redshift fraction $\Delta \lambda/\lambda_{\text{emit}}$ .
Use $v = c(\Delta \lambda/\lambda_{\text{emit}})$ .

Step-by-Step Reasoning

Difference in wavelengths:

\Delta \lambda = (4.91 - 4.62) \times 10^{-7}

Compute the bracket first: $4.91 - 4.62 = 0.29$ .

\Delta \lambda = 0.29 \times 10^{-7} = 2.9 \times 10^{-8}\ \text{m}

Fractional change:

\frac{\Delta \lambda}{\lambda_{\text{emit}}} = \frac{2.9 \times 10^{-8}}{4.62 \times 10^{-7}} \approx 0.0628

Multiply by $c$ :

v = (3.00 \times 10^{8}) \times 0.0628 = 1.88 \times 10^{7}\ \text{m s}^{-1}

To 2 s.f.:

v \approx 1.9 \times 10^{7}\ \text{m s}^{-1}

Key Takeaways

Use the fractional wavelength change to find recessional speed.
For modest redshifts, $v/c \approx \Delta \lambda/\lambda$ is the standard A-level approximation.

Common Mistakes

Using $\lambda_{\text{obs}}$ in the denominator instead of $\lambda_{\text{emit}}$ (usually the definition uses emitted/rest wavelength).
Subtracting the wrong way round and getting a negative speed (sign error).
Forgetting to multiply by $c$ or using $c = 3.0 \times 10^{8}$ with wrong power of ten.

Things to Be Careful About

Keep track of powers of ten when subtracting: $0.29 \times 10^{-7} = 2.9 \times 10^{-8}$ .
Quote the speed in $\text{m s}^{-1}$ and to an appropriate number of significant figures.

Techniques used

calculate fractional wavelength shiftuse the low-speed Doppler/redshift approximation v/c = Δλ/λsubstitute into an equation and compute the speed

(iii)

The wavelength of maximum intensity of emission is used to determine a value for the surface temperature of the star.

Explain how the temperature determined using the observed wavelength compares with the true value of temperature determined using the emitted wavelength.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Using Wien’s law $\lambda_{\max}T = \text{constant}$ , the larger observed peak wavelength gives a smaller temperature.

So the temperature found using $4.91 \times 10^{-7}\ \text{m}$ is lower than the true surface temperature found from $4.62 \times 10^{-7}\ \text{m}$ .

Final answer

It would be underestimated (observed λmax is larger so T is smaller).

Detailed explanation

Background Concept

Wien’s displacement law relates the wavelength at which a blackbody spectrum has maximum intensity, $\lambda_{\max}$ , to the absolute temperature $T$ :

\lambda_{\max} T = b

where $b$ is Wien’s constant. The key relationship is that

T \propto \frac{1}{\lambda_{\max}}

So if the peak wavelength appears larger, the inferred temperature will be smaller.

Understanding the Question

The star’s true surface temperature corresponds to the peak wavelength it emits in its own rest frame ( $4.62 \times 10^{-7}\ \text{m}$ ). However, Earth observes a redshifted peak at $4.91 \times 10^{-7}\ \text{m}$ . The question asks how the temperature calculated from the observed wavelength compares with the true temperature.

Approach

Use the inverse proportionality from Wien’s law: compare $\lambda_{\max}$ values to decide whether the inferred $T$ increases or decreases.

Step-by-Step Reasoning

From Wien’s law:

T = \frac{b}{\lambda_{\max}}

Observed peak wavelength is larger:

\lambda_{\text{obs}} > \lambda_{\text{emit}}

Because $T$ is inversely proportional to $\lambda_{\max}$ :

T_{\text{obs}} = \frac{b}{\lambda_{\text{obs}}} < \frac{b}{\lambda_{\text{emit}}} = T_{\text{true}}

So the temperature calculated using the observed wavelength would be an underestimate.

Key Takeaways

Wien’s law: bigger $\lambda_{\max}$ means lower inferred $T$ .
Redshift pushes the spectrum peak to longer wavelength, biasing temperature measurements low if uncorrected.

Common Mistakes

Claiming the temperature would be higher (getting the inverse relationship wrong).
Mixing up “maximum intensity” wavelength with “maximum frequency” (they do not map one-to-one without care).

Things to Be Careful About

Use kelvin temperature (absolute temperature) with Wien’s law.
The comparison is qualitative here: you do not need to calculate the temperatures, only state the direction of the effect and link it to $\lambda_{\max}T = \text{constant}$ .

Techniques used

use Wien’s displacement law \lambda_max T = constantcompare temperatures using inverse proportionality to wavelength

(c)

A value for the Hubble constant is $2.3 \times 10^{-18}\ \text{s}^{-1}$ .

Use your answer in (b)(ii) to determine the distance of the star in (b) from the Earth.

distance = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Hubble’s law:

v = H_0 d

d = \frac{v}{H_0} = \frac{1.88 \times 10^{7}}{2.3 \times 10^{-18}} = 8.17 \times 10^{24}\ \text{m}

Answer

$8.2 \times 10^{24}\ \text{m}$

Final answer

8.2 × 10^24 m

Detailed explanation

Background Concept

Hubble’s law relates recessional speed $v$ and distance $d$ :

v = H_0 d

Rearranging gives:

d = \frac{v}{H_0}

If $H_0$ is in $\text{s}^{-1}$ and $v$ is in $\text{m s}^{-1}$ , then $d$ comes out in metres.

Understanding the Question

You are given $H_0 = 2.3 \times 10^{-18}\ \text{s}^{-1}$ and told to use the speed found in (b)(ii) to calculate the distance of the star’s galaxy from Earth.

So:

Known: $v$ from (b)(ii), $H_0$
Unknown: $d$

Approach

Use $d = v/H_0$ and substitute the numerical values carefully, especially the negative power of ten in $H_0$ .

Step-by-Step Reasoning

Start with Hubble’s law and rearrange:

v = H_0 d \quad \Rightarrow \quad d = \frac{v}{H_0}

Substitute $v = 1.88 \times 10^{7}\ \text{m s}^{-1}$ and $H_0 = 2.3 \times 10^{-18}\ \text{s}^{-1}$ :

d = \frac{1.88 \times 10^{7}}{2.3 \times 10^{-18}}

Deal with powers of ten:

\frac{10^{7}}{10^{-18}} = 10^{25}

and the numerical part:

\frac{1.88}{2.3} \approx 0.817

So:

d \approx 0.817 \times 10^{25} = 8.17 \times 10^{24}\ \text{m}

To 2 s.f.:

d \approx 8.2 \times 10^{24}\ \text{m}

Key Takeaways

Hubble’s law can be used to estimate cosmic distances from recessional speeds.
Dividing by a very small $H_0$ (about $10^{-18}$ ) gives a very large distance (about $10^{24}$ to $10^{26}$ m).

Common Mistakes

Using $d = H_0/v$ (inverting incorrectly).
Mishandling powers of ten: dividing by $10^{-18}$ should increase the power by 18.
Forgetting to carry forward the answer from (b)(ii) consistently.

Things to Be Careful About

Keep units consistent: $v$ in $\text{m s}^{-1}$ and $H_0$ in $\text{s}^{-1}$ .
Quote the final distance to sensible significant figures based on the data (here 2 s.f. is appropriate).

Techniques used

apply Hubble’s law v = H0 drearrange an equation to make distance the subjectsubstitute numerical values and compute with powers of ten

Physics 9702/43 — May/June 2025

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