A circuit contains a power supply that provides a sinusoidal alternating input voltage $V_{IN}$. There is an output voltage $V_{OUT}$ across a load resistor $R$, as shown in Fig. 7.1. (a) State the purpose of the circuit in Fig. 7.1. (b) Fig. 7.2 shows the variation of $V_{OUT}$ with time $t$. (i) The load resistor $R$ has a resistance of $370\ \Omega$. Show that the maximum power dissipated in $R$ is $0.22\ \text{W}$. (ii) On Fig. 7.3, sketch the variation with $t$ of the…

(a) Full-wave rectification to give a unidirectional (pulsating d.c.) output across R. · (b)(i) 0.22 W · (b)(ii) Sketch of P(t): always positive, peaks at 0.22 W, period 0.02 s (sin^2 shape). · (b)(iii) 0.11 W · (c) Same as (b)(iii).

(a) Define half-life of a radioactive isotope. (b) Radioactive isotope X decays to isotope Y. A sample contains only nuclei of X at time $t = 0$. Fig. 9.1 shows the variation with $t$ of the numbers of nuclei of X and of Y as the sample decays. (i) State the name of the quantity represented by the magnitude of the gradient of line X in Fig. 9.1. (ii) State three conclusions about X or Y that may be drawn from Fig. 9.1. The conclusions may be qualitative or quantitative. Use…

(a) Time for the number of undecayed nuclei (or activity) to fall to half its initial value. · (b)(i) Activity (rate of decay), A = −dNX/dt. · (b)(ii) Examples: half-life of X is 14 s; NX+NY constant at 4.0×10^22 so X→Y one-to-one; NX decays exponentially while NY rises to 4.0×10^22. · (c) 11

9702/43 May/June 2024 — Physics Solved Past Paper

Q1MediumGravitational Fields

(a)

Define gravitational potential at a point.

2M Similar

DifficultyEasy

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 that 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 $V$ at a point in a gravitational field is defined using a reference level at infinity:

Take $V = 0$ at $r = \infty$ .
$V$ at a point is the work done per unit mass (by an external agent) in moving a small test mass slowly from infinity to that point.

“Slowly” (or “with no change in kinetic energy”) matters because then all the work done changes the gravitational potential energy rather than the kinetic energy.

Gravitational potential energy $U$ of a mass $m$ at a point is related by

U = mV.

Understanding the Question

You are asked for the definition of gravitational potential at a point, so you must mention:

“per unit mass”,
“work done”,
“from infinity” (the reference where potential is defined as zero),
and typically “test mass” / “no change in kinetic energy”.

Approach

Write the standard Cambridge definition: work done per unit mass by an external agent bringing a test mass from infinity to the point, without changing its kinetic energy.

Step-by-Step Reasoning

Identify that gravitational potential is defined relative to infinity, where $V=0$ .
Since it is “potential”, it is energy-related, so use work done.
Because it is “per unit mass”, divide the work done by the mass (or explicitly say “per unit mass”).
Specify “no change in kinetic energy” to make it clear the work goes into potential energy only.

Key Takeaways

Gravitational potential is an energy-per-unit-mass quantity.
Infinity is the conventional zero of gravitational potential for isolated masses.
$U = mV$ links potential energy and potential.

Common Mistakes

Defining gravitational potential as “force per unit mass” (that is gravitational field strength $g$ ).
Missing “per unit mass”.
Missing the reference to “infinity”.
Not stating “no change in kinetic energy / moved slowly”.

Things to Be Careful About

Use “external agent” (or equivalent wording) to avoid ambiguity about who does the work.
Do not confuse gravitational potential $V$ with gravitational potential energy $U$ (they differ by a factor of mass).

Techniques used

state the definition of gravitational potential using work done per unit massreference infinity as the zero of potential

(b)

A satellite X, of mass $M$ , orbits a planet at a constant distance $4R$ from the centre of the planet, as shown in Fig. 1.1.

Fig. 1.1

A second satellite Y, of mass $2M$ , orbits the planet with orbital radius $R$ .

The gravitational potential at X due to the planet is $-\Phi$ . The planet is a uniform sphere.

8M

(i)

Explain why the gravitational potential at X is negative.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Gravitational potential is defined to be zero at infinity. Since the gravitational force is attractive, bringing a mass from infinity to X releases energy (the field does positive work), so the potential energy per unit mass at X is less than zero; hence the potential at X is negative.

Final answer

Because potential is zero at infinity and gravity is attractive, energy is released moving from infinity to X so potential (energy per unit mass) at X is below zero, i.e. negative.

Detailed explanation

Background Concept

By definition, gravitational potential $V$ is the work done per unit mass by an external agent in bringing a test mass from infinity to the point.

For an isolated spherical mass, the potential is

V = -\frac{GM}{r}

for points outside the sphere (taking $V=0$ at $r=\infty$ ).

The negative sign is not “optional”: it comes from the convention that potential is zero at infinity and the fact that gravitational forces are attractive.

Understanding the Question

You are told that the potential at X is $-\Phi$ and asked to explain why it is negative. This is a conceptual question about sign convention and energy/work, not a numerical calculation.

Approach

Start from the definition $V=0$ at infinity. Then argue that as a mass moves from infinity towards the planet, the gravitational field does positive work (the mass speeds up if released), so the potential energy decreases; therefore $V$ must be negative at any finite distance.

Step-by-Step Reasoning

Set the reference: at infinity, the gravitational interaction becomes negligible, so we define

V(\infty) = 0.

Gravity is attractive, so a mass released from rest far away would accelerate towards the planet. That means the gravitational force does positive work on the mass.
If the gravitational field does positive work, the gravitational potential energy decreases (becomes more negative) compared with its value at infinity.
Since potential is potential energy per unit mass, it also becomes negative at finite $r$ .

Key Takeaways

The choice $V(\infty)=0$ together with attraction forces $V<0$ at finite distance.
Negative potential indicates a bound system: energy must be supplied to escape to infinity.

Common Mistakes

Saying “negative because gravity is negative” without mentioning the zero at infinity.
Confusing potential $V$ with field strength $g$ .
Claiming potential is negative because “work is done against gravity” without stating who does the work (field vs external agent).

Things to Be Careful About

Be clear: when an object falls inwards, the field does positive work; an external agent bringing it in slowly would do negative work.
The sign comes from the reference choice and the direction of the gravitational force.

Techniques used

use the reference level of potential at infinityrelate the sign of potential to attractive forces and work done

(ii)

State an expression, in terms of $\Phi$ , for the gravitational potential at Y due to the planet.

gravitational potential = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For $r \ge R$ , gravitational potential

V = -\frac{GM_p}{r}.

At X, $r=4R$ :

-\Phi = -\frac{GM_p}{4R} \Rightarrow \frac{GM_p}{R} = 4\Phi.

At Y, $r=R$ :

V_Y = -\frac{GM_p}{R} = -4\Phi.

Answer

$-4\Phi$

Final answer

-4\Phi

Detailed explanation

Background Concept

For a spherically symmetric mass (including a uniform sphere), the gravitational field outside the sphere is the same as if all the mass were concentrated at the centre. Therefore, for $r \ge R$ :

V(r) = -\frac{GM_p}{r}

where $M_p$ is the planet’s mass. This uses the convention $V(\infty)=0$ .

A key consequence is that gravitational potential varies as $1/r$ .

Understanding the Question

You are told:

satellite X is at distance $4R$ from the centre,
satellite Y is at distance $R$ from the centre,
the potential at X is $V_X = -\Phi$ .

You must find the potential at Y in terms of $\Phi$ .

Approach

Use $V \propto -1/r$ . Since Y is four times closer to the centre than X, the magnitude of the potential is four times larger (more negative).

Step-by-Step Reasoning

Write the potential at X using the point-mass form:

V_X = -\frac{GM_p}{4R}.

Set this equal to the given value $-\Phi$ :

-\frac{GM_p}{4R} = -\Phi.

Multiply both sides by $-1$ and rearrange to express $GM_p$ in terms of $\Phi$ :

\frac{GM_p}{4R} = \Phi \Rightarrow GM_p = 4R\Phi.

Now compute the potential at Y ( $r=R$ ):

V_Y = -\frac{GM_p}{R} = -\frac{4R\Phi}{R} = -4\Phi.

Key Takeaways

Outside a spherical planet, $V=-GM/r$ and scales as $1/r$ .
Moving closer to the planet makes the potential more negative.

Common Mistakes

Using an inverse-square dependence (that is for field strength $g$ , not potential $V$ ).
Forgetting the negative sign in $V=-GM/r$ .
Using the satellite’s mass in the potential (potential is property of the field only).

Things to Be Careful About

The planet is a uniform sphere: this justifies using the point-mass formula for $r \ge R$ .
Keep track that $\Phi$ is defined via $V_X=-\Phi$ (so $\Phi$ itself is positive).

Techniques used

use the gravitational potential formula for a spherical massapply inverse proportionality of potential with orbital radius

(iii)

Complete Table 1.1 by giving expressions, in terms of some or all of $M$ , $R$ and $\Phi$ , for the quantities indicated for each of the satellites X and Y.

Table 1.1

	satellite X	satellite Y
gravitational field strength at satellite due to planet
gravitational potential energy of satellite

4M Similar

DifficultyMedium

Worked solution

Working

From part (ii), at $r=4R$ :

-\Phi=-\frac{GM_p}{4R} \Rightarrow GM_p = 4R\Phi.

Gravitational field strength:

g = \frac{GM_p}{r^2}.

At X ( $r=4R$ ):

g_X = \frac{4R\Phi}{(4R)^2}=\frac{\Phi}{4R}.

At Y ( $r=R$ ):

g_Y = \frac{4R\Phi}{R^2}=\frac{4\Phi}{R}.

Gravitational potential energy $U = mV$ .

At X: $V_X=-\Phi$ , $m=M$ :

U_X = M(-\Phi) = -M\Phi.

At Y: $V_Y=-4\Phi$ , $m=2M$ :

U_Y = 2M(-4\Phi) = -8M\Phi.

Answer

$g_X = \dfrac{\Phi}{4R}$ , $g_Y = \dfrac{4\Phi}{R}$
$U_X = -M\Phi$ , $U_Y = -8M\Phi$

Final answer

g_X = \Phi/(4R), g_Y = 4\Phi/R; U_X = -M\Phi, U_Y = -8M\Phi

Detailed explanation

Background Concept

For a planet of mass $M_p$ that is spherically symmetric:

Gravitational field strength (magnitude) at distance $r$ is

g = \frac{GM_p}{r^2}.

Gravitational potential at distance $r$ is

V = -\frac{GM_p}{r}.

Gravitational potential energy $U$ of a mass $m$ at that point is

U = mV.

So:

$g$ depends on $1/r^2$ ,
$V$ depends on $1/r$ ,
$U$ depends on both the mass $m$ and $V$ .

Understanding the Question

You must complete a table for two satellites:

Satellite X: mass $M$ , radius $r_X=4R$ , potential $V_X=-\Phi$ .
Satellite Y: mass $2M$ , radius $r_Y=R$ .

You need expressions (not numbers) for:

gravitational field strength at each satellite due to the planet,
gravitational potential energy of each satellite.

All answers must be in terms of some/all of $M$ , $R$ , and $\Phi$ .

Approach

First eliminate the unknown planet mass $M_p$ by using the given potential at X:
$-\Phi = -GM_p/(4R) \Rightarrow GM_p = 4R\Phi$ .
Substitute this into $g=GM_p/r^2$ for each radius.
Use $U=mV$ with the correct satellite masses and potentials at their radii.

Step-by-Step Reasoning

Use the potential at X to relate $GM_p$ and $\Phi$ :

V_X=-\frac{GM_p}{4R}=-\Phi \Rightarrow GM_p = 4R\Phi.

Field strength at X (radius $4R$ ):

g_X=\frac{GM_p}{(4R)^2}=\frac{4R\Phi}{16R^2}=\frac{\Phi}{4R}.

Notice how the inverse-square makes the field much smaller at the larger radius.

Field strength at Y (radius $R$ ):

g_Y=\frac{GM_p}{R^2}=\frac{4R\Phi}{R^2}=\frac{4\Phi}{R}.

Potential energy at X:

Potential there is given: $V_X=-\Phi$ .
Satellite mass is $M$ .

U_X = mV = M(-\Phi) = -M\Phi.

Potential energy at Y:

From part (ii), $V_Y=-4\Phi$ .
Satellite mass is $2M$ .

U_Y = 2M(-4\Phi) = -8M\Phi.

The bigger mass and the more negative potential both make $U_Y$ much more negative.

Key Takeaways

Outside a spherical planet: $V \propto -1/r$ but $g \propto 1/r^2$ .
Potential energy uses the satellite’s mass: $U=mV$ .
A useful technique is to eliminate $GM_p$ using given information about potential.

Common Mistakes

Using $V \propto 1/r^2$ (confusing potential with field strength).
Forgetting that $U$ depends on the satellite mass, while $V$ does not.
Missing the factor $2$ for satellite Y’s mass ( $2M$ ).
Losing the negative sign for potential energy (bound systems have negative $U$ relative to infinity).

Things to Be Careful About

The question asks for gravitational field strength “at satellite due to planet”: use the planet’s field at that radius, not something involving the satellite mass.
Keep $\Phi$ as a positive symbol: the potential at X is $-\Phi$ , not $\Phi$ .
When substituting $r=4R$ into $r^2$ , remember $(4R)^2=16R^2$ .

Techniques used

use inverse-square law to determine gravitational field strength at a radiususe gravitational potential energy as mass times gravitational potentialsubstitute for the planet's mass parameter using a given potential

Q2Medium-EasyTemperature

(a)

2M

(i)

State the magnitude and unit of absolute zero on the thermodynamic temperature scale.

1M Similar

(ii)

Explain why temperature measured using a laboratory liquid-in-glass thermometer does not give a measurement of thermodynamic temperature.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

A liquid-in-glass thermometer uses a thermometric property (e.g. length/volume of liquid) that is not exactly proportional to thermodynamic temperature, so its scale depends on calibration points and the liquid used and is not the thermodynamic temperature.

Final answer

Because its thermometric property is not exactly proportional to thermodynamic temperature; it depends on calibration/liquid.

Detailed explanation

Background Concept

Thermodynamic temperature $T$ (kelvin) is defined in a way that does not depend on the material of the thermometer (in practice it can be related to an ideal gas scale or absolute thermodynamic definitions).

A practical thermometer works by measuring some thermometric property (a physical property that varies with temperature), such as:

length of a mercury/alcohol column
resistance of a metal wire
e.m.f. of a thermocouple

To turn the property into a temperature reading, the thermometer is calibrated using fixed points (e.g. ice point and steam point) and assuming a relationship between the property and temperature (often assumed linear over a limited range).

Understanding the Question

The question asks why a laboratory liquid-in-glass thermometer does not directly measure thermodynamic temperature. It’s asking you to recognise that it is a practical scale based on a particular liquid and calibration, not the fundamental thermodynamic scale.

Approach

State that:

the liquid’s expansion (or column length) is the thermometric property;
this property is not exactly proportional/linear with thermodynamic temperature over all temperatures;
therefore the reading depends on the substance and calibration, so it is not a true thermodynamic temperature.

Step-by-Step Reasoning

In a liquid-in-glass thermometer, the measured quantity is the length (or volume) of the liquid column.
The conversion from column length to “temperature” assumes a calibration (fixed points) and often assumes linear behaviour between them.
Real liquids do not expand perfectly linearly with thermodynamic temperature, and the glass bulb/stem also expands.
Therefore, the same thermodynamic temperature could correspond to slightly different column lengths depending on the liquid and construction.
Hence the thermometer does not give a direct measurement of thermodynamic temperature; it gives a temperature on its own practical scale.

Key Takeaways

Thermodynamic temperature is substance-independent.
Practical thermometers rely on a particular thermometric property and calibration assumptions.

Common Mistakes

Saying only “it is inaccurate”: the key idea is dependence on material/calibration, not just random error.
Forgetting that both the liquid and the glass expand (so the property is not purely due to the liquid).

Things to Be Careful About

You do not need to mention specific fixed points, but you must link the issue to the fact that the scale is based on a thermometric property that is not exactly proportional to $T$ .

Techniques used

distinguish thermodynamic temperature from a practical temperature scalelink thermometer readings to calibration assumptionsexplain limitations due to non-linearity of thermometric properties

(b)

Fig. 2.1 shows a simplified diagram of a type of thermometer called a platinum resistance thermometer.

Fig. 2.1

The glass tube is immersed in the environment for which the temperature is to be determined. The resistance between the terminals X and Y is measured.

Fig. 2.2 shows the variation of the resistivity $\rho$ of platinum with thermodynamic temperature $T$ .

Fig. 2.2

4M

(i)

Explain how Fig. 2.2 shows that platinum is a suitable metal for use in a resistance thermometer.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Fig. 2.2 shows that $\rho$ varies (approximately) linearly with $T$ and increases steadily with $T$ (single-valued/monotonic), so resistance can be calibrated to give temperature accurately over the range.

Final answer

Because resistivity increases monotonically and approximately linearly with temperature, allowing calibration.

Detailed explanation

Background Concept

A resistance thermometer uses electrical resistance as the thermometric property.

For a uniform wire of length $L$ and cross-sectional area $A$ :

R = \rho \frac{L}{A}

If $L$ and $A$ are essentially constant, then changes in resistance $R$ directly follow changes in resistivity $\rho$ . For temperature measurement, we want a thermometric property that:

changes significantly with temperature (good sensitivity)
changes in a predictable way (ideally linear)
is single-valued and monotonic (each temperature corresponds to one value)

Understanding the Question

You are given a graph of platinum resistivity $\rho$ against thermodynamic temperature $T$ . You must use features of the graph to argue that platinum is suitable for a resistance thermometer.

Two marks means you should give two distinct reasons based on the graph shape.

Approach

Look for the key features that make calibration easy and readings unambiguous:

straight line (or close to it) (\Rightarrow) linear relationship
increasing consistently (\Rightarrow) monotonic and predictable
Optionally, mention that a noticeable gradient implies good sensitivity.

Step-by-Step Reasoning

The graph is a straight line with positive gradient, so $\rho$ is approximately proportional to $T$ over the plotted range.
Because it rises steadily (no turning points), each value of $\rho$ corresponds to only one value of $T$ , so temperature readings are unambiguous.
Therefore, measuring $R$ (and hence $\rho$ ) allows a straightforward calibration to determine $T$ .

Key Takeaways

Suitability comes from predictable, monotonic variation of a thermometric property with temperature.
Approximate linearity is especially useful because it simplifies calibration.

Common Mistakes

Saying only “it is a straight line”: you should connect that to easy calibration.
Claiming it must pass through the origin: real metals often have a non-zero intercept due to residual resistivity; this does not prevent use as a thermometer.

Things to Be Careful About

The question is about suitability for a resistance thermometer, so your explanation must link the graph to resistance measurement (via $R \propto \rho$ if geometry is fixed).

Techniques used

interpret a resistivity–temperature graphidentify linearity and monotonic behaviourlink thermometric property behaviour to suitability for measurement

(ii)

Suggest a reason why a platinum resistance thermometer is not suitable for measuring a rapidly changing temperature.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

It has a relatively large thermal mass/slow response time, so the platinum wire takes time to reach thermal equilibrium with the surroundings and cannot follow rapid temperature changes.

Final answer

Slow response (large thermal mass); wire not in equilibrium quickly enough.

Detailed explanation

Background Concept

A thermometer measures temperature by coming into thermal equilibrium with its surroundings. The time it takes to respond is related to:

thermal mass (heat capacity) of the sensing element and its housing
thermal contact with the environment

A large thermal mass and/or poor thermal contact gives a long time constant, so the thermometer “lags behind” a rapidly changing temperature.

Understanding the Question

The platinum resistance thermometer shown has a platinum wire coil mounted inside a glass tube. The question asks why this design is unsuitable when temperature changes quickly.

Approach

Identify that the sensor must exchange energy with the environment and reach equilibrium. Then point to physical reasons it might do that slowly (glass tube, coil, overall mass).

Step-by-Step Reasoning

The environment temperature changes rapidly.
The platinum wire and its support/glass tube need time for heat transfer to occur.
Because the assembly has non-negligible heat capacity (thermal mass), its temperature cannot change instantly.
Therefore the measured resistance corresponds to the sensor’s temperature, which lags behind the true (instantaneous) environment temperature.

Key Takeaways

Rapidly changing temperatures require a sensor with small thermal mass and good thermal contact.
Resistance thermometers in protective housings often respond too slowly.

Common Mistakes

Saying “resistance changes slowly”: the resistance changes as fast as the wire temperature changes; the limiting factor is heat transfer/thermal inertia.
Vague answers like “not accurate”: the issue is specifically the time lag.

Things to Be Careful About

Mention either “slow to reach thermal equilibrium” or “large thermal mass/slow response time” explicitly to target the mark-scheme idea.

Techniques used

apply the idea of thermal response time and thermal inertialink thermometer construction to rate of reaching thermal equilibrium

(iii)

Suggest a type of thermometer that is suitable for measuring a rapidly changing temperature.

1M Similar

(c)

A negative temperature coefficient thermistor may be used as a type of resistance thermometer.

State one way in which the variation with temperature of the resistance of a thermistor differs from that of a platinum wire.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

A thermistor has a negative temperature coefficient: its resistance decreases as temperature increases (and the variation is non-linear), whereas a platinum wire’s resistance increases approximately linearly with temperature.

Final answer

Thermistor resistance decreases with increasing temperature (non-linear), unlike platinum which increases roughly linearly.

Detailed explanation

Background Concept

Different materials show different resistance–temperature behaviour:

Metals (like platinum) generally have a positive temperature coefficient: resistance increases with temperature, often approximately linearly over a useful range.
Negative temperature coefficient (NTC) thermistors are semiconductors: as temperature rises, more charge carriers become available, so resistance decreases. This decrease is typically strongly non-linear (often close to exponential).

Understanding the Question

You are told an NTC thermistor can be used as a resistance thermometer. You must state one difference between how its resistance varies with temperature compared with platinum wire.

Only one clear statement is needed for 1 mark.

Approach

Give the most direct contrast:

thermistor: $R$ decreases as $T$ increases (negative temperature coefficient)
platinum: $R$ increases with $T$ (positive temperature coefficient)
Optionally include that thermistors are non-linear.

Step-by-Step Reasoning

For platinum wire, increasing $T$ increases lattice vibrations and electron scattering, so $R$ increases (approximately linearly).
For an NTC thermistor, increasing $T$ increases the number of charge carriers, so conductivity increases and resistance decreases (and not in a straight-line way).

Key Takeaways

Metals: $R$ increases with $T$ (PTC).
NTC thermistors: $R$ decreases with $T$ , usually non-linearly.

Common Mistakes

Saying “thermistor resistance increases”: that would describe a PTC thermistor, not an NTC one.
Giving a construction difference (size/material) rather than the required variation with temperature.

Things to Be Careful About

The question specifies negative temperature coefficient; use that cue to state the direction of change correctly.

Techniques used

compare temperature dependence of resistance for different sensorsidentify sign of temperature coefficientdescribe qualitative graph shape (linear vs non-linear)

Q3Medium-EasyIdeal GasesThermodynamics

(a)

4M

(i)

State what is meant by an ideal gas.

2M Similar

DifficultyEasy

Worked solution

Answer

An ideal gas is a gas that obeys

pV = nRT

(or $pV = NkT$ ) at all temperatures and pressures (equivalently: molecules have negligible volume and no intermolecular forces except during elastic collisions).

Final answer

A gas that obeys pV = nRT (or pV = NkT) at all temperatures and pressures.

Detailed explanation

Background Concept

An ideal gas is a model used to describe real gases when they are dilute. In this model:

the gas obeys the equation of state

pV = nRT

or, in molecular form,

pV = NkT

where $N$ is the number of molecules and $k$ is the Boltzmann constant.

This behaviour comes from the kinetic theory assumptions: molecules are treated as point particles (negligible volume), there are no intermolecular forces except during collisions, and collisions are perfectly elastic.

Understanding the Question

You are asked to state what is meant by an ideal gas. For 2 marks, you typically need a clear statement that either:

it obeys the ideal gas equation at all $p, V, T$ , and/or
it satisfies the kinetic theory assumptions that lead to that equation.

Approach

Give a crisp definition. The safest exam definition is the one tied to the equation of state: “obeys $pV=nRT$ for all temperatures and pressures.” You may add one key assumption (e.g. no intermolecular forces) to secure full credit.

Step-by-Step Reasoning

State the defining mathematical relationship: $pV = nRT$ .
Clarify its meaning: it holds for all states (not only approximately).
Optionally link to the model assumptions (negligible molecular volume, no forces, elastic collisions).

Key Takeaways

“Ideal gas” means perfect obedience to $pV=nRT$ .
The equation is justified by kinetic theory assumptions.

Common Mistakes

Saying “a gas that expands when heated” (too vague).
Stating only one assumption without linking to ideal-gas behaviour.
Saying it obeys $pV=nRT$ only at low pressure (that describes real gases approximately, not the definition of ideal).

Things to Be Careful About

The question is state: keep it short and definition-like.
If you include assumptions, make sure they are correct (no intermolecular forces except during collisions; collisions elastic; random motion).

Techniques used

state the defining property of an ideal gaslink the definition to kinetic theory assumptions

(ii)

Use one of the basic assumptions of the kinetic theory to explain what can be deduced about the potential energy associated with the random motion of molecules in an ideal gas.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

For an ideal gas we assume no intermolecular forces (except during collisions). Hence the intermolecular potential energy is zero/constant, so there is no potential-energy term associated with the random motion; the internal energy is due only to random kinetic energy.

Final answer

No intermolecular forces ⇒ potential energy is zero/constant, so internal energy is only random kinetic energy.

Detailed explanation

Background Concept

The internal energy $U$ of a substance is the total microscopic energy of its particles. In general it includes:

random kinetic energy (translational, and possibly rotational/vibrational), and
potential energy due to intermolecular forces (attraction/repulsion between molecules).

Kinetic theory’s ideal-gas model assumes molecules interact only by brief elastic collisions and otherwise experience no forces.

Understanding the Question

You must use one basic assumption of kinetic theory to say what can be deduced about the potential energy associated with the random motion of molecules in an ideal gas.

The key clue is: potential energy between molecules exists only if there are forces between them when they are separated.

Approach

Choose the assumption “no intermolecular forces (except during collisions).” Then argue:

with no forces, there is no change of potential energy with separation,
so the potential energy term can be taken as zero (or constant),
therefore internal energy changes are due only to kinetic energy.

Step-by-Step Reasoning

Kinetic theory assumption for an ideal gas: molecules exert no forces on each other except during collisions.
Intermolecular potential energy arises from work done against intermolecular forces as molecules move relative to each other.
If the force is zero (between collisions), then no work is done in separating/approaching molecules, so the intermolecular potential energy does not change and can be taken as zero/constant.
Therefore, in an ideal gas there is no potential-energy contribution to the internal energy; $U$ depends only on the random kinetic energy of the molecules.

Key Takeaways

“No intermolecular forces” is the assumption that removes potential energy from the model.
Ideal-gas internal energy is purely kinetic (in the simplest treatment: translational).

Common Mistakes

Saying “potential energy is negligible because molecules are far apart” without explicitly linking to the assumption of zero forces.
Confusing potential energy in a gravitational field with intermolecular potential energy.
Claiming the potential energy is always exactly zero in real gases (it is a model assumption).

Things to Be Careful About

Wording: examiners accept “zero” or “constant” potential energy; the key point is it does not change and so does not contribute to changes in $U$ .
Make clear you are referring to intermolecular potential energy.

Techniques used

use a kinetic theory assumption to infer energy contributionsdeduce what happens to intermolecular potential energy in the model

(b)

A sample of $0.26\ \text{m}^3$ of an ideal gas is at pressure $2.0 \times 10^5\ \text{Pa}$ and temperature $290\ \text{K}$ .

Determine:

6M

(i)

the number $N$ of molecules of the gas

$N$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using

pV = NkT

N = \frac{pV}{kT} = \frac{(2.0 \times 10^5)(0.26)}{(1.38 \times 10^{-23})(290)}

N = 1.3 \times 10^{25}

Answer

$N = 1.3 \times 10^{25}$ molecules

Final answer

1.3 × 10^25

Detailed explanation

Background Concept

For an ideal gas, the equation of state can be written in molecular form:

pV = NkT

where:

$p$ is pressure (Pa),
$V$ is volume (m $^3$ ),
$N$ is number of molecules,
$k$ is Boltzmann constant ( $1.38 \times 10^{-23}\ \text{J K}^{-1}$ ),
$T$ is absolute temperature (K).

This is the most direct way to find $N$ when $p, V, T$ are given.

Understanding the Question

Given: $V = 0.26\ \text{m}^3$ , $p = 2.0 \times 10^5\ \text{Pa}$ , $T = 290\ \text{K}$ .

You must determine the number of molecules $N$ in the sample. This is a 2-mark calculation: method + answer.

Approach

Rearrange $pV = NkT$ to make $N$ the subject:

N = \frac{pV}{kT}

Then substitute values in SI units (they already are).

Step-by-Step Reasoning

Start from the molecular ideal-gas equation:

pV = NkT

Rearrange for $N$ :

N = \frac{pV}{kT}

Substitute:

$pV = (2.0 \times 10^5)(0.26) = 5.2 \times 10^4$
$kT = (1.38 \times 10^{-23})(290) \approx 4.00 \times 10^{-21}$

Divide:

N = \frac{5.2 \times 10^4}{4.00 \times 10^{-21}} \approx 1.3 \times 10^{25}

Key Takeaways

Use $pV = NkT$ when the question asks for number of molecules.
Always keep units in SI so powers of ten come out correctly.

Common Mistakes

Using $R$ instead of $k$ without converting to moles.
Forgetting the $\times 10^{-23}$ in $k$ .
Giving $N$ to an unrealistic number of significant figures (data are mostly 2 s.f.).

Things to Be Careful About

$N$ is dimensionless (a count), so no unit is written.
Use standard form for such a large number.

Techniques used

apply the ideal gas equation in molecular formrearrange an equation to make the required quantity the subjectsubstitute values with consistent SI units

(ii)

the average translational kinetic energy $E_K$ of one molecule of the gas

$E_K$ = ______ $\text{J}$

2M Similar

DifficultyEasy

Worked solution

Working

Average translational kinetic energy per molecule:

E_K = \frac{3}{2}kT = \frac{3}{2}(1.38 \times 10^{-23})(290)

E_K = 6.0 \times 10^{-21}\ \text{J}

Answer

$E_K = 6.0 \times 10^{-21}\ \text{J}$

Final answer

6.0 × 10^-21 J

Detailed explanation

Background Concept

Kinetic theory links temperature to the average translational kinetic energy of a molecule:

\langle E_K \rangle = \frac{3}{2}kT

This is an average over many molecules in random motion. It depends only on absolute temperature $T$ .

Understanding the Question

You are given $T = 290\ \text{K}$ and asked for the average translational kinetic energy of one molecule. So you should use $\frac{3}{2}kT$ directly.

Approach

Insert $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ and $T = 290\ \text{K}$ into:

E_K = \frac{3}{2}kT

Step-by-Step Reasoning

Write the formula:

E_K = \frac{3}{2}kT

Substitute values:

E_K = 1.5 \times (1.38 \times 10^{-23}) \times 290

Calculate:

$(1.38 \times 290) \approx 4.00 \times 10^{-21}$
multiply by $1.5$ gives $\approx 6.00 \times 10^{-21}$

So

E_K \approx 6.0 \times 10^{-21}\ \text{J}

Key Takeaways

Average translational kinetic energy per molecule is proportional to $T$ .
Use kelvin; do not use Celsius.

Common Mistakes

Using $E_K = \frac{1}{2}mv^2$ without having $v$ .
Forgetting the factor $\frac{3}{2}$ and using $kT$ .
Using $T=290^\circ\text{C}$ (temperature must be in kelvin).

Things to Be Careful About

Quote the answer to appropriate significant figures (2 s.f. is consistent here).
Ensure the unit is joule (J).

Techniques used

use the average translational kinetic energy relationsubstitute temperature in kelvin into a proportionality

(iii)

the internal energy of the gas. Explain your reasoning.

internal energy = ______ $\text{J}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For an ideal gas, intermolecular potential energy is negligible, so internal energy is the total random kinetic energy:

U = NE_K

Using $N = 1.3 \times 10^{25}$ and $E_K = 6.0 \times 10^{-21}\ \text{J}$ ,

U = (1.3 \times 10^{25})(6.0 \times 10^{-21}) = 7.8 \times 10^4\ \text{J}

Answer

internal energy $= 7.8 \times 10^4\ \text{J}$

Final answer

7.8 × 10^4 J

Detailed explanation

Background Concept

Internal energy $U$ is the microscopic energy stored in a system. For a gas this is usually thought of as:

random kinetic energy of molecules, and
intermolecular potential energy.

For an ideal gas, there are no intermolecular forces (except during collisions), so intermolecular potential energy is taken as zero/constant. Therefore the internal energy is just the total random kinetic energy.

In the simplest ideal-gas model (treated as monatomic), the random kinetic energy is purely translational, giving:

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

Using $NkT = pV$ , an equivalent useful form is:

U = \frac{3}{2}pV

Understanding the Question

You have already found:

$N$ (number of molecules),
$E_K$ (average translational kinetic energy per molecule).

Now you must find the internal energy of the whole sample and explain why the method is valid for an ideal gas.

Approach

State the key idea: for an ideal gas, internal energy equals total kinetic energy because potential energy is negligible/constant.

Then calculate either:

$U = NE_K$ , using your answers from (i) and (ii), or
$U = \frac{3}{2}pV$ directly.

Step-by-Step Reasoning

For an ideal gas, no intermolecular forces means no changing potential energy term, so total internal energy is kinetic:

U = NE_K

Use previous results:

N = 1.3 \times 10^{25}, \quad E_K = 6.0 \times 10^{-21}\ \text{J}

Multiply:

U = (1.3 \times 10^{25})(6.0 \times 10^{-21})

Combine powers of ten: $10^{25} \times 10^{-21} = 10^4$ .

Multiply the numbers: $1.3 \times 6.0 = 7.8$ .

So:

U = 7.8 \times 10^4\ \text{J}

(Consistency check: since $pV = 5.2 \times 10^4\ \text{J}$ , $\frac{3}{2}pV = 7.8 \times 10^4\ \text{J}$ , same result.)

Key Takeaways

Ideal gas: internal energy is due only to random kinetic energy (no potential energy contribution).
Total internal energy is number of molecules times average energy per molecule.
Useful shortcut: $U = \frac{3}{2}pV$ for a monatomic ideal gas.

Common Mistakes

Adding a potential energy term (contradicts ideal-gas assumption).
Using $U = pV$ instead of $\frac{3}{2}pV$ .
Using $U = \frac{3}{2}nRT$ but taking $n$ as $N$ (mixing moles and molecules).

Things to Be Careful About

Explain the reasoning: “no intermolecular forces ⇒ no potential energy change ⇒ internal energy is kinetic.”
Significant figures: inputs are mostly 2 s.f., so $U$ should be 2 s.f.
$pV$ has units of joule because $\text{Pa} \cdot \text{m}^3 = \text{N m} = \text{J}$ .

Techniques used

relate internal energy to total molecular kinetic energy for an ideal gasuse U = N E_K or U = 3/2 pVuse a previously calculated quantity to find a total

(c)

The volume $V$ of the gas in (b) is now varied, keeping its pressure constant.

On Fig. 3.1, sketch the variation with $V$ of the internal energy $U$ of the gas.

Fig. 3.1

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

At constant $p$ , from $pV = NkT$ , $T \propto V$ . For an ideal gas $U = \frac{3}{2}NkT \propto T$ , so $U \propto V$ .

diagram-3c

Final answer

U increases linearly with V (straight line through origin).

Detailed explanation

Background Concept

For an ideal gas (in the simple kinetic theory model):

U = \frac{3}{2}NkT

So for a fixed amount of gas ( $N$ constant), internal energy is proportional to absolute temperature:

U \propto T

The ideal gas equation is:

pV = NkT

If the pressure $p$ is held constant and $N$ is fixed, then:

T \propto V

Combine these proportionalities to relate $U$ to $V$ .

Understanding the Question

You vary the volume $V$ while keeping pressure constant, and you must sketch how internal energy $U$ varies with $V$ .

So you need a qualitative graph shape, not numerical points.

Approach

Use $pV = NkT$ with $p$ constant to connect $V$ to $T$ .
Use $U \propto T$ for an ideal gas to connect $U$ to $V$ .
Sketch the resulting relationship.

Step-by-Step Reasoning

Start from the ideal gas equation:

pV = NkT

With $p$ constant and $N$ constant, rearrange:

T = \frac{pV}{Nk}

So $T$ is directly proportional to $V$ .

Internal energy for an ideal gas is:

U = \frac{3}{2}NkT

With $N$ constant, $U \propto T$ .

Since $T \propto V$ , it follows that $U \propto V$ . Therefore the $U$ vs $V$ graph is a straight line with positive gradient.

diagram-3c

Key Takeaways

For an ideal gas, $U$ depends only on $T$ (for fixed amount of gas).
At constant pressure, $T \propto V$ (Charles’ law form).
Combining proportionalities lets you sketch the correct graph quickly.

Common Mistakes

Drawing $U$ as constant with $V$ (that would be true only if $T$ were constant, i.e. isothermal, not constant pressure).
Drawing an inverse curve like $1/V$ (confusing with Boyle’s law, which applies when $T$ is constant).
Forgetting that changing $V$ at constant $p$ implies the temperature changes.

Things to Be Careful About

The axes start at 0, so the expected sketch is typically a straight line through the origin (even though $V=0$ is not physically achievable for a gas).
You are not asked for a numerical gradient; just show the correct trend and shape (linear increase).

Techniques used

use the ideal gas law to relate temperature and volume at constant pressurelink internal energy to temperature for an ideal gassketch a qualitative graph from proportional reasoning

Q4Medium-EasyOscillations

(a)

State what is meant by resonance.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Resonance occurs when a system is driven at a frequency equal to its natural frequency, so that energy transfer is maximum and the amplitude of oscillation becomes a maximum (very large).

Final answer

Resonance occurs when a system is driven at a frequency equal to its natural frequency, giving maximum energy transfer and hence maximum (very large) amplitude.

Detailed explanation

Background Concept

Resonance is a feature of forced oscillations, where an external periodic driving force makes a system oscillate. Every oscillating system has a natural frequency (the frequency it would oscillate at if disturbed and then left to itself). The response (amplitude) depends on the driving frequency.

At a particular driving frequency, the system absorbs energy most effectively from the driver. This happens when the driving frequency matches the natural frequency, producing a peak in amplitude (limited in practice by damping).

Understanding the Question

You are asked to state what resonance means (a definition). For 2 marks, you typically need:

the frequency condition (driving frequency equals natural frequency), and
the consequence (maximum energy transfer and hence maximum amplitude).

Approach

Write a concise definition that includes both the condition and the outcome. Avoid vague phrases like “it vibrates more” without specifying the frequency condition.

Step-by-Step Reasoning

Identify that resonance is about a driven (forced) oscillator.
State the key condition:

\text{driving frequency} = \text{natural frequency}

State the effect: energy transfer per cycle is greatest, so the amplitude becomes maximum (very large, limited by damping).

Key Takeaways

Resonance is a forced oscillation phenomenon.
It occurs at driving frequency = natural frequency.
It produces maximum amplitude because energy transfer is maximum.

Common Mistakes

Defining resonance without mentioning a driving force.
Saying “frequency increases” rather than “amplitude increases”.
Missing the idea of maximum energy transfer / maximum amplitude.

Things to Be Careful About

In real systems, amplitude does not become infinite because damping limits the peak.
Use the term natural frequency (not just “frequency”).

Techniques used

state the condition for resonance in terms of driving and natural frequencylink resonance to maximum energy transfer and maximum amplitude

(b)

A small ball is held in place using a stretched string. One end of the string is fixed to a wall and the other end is attached to a vibration generator, as shown in Fig. 4.1.

Fig. 4.1

Initially, the vibration generator is switched off.

A student displaces the ball vertically and then releases it. Fig. 4.2 shows the variation of the displacement of the ball with time after it is released.

Fig. 4.2

4M

(i)

State the name of the phenomenon illustrated by the decrease in the amplitude of the oscillations in Fig. 4.2.

1M Similar

(ii)

Explain the decrease with time of the amplitude of the oscillations of the ball.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Resistive forces (e.g. air resistance / friction in the string) do work on the ball so energy is transferred to the surroundings (thermal / sound). Hence the total energy of the oscillation decreases and the amplitude falls with time.

Final answer

Resistive forces cause energy loss to the surroundings (e.g. as heat/sound), so the oscillator’s energy decreases and the amplitude falls.

Detailed explanation

Background Concept

For an oscillator, the amplitude is related to its total mechanical energy. For SHM (and approximately for small oscillations), the total energy is proportional to the square of the amplitude:

E \propto A^2

If energy is removed from the system each cycle by resistive forces, the mechanical energy decreases, so the amplitude must decrease.

Understanding the Question

The ball oscillates after being released, and the graph shows that the peaks get smaller. The question asks you to explain why this happens, so you must describe the mechanism of energy loss.

Approach

State that:

there are resistive forces acting (air resistance, internal friction, friction at supports),
these forces do work against the motion and transfer energy out of the oscillation,
therefore mechanical energy and amplitude decrease with time.

Step-by-Step Reasoning

As the ball moves, it experiences air resistance and there may be internal friction/losses in the string and supports.
These resistive forces act opposite to the direction of motion, so they do negative work on the oscillating system.
The energy removed appears as thermal energy in the surroundings (and possibly sound).
With less mechanical energy available, the maximum displacement from equilibrium becomes smaller, so the amplitude decreases with time.

Key Takeaways

Damping is caused by energy loss due to resistive forces.
Less energy in the oscillator means smaller amplitude.

Common Mistakes

Saying only “friction” without linking it to energy transfer.
Saying “amplitude decreases because it slows down” (frequency and amplitude are different ideas).

Things to Be Careful About

The key mark-winning phrasing is that resistive forces do work and energy is transferred to the surroundings, causing amplitude to fall.

Techniques used

link amplitude decrease to energy loss from resistive forcesidentify forms of energy transfer to the surroundings

(iii)

Determine the frequency of the oscillations of the ball.

frequency = ______ $\text{Hz}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 4.2, the time between successive identical points (e.g. trough to trough) is about

T \approx 0.20\ \text{s}

So

f = \frac{1}{T} = \frac{1}{0.20} = 5.0\ \text{Hz}

Answer

$5.0\ \text{Hz}$

Final answer

5.0 Hz

Detailed explanation

Background Concept

The period $T$ is the time for one complete oscillation. The frequency $f$ is the number of oscillations per second:

f = \frac{1}{T}

On a displacement–time graph, one period can be found by measuring the time between two successive points in the motion that are in the same state (e.g. peak to peak, trough to trough, or identical zero-crossings with the same direction of motion).

Understanding the Question

You are given a displacement–time graph of the ball after it is released. You must determine the frequency of the oscillations using the time axis.

Approach

Choose two clear, repeated points (e.g. troughs).
Read the time difference to get $T$ (use multiple cycles if possible for accuracy).
Use $f = 1/T$ .

Step-by-Step Reasoning

Identify two successive troughs (or peaks) on the graph.
Read their times from the horizontal axis. The separation is about $0.20\ \text{s}$ , so:

T \approx 0.20\ \text{s}

(Alternatively, if you can see 3 cycles in $0.60\ \text{s}$ , then $T = 0.60/3 = 0.20\ \text{s}$ .)

Convert period to frequency:

f = \frac{1}{T} = \frac{1}{0.20\ \text{s}} = 5.0\ \text{Hz}

Key Takeaways

Read a period from repeating features on the graph.
Use $f = 1/T$ .
Using several cycles reduces percentage reading error.

Common Mistakes

Measuring half a period (peak to trough) and using it as $T$ .
Counting squares incorrectly on the time axis.
Giving $T$ as the answer instead of $f$ .

Things to Be Careful About

Use two equivalent points (e.g. trough-to-trough), not just any two points.
Make sure the final unit is $\text{Hz}$ (not seconds).

Techniques used

read the period from an oscillation-time graphuse $f = 1/T$

(c)

The vibration generator in (b) is switched on and its frequency $f$ of vibration is gradually increased from 0 to $10\ \text{Hz}$ .

On Fig. 4.3, sketch the variation with $f$ of the amplitude of the oscillations of the ball.

Fig. 4.3

2M Similar

DifficultyMedium-Easy

Worked solution

diagram-c-resonance-curve

Answer

A resonance curve with small amplitude at low and high $f$ , and a maximum amplitude at $f \approx 5\ \text{Hz}$ .

Final answer

Resonance curve with a peak at f ≈ 5 Hz.

Detailed explanation

Background Concept

In forced oscillations, a periodic driving force of frequency $f$ makes a system oscillate. The steady-state amplitude depends on how close $f$ is to the system’s natural frequency $f_0$ .

When $f \ll f_0$ or $f \gg f_0$ , the energy transfer from the driver is relatively inefficient, so the amplitude is small.
When $f \approx f_0$ , energy is transferred most efficiently each cycle, so the amplitude reaches a maximum: resonance.
Damping prevents the peak from becoming infinite and broadens/lowers the resonance peak.

Understanding the Question

The vibration generator drives the ball-string system while its driving frequency is increased from $0$ to $10\ \text{Hz}$ . You must sketch how the amplitude of the ball’s oscillations varies with $f$ .

From part (b)(iii), the system’s natural frequency is about $5\ \text{Hz}$ , so the maximum amplitude should occur around that frequency.

Approach

Sketch the standard amplitude–frequency response:

Start near zero amplitude at $f = 0$ .
Rise to a peak at $f \approx f_0$ .
Fall again as $f$ increases beyond $f_0$ .

Step-by-Step Reasoning

Use the result $f_0 \approx 5\ \text{Hz}$ .
Place the highest point of the curve above $f = 5\ \text{Hz}$ .
Ensure the curve is smooth with lower amplitude on both sides of the peak (a “bell-shaped” resonance curve).

diagram-c-resonance-curve

Key Takeaways

Resonance response is a peak in amplitude at the natural frequency.
Damping limits the maximum amplitude and sets the width of the peak.

Common Mistakes

Drawing amplitude increasing steadily with frequency (no peak).
Putting the peak at an arbitrary frequency not linked to the natural frequency from (b)(iii).
Drawing a sharp spike with vertical sides rather than a smooth curve.

Things to Be Careful About

Axes: horizontal is $f/\text{Hz}$ , vertical is amplitude (no unit given).
Peak should be around $5\ \text{Hz}$ (from the earlier graph-based frequency).

Techniques used

sketch an amplitude-frequency resonance curveidentify the resonant frequency from the natural frequency

Q5Medium-EasyElectric FieldsMagnetic Fields

(a)

Define electric field.

2M Similar

DifficultyEasy

Worked solution

Answer

Electric field strength at a point is the force per unit positive charge at that point:

E = \frac{F}{q}

Final answer

Force per unit positive charge at a point (E = F/q).

Detailed explanation

Background Concept

An electric field is a way of describing how charges exert forces on other charges at a distance. At any point in space, the electric field strength $E$ is defined so that it tells you the force $F$ that would act on a charge $q$ placed at that point:

F = qE

This is a vector relationship: the direction of $E$ is defined as the direction of the force on a positive test charge.

Understanding the Question

The question asks for the definition of electric field (meaning electric field strength). For full credit you must mention:

force per unit charge
the charge is positive (test charge)
optionally, the formula $E = F/q$ .

Approach

Recall the standard definition and present it in words and/or as an equation.

Step-by-Step Reasoning

Take a small positive test charge $q$ placed at a point in the field.
If the field exerts a force $F$ on it, then define the field strength as force per unit charge:

E = \frac{F}{q}

Mention that the direction of $E$ is the direction of the force on a positive charge (this is often implied but helps clarity).

Key Takeaways

Electric field strength is defined by what force it produces per coulomb on a positive test charge.
The defining equations are $E = F/q$ and $F = qE$ .

Common Mistakes

Defining it using a negative charge (direction would be opposite).
Saying “force per unit mass” (that is gravitational field strength).
Missing the idea of “at a point” (field is defined point-by-point).

Things to Be Careful About

State “positive test charge” to fix the direction convention.
Use correct units: $E$ has units $\text{N C}^{-1}$ (equivalently $\text{V m}^{-1}$ ).

Techniques used

state the definition in terms of force per unit positive test chargeexpress the definition using the relation $E = F/q$

(b)

Fig. 5.1 shows two parallel conducting plates that are in a vacuum. The plates are separated by a distance of $6.7\ \text{cm}$ and have a potential difference (p.d.) of $430\ \text{V}$ between them.

Fig. 5.1

6M

(i)

On Fig. 5.1, draw four field lines to represent the electric field between the plates.

2M Similar

DifficultyEasy

Worked solution

diagram-b-i

Answer

Four straight, parallel, equally spaced field lines between the plates, with arrows from the +430 V plate to the 0 V plate.

Final answer

Four straight, parallel, equally spaced lines with arrows from +430 V plate to 0 V plate.

Detailed explanation

Background Concept

Electric field lines are a visual way to show the direction and nature of an electric field:

The direction of the electric field at a point is the direction a positive test charge would accelerate.
Field lines go from positive to negative.
Between two large, parallel plates, the field in the central region is approximately uniform, so field lines are straight, parallel and equally spaced.

Understanding the Question

You are given two parallel conducting plates: the top at $+430\ \text{V}$ and the bottom at $0\ \text{V}$ . You must draw four electric field lines representing the field between them.

Approach

Use the fact that electric field lines start on the positive plate and end on the negative (lower potential) plate. Because it is a uniform field region, draw straight, parallel, evenly spaced lines.

Step-by-Step Reasoning

Identify polarity: top plate is at higher potential ( $+430\ \text{V}$ ), bottom plate at lower potential ( $0\ \text{V}$ ).
Electric field direction is from higher potential/positive plate to lower potential/negative plate, so downwards.
Draw four straight, parallel lines in the gap, spaced evenly.
Put arrowheads on each line pointing downwards.

diagram-b-i

Key Takeaways

Field direction is defined by force on a positive charge.
A uniform field is shown by straight, parallel, equally spaced lines.

Common Mistakes

Drawing lines with arrows in the wrong direction (from 0 V to +430 V).
Drawing curved lines in the central region (suggests non-uniform field).
Leaving out arrowheads (direction is required).

Things to Be Careful About

Keep the lines inside the plate region (edge effects are usually ignored unless shown).
Ensure the lines do not cross and are approximately equally spaced.

Techniques used

identify the direction of the electric field from the potential difference polarityrepresent a uniform electric field using straight, parallel, equally spaced field linesadd arrows showing field direction from positive to negative potential

(ii)

Determine the strength $E$ of the electric field between the plates.

$E$ = ______ $\text{N C}^{-1}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Plate separation

d = 6.7\ \text{cm} = 0.067\ \text{m}

Uniform field:

E = \frac{V}{d} = \frac{430}{0.067} = 6.4 \times 10^{3}\ \text{N C}^{-1}

Answer

$E = 6.4 \times 10^{3}\ \text{N C}^{-1}$

Final answer

6.4 × 10^3 N C^-1

Detailed explanation

Background Concept

Between large parallel plates, the electric field is approximately uniform. For a uniform field, the field strength $E$ is related to potential difference $V$ and plate separation $d$ by:

E = \frac{V}{d}

This is consistent with the unit equivalence $1\ \text{N C}^{-1} = 1\ \text{V m}^{-1}$ .

Understanding the Question

You are told:

potential difference between plates $V = 430\ \text{V}$
separation $d = 6.7\ \text{cm}$

You must find the uniform electric field strength $E$ in $\text{N C}^{-1}$ .

Approach

Convert $d$ into SI units (metres), then substitute into $E = V/d$ .

Step-by-Step Reasoning

Convert the distance:

d = 6.7\ \text{cm} = 6.7 \times 10^{-2}\ \text{m} = 0.067\ \text{m}

Use the uniform field formula:

E = \frac{V}{d}

Substitute values:

E = \frac{430}{0.067} = 6417.9...\ \text{V m}^{-1}

Round to appropriate significant figures (limited by $6.7$ and $430$ ):

E \approx 6.4 \times 10^{3}\ \text{N C}^{-1}

Key Takeaways

For parallel plates: $E = V/d$ .
Always convert centimetres to metres before substituting.

Common Mistakes

Using $d = 6.7$ instead of $0.067$ (gives answer $100$ times too small).
Giving units as $\text{N m}^{-1}$ or leaving off units.
Excessive significant figures not justified by data.

Things to Be Careful About

Keep SI units throughout.
$E$ can be written as $\text{V m}^{-1}$ or $\text{N C}^{-1}$ ; the question requests $\text{N C}^{-1}$ .

Techniques used

convert the plate separation into metresapply the uniform-field relation $E = V/d$quote the answer with appropriate significant figures and units

(iii)

An electron travels at a speed of $2.6 \times 10^7\ \text{m s}^{-1}$ towards the region between the plates, as shown in Fig. 5.1.

On Fig. 5.1, draw the path of the electron as it moves between and beyond the plates.

2M Similar

DifficultyMedium-Easy

Worked solution

diagram-b-iii

Answer

Electron curves towards the +430 V plate (upwards) between the plates, then continues beyond the plates in a straight line along the tangent to the exit direction.

Final answer

Curves upward toward the + plate between plates; then straight-line tangent after leaving.

Detailed explanation

Background Concept

A charge in an electric field experiences a force

F = qE

For a positive charge, the force is in the same direction as $E$ .
For a negative charge (electron), the force is in the opposite direction to $E$ .

Between parallel plates, $E$ is uniform, so the force is constant in magnitude and direction. If the electron enters with horizontal velocity, it has:

constant horizontal velocity (no horizontal force)
constant vertical acceleration (due to constant vertical force)

This produces a parabolic path within the plates, like projectile motion.

Understanding the Question

An electron enters the region between the plates moving horizontally from left to right. You must sketch its path while it is between the plates and after it leaves.

The diagram shows the top plate at $+430\ \text{V}$ and the bottom plate at $0\ \text{V}$ , so the electric field points downward.

Approach

Decide direction of $E$ (from + to 0 V).
Use $F = qE$ with $q$ negative to get force direction on the electron.
Sketch: curve in the field region, then straight line after leaving (no force outside).

Step-by-Step Reasoning

Electric field direction: from the + plate to the 0 V plate, so downward.
Electron has charge $q=-e$ $q = - e$ , so force direction is opposite to $E$ $E$ :
- force is upward, towards the positive plate.
While between plates:
- horizontal motion continues,
- electron accelerates upward,
- path bends upward, forming a curve (parabola).
Once the electron leaves the plate region:
- the electric field is (assumed) zero,
- no force acts,
- it continues in a straight line in the direction it had at the exit (tangent).

diagram-b-iii

Key Takeaways

Field direction is from positive to negative; electron force is opposite to field.
Uniform electric field gives constant acceleration perpendicular to initial motion, producing a curved path.
Outside the field region, motion is straight at constant velocity.

Common Mistakes

Curving the electron downwards (forgetting electron is negative).
Drawing the electron continuing to curve after leaving the plates (no field assumed outside).
Drawing the path as a sharp kink rather than smooth curve-to-tangent.

Things to Be Careful About

The electron is attracted to the positive plate.
The path inside should be smooth, not piecewise straight segments.

Techniques used

determine the force direction using $F = qE$ and the sign of chargededuce the resulting trajectory of a charged particle in a uniform electric fieldsketch the path as a curve inside the field and a straight line outside

(c)

A uniform magnetic field is now applied in the region of the electric field in Fig. 5.1, so that the electron in (b)(iii) travels undeviated through the region.

5M

(i)

Determine the direction of the uniform magnetic field.

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Electric field is downward, so force on electron (negative) is upward. Magnetic force must therefore be downward.

For an electron,

\vec{F}_B = q(\vec{v} \times \vec{B})

With $\vec{v}$ to the right, $\vec{B}$ must be into the page so that $\vec{F}_B$ is downward.

Answer

Magnetic field is into the page.

Final answer

Into the page.

Detailed explanation

Background Concept

A moving charge in a magnetic field experiences a force

\vec{F}_B = q(\vec{v} \times \vec{B})

The direction is given by $\vec{v} \times \vec{B}$ for a positive charge.
For an electron ( $q=-e$ ), the force direction is opposite to $\vec{v} \times \vec{B}$ .

In an arrangement where electric and magnetic forces cancel (a velocity selector), the particle travels undeviated.

Understanding the Question

The electron travels to the right between plates. The electric field is downward (from +430 V to 0 V), so the electric force on the electron is upward. A magnetic field is applied so that the electron is undeviated, meaning the magnetic force must be downward.

Approach

Determine direction of electric force on the electron.
Require magnetic force to be equal and opposite.
Use $\vec{v} \times \vec{B}$ and then reverse for the electron’s negative charge.

Step-by-Step Reasoning

Electric field is downward; electron has negative charge, so electric force is upward.
For no deflection, magnetic force must be downward.
Choose $\vec{B}$ so that $\vec{v} \times \vec{B}$ is upward (because the electron reverses the direction, making $\vec{F}_B$ downward).
With $\vec{v}$ to the right, $\vec{B}$ into the page gives $\vec{v} \times \vec{B}$ upward for a positive charge, hence downward for an electron.

diagram-c-i

Key Takeaways

Use cancellation: $\vec{F}_E + \vec{F}_B = 0$ for undeviated motion.
Remember an electron reverses the direction from the right-hand rule for $\vec{v} \times \vec{B}$ .

Common Mistakes

Forgetting the sign of charge and giving the opposite field direction.
Stating “out of the page” when the required force direction is downward.

Things to Be Careful About

Clearly relate directions: $E$ direction, then $F_E$ direction (opposite for electron), then choose $B$ to oppose $F_E$ .
If you use Fleming’s left-hand rule, be consistent about using conventional current vs electron motion.

Techniques used

use the requirement of zero deflection to set the magnetic force opposite to the electric forceapply the right-hand rule to the cross product $\vec{v} \times \vec{B}$account for the negative sign of electron charge when determining force direction

(ii)

Explain, with reference to the forces exerted by the two fields on the electron, why the path of the electron is undeviated.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Electric force on electron:

F_E = eE

Magnetic force on electron:

F_B = evB

The magnetic force is opposite in direction to the electric force and equal in magnitude, so resultant force is zero and the electron is undeviated.

Final answer

Magnetic force equals and opposes electric force, so resultant force is zero and the electron is undeviated.

Detailed explanation

Background Concept

Two different fields can exert forces on a charged particle:

Electric field:

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

Magnetic field (when the particle moves with velocity $\vec{v}$ ):

\vec{F}_B = q(\vec{v} \times \vec{B})

If the net force is zero, then the particle has no acceleration, so its velocity (both magnitude and direction) stays constant and its path is a straight line.

Understanding the Question

A magnetic field is applied so that the electron passes through the region of electric field without any deviation. You must explain this by referring to the forces from the electric and magnetic fields.

Approach

State the two forces and explain that they are in opposite directions with equal magnitudes, giving zero resultant force.

Step-by-Step Reasoning

The electric field exerts a force on the electron. Since the electron has charge $-e$ , the force is opposite to $\vec{E}$ , with magnitude $F_E = eE$ .
The magnetic field exerts a force because the electron is moving. With $\vec{B}$ chosen appropriately, the magnetic force is opposite in direction to the electric force, with magnitude $F_B = evB$ (here $\sin\theta = 1$ because $\vec{v}$ is perpendicular to $\vec{B}$ ).
For the electron to be undeviated, these forces must balance:

\vec{F}_E + \vec{F}_B = 0

Zero resultant force means zero acceleration, so the electron continues in a straight line at constant velocity through the region.

Key Takeaways

Undeviated motion requires zero resultant force.
Electric and magnetic forces can be arranged to cancel (velocity selector principle).

Common Mistakes

Saying “the fields cancel” (fields do not cancel; forces cancel).
Forgetting that the electron’s electric force is opposite the field direction.
Missing the requirement that forces are equal in magnitude as well as opposite in direction.

Things to Be Careful About

Be explicit: mention both forces and their directions.
If using $F_B = qvB\sin\theta$ , ensure $\theta = 90^\circ$ is understood here (velocity perpendicular to magnetic field).

Techniques used

state the electric force on a charge in an electric fieldstate the magnetic force on a moving charge in a magnetic fieldargue that equal and opposite forces give zero resultant force and hence no deflection

(iii)

Determine the flux density $B$ of the uniform magnetic field. Give a unit with your answer.

$B$ = ______ unit ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For undeviated motion:

F_E = F_B

eE = evB \Rightarrow B = \frac{E}{v}

B = \frac{6.4 \times 10^{3}}{2.6 \times 10^{7}} = 2.5 \times 10^{-4}\ \text{T}

Answer

$B = 2.5 \times 10^{-4}\ \text{T}$

Final answer

2.5 × 10^-4 T

Detailed explanation

Background Concept

If a charged particle moves through crossed electric and magnetic fields and is undeviated, the net force must be zero. Using magnitudes (with directions already chosen to oppose):

F_E = F_B

with

F_E = eE, \qquad F_B = evB \sin\theta

Here $\theta = 90^\circ$ so $\sin\theta = 1$ .

Understanding the Question

You already have the electron speed $v = 2.6 \times 10^{7}\ \text{m s}^{-1}$ and (from part (b)(ii)) the electric field strength $E$ . You must find the magnetic flux density $B$ that makes the electron travel straight through.

Approach

Balance the forces: $eE = evB$ , cancel $e$ , then calculate $B = E/v$ and give the unit tesla.

Step-by-Step Reasoning

For no deflection, magnitudes balance:

eE = evB

Cancel the common factor $e$ :

E = vB

Rearrange:

B = \frac{E}{v}

Substitute values (using $E = 6.4 \times 10^{3}\ \text{N C}^{-1}$ ):

B = \frac{6.4 \times 10^{3}}{2.6 \times 10^{7}} = 2.46... \times 10^{-4}\ \text{T}

Round suitably:

B \approx 2.5 \times 10^{-4}\ \text{T}

Key Takeaways

Undeviated path in crossed fields implies $qE = qvB$ (for perpendicular geometry).
The required magnetic field is proportional to $E$ and inversely proportional to $v$ .

Common Mistakes

Forgetting $\sin\theta$ and using an incorrect angle.
Not cancelling the charge $e$ (it cancels because both forces are proportional to charge magnitude).
Omitting the unit; $B$ must be in tesla (T).

Things to Be Careful About

Use the correct $E$ from (b)(ii), with metres not centimetres.
Keep powers of ten consistent when dividing in standard form.

Techniques used

set electric and magnetic forces equal for no deflectionrearrange to obtain $B = E/v$substitute values and quote a correct unit

Q6MediumCapacitance

Fig. 6.1 shows a capacitor of capacitance $C$ connected in series with a resistor of resistance $R$ .

Fig. 6.1

Initially the switch is open and there is a p.d. of $12\ \text{V}$ across the capacitor.

At time $t = 0$ , the switch is closed so that there is a current $I$ in the resistor.

Fig. 6.2 shows the variation of $I$ with $t$ .

Fig. 6.2

(a)

Explain the shape of the line in Fig. 6.2.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

When the switch is closed the capacitor discharges through $R$ , so the p.d. across $R$ equals the capacitor p.d.

Initially the capacitor p.d. is maximum ( $12\ \text{V}$ ), so the current is maximum.

As charge leaves the capacitor, its p.d. falls, so $I = V_R/R$ falls; the rate of fall decreases with time, giving an exponential decay towards zero current.

Final answer

Exponential decay of current as the capacitor discharges and its p.d. falls.

Detailed explanation

Background Concept

For a capacitor discharging through a resistor (an $RC$ circuit), the capacitor’s charge $Q$ and p.d. $V_C$ decrease with time. The key relations are

Q = CV_C

and, for the resistor,

I = \frac{V_R}{R}.

In a simple series discharge loop, the resistor p.d. is the same as the capacitor p.d. in magnitude (no supply present), so $V_R = V_C$ . The discharge obeys the exponential law

I = I_0 e^{-t/RC}.

The time constant is $\tau = RC$ , and it sets how quickly the current falls.

Understanding the Question

The capacitor is initially charged to $12\ \text{V}$ and then, at $t=0$ , the switch is closed so the capacitor discharges through the resistor. The graph shows current $I$ (through the resistor) against time. You are asked to explain why the graph is curved and why it approaches zero.

Approach

Connect the circuit behaviour to the graph:

Identify what sets the initial current at $t=0$ .
Explain why the current decreases as time increases.
Explain why the curve is not a straight line (why the decrease slows down).

Step-by-Step Reasoning

At $t=0$ , the capacitor has its maximum p.d. ( $12\ \text{V}$ ). With the switch closed, that p.d. appears across the resistor, so the current is initially

I_0 = \frac{12}{R}.

This explains why the graph starts at a non-zero maximum value.

As current flows, charge leaves the capacitor plates, so $Q$ decreases. Since $V_C = Q/C$ , the capacitor p.d. must also decrease.
Because $V_R = V_C$ during discharge, the driving p.d. across the resistor decreases, so by Ohm’s law $I = V_R/R$ also decreases.
The crucial shape point: the current at any instant is proportional to the p.d. at that instant, and that p.d. is itself falling. When $V_C$ is large (early times), the current is larger so charge is removed quickly; later, $V_C$ is smaller so the current is smaller and charge is removed more slowly. This produces a decay that gets less steep with time (an exponential), approaching zero asymptotically.

Key Takeaways

In an $RC$ discharge, $I$ is largest at $t=0$ and then decays exponentially.
The physical reason is that the capacitor’s p.d. provides the driving p.d., and it falls as the capacitor loses charge.
The curve approaches zero because eventually there is almost no p.d. left to drive a current.

Common Mistakes

Saying the current decreases “because resistance increases” (the resistance is constant).
Describing the graph as “linear decrease” rather than exponential.
Forgetting to link the shape to the decreasing capacitor p.d. (or charge).

Things to Be Careful About

Distinguish between the value of the current and the rate of change of the current: the rate of decrease is largest at the start and becomes smaller with time.
The current approaches zero but does not reach exactly zero on an ideal exponential curve; it tends to zero as $t \to \infty$ .

Techniques used

relate current to potential difference across the resistor using Ohm's lawuse the exponential discharge model for a capacitor in an RC circuitlink the decreasing p.d. across the capacitor to a decreasing current

(b)

Use Fig. 6.2 to determine:

5M

(i)

resistance $R$

$R$ = ______ $\Omega$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 6.2, initial current at $t=0$ is $I_0 \approx 0.13\ \text{mA} = 1.3 \times 10^{-4}\ \text{A}$ .

At $t=0$ , $V_R = 12\ \text{V}$ , so

R = \frac{V}{I_0} = \frac{12}{1.3 \times 10^{-4}} \approx 9.2 \times 10^{4}\ \Omega.

Answer

$R \approx 9.2 \times 10^{4}\ \Omega$

Final answer

9.2 × 10^4 Ω

Detailed explanation

Background Concept

At the moment the discharge begins ( $t=0$ ), the capacitor is still charged to its initial p.d. $V_0$ (here $12\ \text{V}$ ). In a series loop containing only the capacitor and resistor, that capacitor p.d. appears across the resistor, so the initial current is set by Ohm’s law:

I_0 = \frac{V_0}{R} \quad \Rightarrow \quad R = \frac{V_0}{I_0}.

Understanding the Question

You are given a graph of current $I$ against time $t$ for a discharging capacitor. The question asks you to find $R$ . The graph lets you find the initial current $I_0$ at $t=0$ . With $V_0 = 12\ \text{V}$ , you can then calculate $R$ .

Approach

Read $I_0$ from the vertical intercept of the graph at $t=0$ .
Convert $I_0$ from mA to A.
Use $R = V_0/I_0$ .

Step-by-Step Reasoning

From the graph, the initial current is about $0.13\ \text{mA}$ .
Convert units:

0.13\ \text{mA} = 0.13 \times 10^{-3}\ \text{A} = 1.3 \times 10^{-4}\ \text{A}.

Apply Ohm’s law at $t=0$ :

R = \frac{12\ \text{V}}{1.3 \times 10^{-4}\ \text{A}} \approx 9.2 \times 10^{4}\ \Omega.

Any small variation depends on how precisely $I_0$ is read from the graph.

Key Takeaways

Use the intercept at $t=0$ to get $I_0$ for a discharging capacitor.
At $t=0$ , $R = V_0/I_0$ because the capacitor still has its full initial p.d.

Common Mistakes

Forgetting to convert mA to A (leads to an answer $1000$ times too small).
Using a later value of $I$ instead of the initial value.
Using $V=12\ \text{V}$ with the wrong current reading (e.g. at $t=1\ \text{s}$ ).

Things to Be Careful About

Read $I_0$ carefully from the graph at $t=0$ (not from the first gridline after $t=0$ ).
Quote $R$ to a sensible number of significant figures consistent with a graph reading (usually 2 s.f.).

Techniques used

read the initial current from a current–time graphapply Ohm's law at the instant the discharge beginsconvert between milliampere and ampere

(ii)

the time constant $\tau$ of the circuit in Fig. 6.1.

$\tau$ = ______ $\text{s}$

3M Similar

DifficultyMedium

Worked solution

Working

From Fig. 6.2, $I_0 \approx 0.13\ \text{mA}$ .

At $t = \tau$ ,

I = \frac{I_0}{e} \approx \frac{0.13}{2.72} \approx 0.048\ \text{mA}.

From the graph, $I \approx 0.048\ \text{mA}$ at $t \approx 3.0\ \text{s}$ .

Answer

$\tau \approx 3.0\ \text{s}$

Final answer

3.0 s

Detailed explanation

Background Concept

For a capacitor discharging through a resistor,

I = I_0 e^{-t/RC}

where the time constant is

\tau = RC.

A very useful property of the exponential is that when $t = \tau$ ,

I = I_0 e^{-1} = \frac{I_0}{e} \approx 0.368 I_0.

So you can find $\tau$ from a current–time graph by finding the time at which the current has fallen to about $37\%$ of its initial value.

Understanding the Question

You are given the discharge current graph. You must determine the time constant $\tau$ . The graph provides $I_0$ at $t=0$ , then you locate the time when $I = I_0/e$ .

Approach

Read $I_0$ from the graph.
Calculate $I_0/e$ .
Read from the graph the time corresponding to that current value; that time is $\tau$ .

(An alternative often accepted is to draw a tangent at $t=0$ and find where it meets the time axis; that intercept equals $\tau$ .)

Step-by-Step Reasoning

Read initial current: $I_0 \approx 0.13\ \text{mA}$ .
Calculate the target current at one time constant:

I_\tau = \frac{I_0}{e} \approx \frac{0.13}{2.72} \approx 0.048\ \text{mA}.

On the graph, locate $I \approx 0.048\ \text{mA}$ (just below $0.05\ \text{mA}$ ) and read horizontally to the curve and then down to the time axis.
This corresponds to about $t \approx 3.0\ \text{s}$ , so

\tau \approx 3.0\ \text{s}.

Because the graph is read by eye, values in a small range around this (e.g. $2.8$ – $3.2\ \text{s}$ ) would be plausible depending on the exact printed curve.

Key Takeaways

$\tau$ is the time for $I$ (or $V$ or $Q$ ) to fall to $1/e$ of its initial value.
On an exponential decay graph, identifying $37\%$ of the starting value is a fast way to find $\tau$ .

Common Mistakes

Using $I_0/2$ instead of $I_0/e$ .
Reading the time when $I$ reaches $0.048\ \text{mA}$ but mixing up the axes (time is on the horizontal axis).
Using $e \approx 2.7$ incorrectly (e.g. multiplying by $e$ instead of dividing).

Things to Be Careful About

Keep units consistent: the current can stay in mA for the $I_0/e$ step since it is a ratio.
Be precise about the initial value $I_0$ you use; if you read a slightly different $I_0$ , your $I_0/e$ level changes.
When reading from the graph, avoid using the very end of the curve where it flattens close to zero (hard to read accurately).

Techniques used

use the exponential decay property that $I = I_0/e$ at $t = \tau$read coordinates from an exponential decay graphcalculate a time from a chosen current level

(c)

Use your answers in (b) to determine capacitance $C$ .

$C$ = ______ $\text{F}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $\tau = RC$ ,

C = \frac{\tau}{R} = \frac{3.0}{9.2 \times 10^{4}} \approx 3.3 \times 10^{-5}\ \text{F}.

Answer

$C \approx 3.3 \times 10^{-5}\ \text{F}$

Final answer

3.3 × 10^-5 F

Detailed explanation

Background Concept

The time constant of a series $RC$ circuit is

\tau = RC.

It has units of seconds because $\Omega \cdot \text{F} = \text{s}$ . Rearranging gives

C = \frac{\tau}{R}.

Understanding the Question

You have already found $R$ and $\tau$ from the graph. This part asks for the capacitance $C$ by combining those results using $\tau = RC$ .

Approach

Start from $\tau = RC$ .
Rearrange to $C = \tau/R$ .
Substitute your numerical values (with units).

Step-by-Step Reasoning

With $R \approx 9.2 \times 10^{4}\ \Omega$ and $\tau \approx 3.0\ \text{s}$ :

C = \frac{3.0\ \text{s}}{9.2 \times 10^{4}\ \Omega}.

Compute the value:

C \approx 3.26 \times 10^{-5}\ \text{F} \approx 3.3 \times 10^{-5}\ \text{F}.

This is about $33\ \mu\text{F}$ , which is a realistic capacitor value for such a slow discharge.

Key Takeaways

Once you know $R$ and $\tau$ , you can immediately find $C$ from $C = \tau/R$ .
Check plausibility: large $R$ and a few-second $\tau$ typically implies a tens-of-microfarads capacitor.

Common Mistakes

Using $C = R/\tau$ (wrong rearrangement).
Mixing units (e.g. using $R$ in $\text{k}\Omega$ without converting).
Quoting too many significant figures given that $R$ and $\tau$ came from a graph.

Things to Be Careful About

Significant figures should reflect the graph readings (often 2 s.f. is appropriate).
If your $\tau$ differs slightly from reading the graph, your $C$ will change proportionally; error carried forward is normally allowed in such questions.

Techniques used

use the time constant relation $\tau = RC$rearrange an equation to solve for capacitancesubstitute measured values with consistent units

Q7MediumAlternating Currents

A circuit contains a power supply that provides a sinusoidal alternating input voltage $V_{IN}$ . There is an output voltage $V_{OUT}$ across a load resistor $R$ , as shown in Fig. 7.1.

Fig. 7.1

(a)

State the purpose of the circuit in Fig. 7.1.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

To full-wave rectify the a.c. input so that the output across $R$ is unidirectional (pulsating d.c.), i.e. current through $R$ is in the same direction for both half-cycles.

Final answer

Full-wave rectification to give a unidirectional (pulsating d.c.) output across R.

Detailed explanation

Background Concept

A diode conducts readily in one direction (forward bias) and does not conduct in the opposite direction (reverse bias). A bridge rectifier uses four diodes arranged so that, for both half-cycles of an a.c. input, current through the load resistor flows in the same direction.

Understanding the Question

Fig. 7.1 shows four diodes in a bridge arrangement, with an a.c. supply labelled $V_{IN}$ connected to one pair of opposite corners and the load resistor $R$ connected to the other pair. The question asks what this circuit is used for.

Approach

Recognise the circuit as a bridge rectifier. State what happens to the output waveform: it becomes full-wave rectified (all positive), so the load experiences a unidirectional voltage/current (pulsating d.c.).

Step-by-Step Reasoning

In the positive half-cycle of $V_{IN}$ , one pair of diodes is forward biased and conducts, driving current through $R$ in a particular direction.
In the negative half-cycle, the other pair of diodes conducts instead.
Because of the bridge arrangement, the direction of current through $R$ is the same in both cases.
Therefore the output is the absolute value of the input sine wave: a full-wave rectified waveform.

Key Takeaways

A bridge rectifier converts a.c. into a unidirectional (pulsating d.c.) output.
The key feature is that current through the load keeps the same direction for both half-cycles.

Common Mistakes

Saying it “smooths” the signal: smoothing requires a capacitor/reservoir; the diagram shows only diodes and $R$ .
Confusing half-wave rectification (one diode) with full-wave rectification (bridge).

Things to Be Careful About

Use the term “full-wave rectification” and mention “unidirectional” or “same direction through $R$ ” to secure both marks.

Techniques used

identify the function of a bridge rectifier circuitdescribe the effect of diodes on current direction through a load

(b)

Fig. 7.2 shows the variation of $V_{OUT}$ with time $t$ .

Fig. 7.2

6M

(i)

The load resistor $R$ has a resistance of $370\ \Omega$ .

Show that the maximum power dissipated in $R$ is $0.22\ \text{W}$ .

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 7.2, $V_{\text{max}} = 9.0\ \text{V}$ .

P_{\text{max}} = \frac{V_{\text{max}}^2}{R}

P_{\text{max}} = \frac{(9.0)^2}{370} = \frac{81}{370} = 2.19 \times 10^{-1}\ \text{W} \approx 0.22\ \text{W}

Answer

$P_{\text{max}} = 0.22\ \text{W}$

Final answer

0.22 W

Detailed explanation

Background Concept

For a resistor, instantaneous power converted to thermal energy is

P = IV

Using Ohm’s law $V = IR$ , this can be written as

P = \frac{V^2}{R}

This form is convenient when the voltage across the resistor is known.

Understanding the Question

The graph in Fig. 7.2 gives $V_{OUT}$ against time. The highest value (peak) of $V_{OUT}$ is $9.0\ \text{V}$ . The resistor has $R = 370\ \Omega$ . The question asks for the maximum power dissipated, i.e. the power at the peak voltage.

Approach

Read $V_{\text{max}}$ from the graph, then use

P_{\text{max}} = \frac{V_{\text{max}}^2}{R}

and round to match the stated result.

Step-by-Step Reasoning

Peak output voltage from Fig. 7.2: $V_{\text{max}} = 9.0\ \text{V}$ .
Substitute into $P = V^2/R$ :

P_{\text{max}} = \frac{(9.0\ \text{V})^2}{370\ \Omega}

Calculate:

P_{\text{max}} = \frac{81}{370}\ \text{W} = 0.219\ \text{W}

Round to two significant figures:

P_{\text{max}} \approx 0.22\ \text{W}

Key Takeaways

For a resistor, $P$ can be found directly from voltage using $P = V^2/R$ .
“Maximum power” corresponds to using the peak voltage.

Common Mistakes

Using $P = IV$ but not finding $I$ consistently.
Using $V_{\text{rms}}$ instead of $V_{\text{max}}$ when asked for maximum power.
Forgetting the unit (W).

Things to Be Careful About

Ensure you use the peak value read from the graph (here $9.0\ \text{V}$ ).
Keep enough figures in intermediate steps so rounding to $0.22\ \text{W}$ is justified.

Techniques used

use the resistor power relation in terms of potential differencesubstitute peak voltage into an equationapply correct units and rounding

(ii)

On Fig. 7.3, sketch the variation with $t$ of the power $P$ dissipated in $R$ .

Fig. 7.3

3M Similar

DifficultyMedium

Worked solution

Working

For a resistor,

P = \frac{V_{OUT}^2}{R}

So $P \ge 0$ always. Peak power occurs at $V_{OUT} = 9.0\ \text{V}$ :

P_{\text{max}} = 0.22\ \text{W}

Zeros of $P$ occur when $V_{OUT}=0$ (same times as in Fig. 7.2). Period is $0.02\ \text{s}$ .

diagram-1

Answer

Sketch as shown: a $\sin^2$ -type waveform, always positive, with peaks at $0.22\ \text{W}$ and repeating every $0.02\ \text{s}$ .

Final answer

Sketch of P(t): always positive, peaks at 0.22 W, period 0.02 s (sin^2 shape).

Detailed explanation

Background Concept

In a resistor, electrical energy is dissipated as thermal energy at a rate (power)

P = IV

Using $V = IR$ gives

P = I^2R = \frac{V^2}{R}

So the power depends on the square of the instantaneous voltage. Squaring has two important effects:

the result is never negative,
the shape becomes “more flattened” near zero because small voltages become even smaller when squared.

Understanding the Question

You are given the output voltage waveform $V_{OUT}(t)$ (full-wave rectified sine wave). You must sketch the corresponding power waveform $P(t)$ for the same resistor $R$ . The axes for $P$ are already provided up to $0.4\ \text{W}$ .

Approach

Use the relationship $P = V_{OUT}^2/R$ .
Use the same time positions for zeros and peaks as in the voltage graph (because $P=0$ when $V_{OUT}=0$ , and $P$ is maximum when $V_{OUT}$ is maximum).
Set the peak value of the power using $P_{\text{max}}$ from part (i).
Draw a smooth curve that looks like the voltage waveform “squared” (a $\sin^2$ shape).

Step-by-Step Reasoning

From Fig. 7.2, $V_{OUT}$ is always positive (full-wave rectified) and reaches a maximum of $9.0\ \text{V}$ .
Instantaneous power:

P(t) = \frac{V_{OUT}(t)^2}{R}

When $V_{OUT} = 0$ (where the voltage wave touches the time axis),

P = 0

So the power graph must also touch the axis at the same times.

When $V_{OUT}$ is at its peak, power is at its peak. Using part (i),

P_{\text{max}} = 0.22\ \text{W}

So the power peaks should reach $0.22\ \text{W}$ .

The period of the given $V_{OUT}$ waveform is $0.02\ \text{s}$ , so the power waveform repeats with the same period (because $V_{OUT}$ is already rectified; squaring does not introduce sign changes that would alter the period).

A good sketch therefore shows four identical positive “humps” from $0$ to $0.08\ \text{s}$ , each hump going from $0$ up to $0.22\ \text{W}$ and back to $0$ , with the characteristic $\sin^2$ curvature.

diagram-1

Key Takeaways

For a resistor, $P \propto V^2$ instantaneously.
Squaring a waveform makes it all positive and changes the curvature (closer to zero near the zeros).
Maxima and zeros occur at the same times as for $V$ .

Common Mistakes

Drawing negative parts of the power graph: power dissipated in a resistor cannot be negative.
Keeping the same shape as $V_{OUT}$ instead of squaring it.
Using the wrong peak value for power (must be $0.22\ \text{W}$ from part (i)).
Putting peaks at the wrong times (peaks must coincide with voltage peaks).

Things to Be Careful About

The power graph must be clearly labelled with the correct maximum on the given scale.
Make sure the curve is smooth and periodic with the correct time spacing (period $0.02\ \text{s}$ from the graph).

Techniques used

relate instantaneous power to instantaneous voltage for a resistortransform a waveform by squaring its ordinateuse peak values and periods from a graph to set the scale

(iii)

Calculate the mean power dissipated in $R$ .

mean power = ______ $\text{W}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For a sinusoidal waveform,

V_{\text{rms}} = \frac{V_{\text{max}}}{\sqrt{2}}

Mean power:

\overline{P} = \frac{V_{\text{rms}}^2}{R} = \frac{V_{\text{max}}^2}{2R} = \frac{P_{\text{max}}}{2}

\overline{P} = \frac{0.22}{2} = 0.11\ \text{W}

Answer

$0.11\ \text{W}$

Final answer

0.11 W

Detailed explanation

Background Concept

The mean power dissipated in a resistor on an a.c. supply is found using the rms voltage:

\overline{P} = \frac{V_{\text{rms}}^2}{R}

For a sinusoidal voltage $V = V_{\text{max}}\sin(\omega t)$ ,

V_{\text{rms}} = \frac{V_{\text{max}}}{\sqrt{2}}

This works because the average of $\sin^2$ over a complete cycle is $1/2$ .

Understanding the Question

You have a full-wave rectified sine voltage across a resistor. Even though the voltage is always positive, the power depends on $V^2$ , so the “sign” of the original sine wave would not change the mean power. You are asked for the mean power dissipated in $R$ .

Approach

Use either of these equivalent methods:

Find $V_{\text{rms}}$ from $V_{\text{max}}$ , then use $\overline{P} = V_{\text{rms}}^2/R$ .
Use the fact that for a (rectified) sine-derived power waveform, the mean is half the maximum: $\overline{P} = P_{\text{max}}/2$ .

Step-by-Step Reasoning

Peak output voltage from the graph: $V_{\text{max}} = 9.0\ \text{V}$ .
rms voltage:

V_{\text{rms}} = \frac{9.0}{\sqrt{2}}\ \text{V}

Mean power:

\overline{P} = \frac{V_{\text{rms}}^2}{R} = \frac{(9.0)^2}{2 \times 370}\ \text{W}

Or, using part (i) $P_{\text{max}} = V_{\text{max}}^2/R = 0.22\ \text{W}$ ,

\overline{P} = \frac{V_{\text{max}}^2}{2R} = \frac{P_{\text{max}}}{2} = 0.11\ \text{W}

Key Takeaways

Mean power in a resistor on a sinusoidal a.c. depends on $V_{\text{rms}}$ , not $V_{\text{max}}$ .
For a sine wave, $V_{\text{rms}} = V_{\text{max}}/\sqrt{2}$ , so $\overline{P} = P_{\text{max}}/2$ .

Common Mistakes

Using $\overline{P} = V_{\text{max}}^2/R$ (that gives maximum, not mean).
Confusing “mean of voltage” with “rms voltage” (mean of a rectified sine is not the rms).

Things to Be Careful About

The result here assumes ideal rectification with no diode voltage drops; the question provides a clean $9.0\ \text{V}$ peak on $V_{OUT}$ , so use that directly.
Quote the answer in watts and to a sensible number of significant figures (here $0.11\ \text{W}$ ).

Techniques used

use the relationship between mean power and rms voltagerelate rms and peak values for a sinusoidal waveformuse proportionality between mean and maximum power for a squared sine wave

(c)

The circuit of Fig. 7.1 is disconnected, and $R$ is connected directly across the power supply.

Explain, without calculation, how the mean power now dissipated in $R$ compares with the answer in (b)(iii).

2M Similar

DifficultyMedium

Worked solution

Answer

The mean power is the same as in (b)(iii), because for a resistor $P \propto V^2$ so changing the voltage from full-wave rectified to a sine wave of the same peak only changes the sign of $V$ for half the time, and squaring removes this sign.

Final answer

Same as (b)(iii).

Detailed explanation

Background Concept

For a resistor,

P(t) = \frac{V(t)^2}{R}

So the power depends on the square of the voltage, not on its sign. A negative voltage gives the same instantaneous power as the corresponding positive voltage of the same magnitude.

Mean power over time is essentially the average value of $V^2/R$ , which is proportional to the mean of $V^2$ (the “mean-square” idea behind rms values).

Understanding the Question

In parts (a) and (b) the resistor had a full-wave rectified voltage across it. Now the rectifier is removed and the resistor is connected directly to the a.c. supply, so the voltage across $R$ becomes a sinusoid that is positive for half a cycle and negative for half a cycle. You must compare the mean power in this new situation with your previous mean power, without doing numerical work.

Approach

Compare the two situations by focusing on what determines mean power in a resistor:

it depends on $V^2$ (or $V_{\text{rms}}^2$ ),
changing a voltage waveform by flipping the sign of parts of it does not change $V^2$ .

Step-by-Step Reasoning

With the bridge rectifier, the voltage across $R$ is approximately $V_{OUT}(t) = |V_{IN}(t)|$ (ignoring any diode drops).
Without the rectifier, the voltage across $R$ is $V(t) = V_{IN}(t)$ .
At every instant,

\left(|V_{IN}(t)|\right)^2 = \left(V_{IN}(t)\right)^2

So the instantaneous power $P(t) = V(t)^2/R$ is the same in both cases at every time.

Therefore the mean (average) power over a cycle is unchanged.

So the mean power with $R$ directly across the supply is the same as in (b)(iii).

Key Takeaways

Power in a resistor depends on $V^2$ , so the sign of the voltage is irrelevant.
Full-wave rectification changes the sign of the voltage but not the mean-square value (for ideal diodes), so it does not change the mean power delivered to a purely resistive load.

Common Mistakes

Claiming the mean power “increases because the voltage is now sometimes negative” (negative voltage does not mean negative power in a resistor).
Confusing mean voltage with mean power.
Stating it “halves” because rectification “uses both halves”: full-wave rectification already uses both halves; direct connection also has both halves present.

Things to Be Careful About

This comparison assumes the same peak magnitude across $R$ in both cases (i.e. ideal diodes or that diode drops are neglected, as is typical unless stated).
Keep the explanation qualitative as requested: focus on $P \propto V^2$ and the sign cancellation when squaring.

Techniques used

compare waveforms by considering how power depends on voltage squaredreason qualitatively using rms/mean-square ideasidentify that voltage sign does not affect power in a resistor

Q8MediumQuantum PhysicsMedical Physics

(a)

State what is meant by a photon.

2M Similar

(b)

Fig. 8.1 shows a tube in which X-rays are produced at a metal target.

Fig. 8.1

Particles are accelerated from the filament to the target by a constant high voltage applied across the terminals X and Y.

2M

(i)

State the name of the particles.

1M Similar

(ii)

On Fig. 8.1, use + and – signs to label terminals X and Y to indicate the polarity of the high voltage.

1M Similar

(c)

For an accelerating voltage of $32\ \text{kV}$ in Fig. 8.1, determine:

6M

(i)

the maximum energy, in MeV, of an X-ray photon produced at the target

maximum photon energy = ______ $\text{MeV}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Maximum photon energy equals maximum electron energy:

E_{\max} = eV = 32\ \text{keV} = 0.032\ \text{MeV}.

Answer

$0.032\ \text{MeV}$

Final answer

0.032 MeV

Detailed explanation

Background Concept

In an X-ray tube, electrons are accelerated through a potential difference $V$ . The electrical work done on each electron is

E = eV,

where $e = 1.60 \times 10^{-19}\ \text{C}$ .

The maximum possible X-ray photon energy occurs if one electron transfers all of its kinetic energy to a single photon (this is the limiting case that gives the cut-off energy).

Understanding the Question

The accelerating voltage is $32\ \text{kV}$ . You are asked for the maximum X-ray photon energy in $\text{MeV}$ .

Approach

Use $E_{\max} = eV$ .
Express the result in electronvolts, then convert $\text{keV}$ to $\text{MeV}$ .

Step-by-Step Reasoning

An electron accelerated through $32\ \text{kV}$ gains energy

E = eV = 32\ \text{keV}

because 1 electronvolt is the energy gained by charge $e$ through 1 volt.

Convert:

32\ \text{keV} = 32 \times 10^3\ \text{eV} = 0.032\ \text{MeV}

since $1\ \text{MeV} = 10^6\ \text{eV}$ .

Key Takeaways

Maximum photon energy in an X-ray tube is limited by the electron energy: $E_{\max}=eV$ .
Be fluent converting $\text{keV} \leftrightarrow \text{MeV}$ .

Common Mistakes

Writing $32\ \text{MeV}$ instead of $0.032\ \text{MeV}$ (incorrect prefix conversion).
Converting to joules when the question explicitly asks for MeV.

Things to Be Careful About

$32\ \text{kV}$ corresponds to $32\ \text{keV}$ for an electron (not $32\ \text{eV}$ ).
State the unit clearly as $\text{MeV}$ .

Techniques used

use energy gained by a charge accelerated through a potential differenceconvert between eV, keV and MeV

(ii)

the maximum momentum of an X-ray photon produced at the target

maximum photon momentum = ______ $\text{N s}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

E_{\max} = 32\ \text{keV} = 32\times 10^3 \times 1.60\times 10^{-19} = 5.12\times 10^{-15}\ \text{J}.

p_{\max} = \frac{E_{\max}}{c} = \frac{5.12\times 10^{-15}}{3.00\times 10^8} = 1.71\times 10^{-23}\ \text{N s}.

Answer

$1.7\times 10^{-23}\ \text{N s}$

Final answer

1.7 × 10^−23 N s

Detailed explanation

Background Concept

For a photon:

E = pc

so

p = \frac{E}{c}

where $E$ is photon energy in joules, $p$ is photon momentum in $\text{kg m s}^{-1}$ (equivalently $\text{N s}$ ), and $c$ is the speed of light.

Understanding the Question

From the same $32\ \text{kV}$ accelerating voltage, you previously found the maximum photon energy. Now you must find the maximum momentum of that photon, in SI units ( $\text{N s}$ ).

Approach

Convert $32\ \text{keV}$ into joules.
Use $p = E/c$ .

Step-by-Step Reasoning

Convert energy:

E_{\max} = 32\ \text{keV} = 32\times 10^3\ \text{eV}.

Using $1\ \text{eV} = 1.60\times 10^{-19}\ \text{J}$ ,

E_{\max} = 32\times 10^3 \times 1.60\times 10^{-19} = 5.12\times 10^{-15}\ \text{J}.

Momentum:

p_{\max} = \frac{E_{\max}}{c} = \frac{5.12\times 10^{-15}}{3.00\times 10^8} = 1.71\times 10^{-23}\ \text{kg m s}^{-1} = 1.71\times 10^{-23}\ \text{N s}.

Key Takeaways

Photon momentum can be found directly from energy using $p=E/c$ .
For momentum in $\text{N s}$ , energy must be in joules.

Common Mistakes

Using $p = mc$ (photons have zero rest mass).
Leaving energy in $\text{eV}$ and dividing by $c$ (units then are not SI).
Using $p = E/c^2$ (incorrect).

Things to Be Careful About

Keep $c = 3.00\times 10^8\ \text{m s}^{-1}$ .
$\text{N s}$ is the same as $\text{kg m s}^{-1}$ ; either may appear but the question requests $\text{N s}$ .

Techniques used

convert photon energy from electronvolts to joulesuse the photon momentum relation p = E/c

(iii)

the minimum wavelength of X-rays produced at the target.

minimum wavelength = ______ $\text{m}$

3M Similar

DifficultyMedium

Worked solution

Working

E_{\max} = eV = 32\ \text{keV} = 5.12\times 10^{-15}\ \text{J}.

E = \frac{hc}{\lambda} \;\Rightarrow\; \lambda_{\min} = \frac{hc}{E_{\max}} = \frac{6.63\times 10^{-34}\times 3.00\times 10^8}{5.12\times 10^{-15}} = 3.88\times 10^{-11}\ \text{m}.

Answer

$3.9\times 10^{-11}\ \text{m}$

Final answer

3.9 × 10^−11 m

Detailed explanation

Background Concept

The shortest (minimum) wavelength X-rays correspond to the highest-energy photons, because photon energy and wavelength are related by

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

So for a fixed maximum photon energy $E_{\max}$ , the minimum wavelength is

\lambda_{\min} = \frac{hc}{E_{\max}}.

In an X-ray tube, $E_{\max}$ is limited by the accelerating voltage:

E_{\max} = eV.

Understanding the Question

With accelerating voltage $V = 32\ \text{kV}$ , find the minimum wavelength of X-rays produced at the target. This is the cut-off wavelength of the bremsstrahlung spectrum.

Approach

Find $E_{\max}$ from $eV$ .
Convert to joules (because $h$ is in $\text{J s}$ ).
Use $\lambda_{\min} = hc/E_{\max}$ .

Step-by-Step Reasoning

Maximum electron energy (and hence maximum photon energy):

E_{\max} = eV = 32\ \text{keV}.

Convert to joules:

E_{\max} = 32\times 10^3 \times 1.60\times 10^{-19} = 5.12\times 10^{-15}\ \text{J}.

Apply the wavelength-energy relation:

\lambda_{\min} = \frac{hc}{E_{\max}}.

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

\lambda_{\min} = \frac{6.63\times 10^{-34}\times 3.00\times 10^8}{5.12\times 10^{-15}} = 3.88\times 10^{-11}\ \text{m}.

Key Takeaways

Higher photon energy means shorter wavelength.
The minimum X-ray wavelength is set by the accelerating voltage: $\lambda_{\min} = hc/(eV)$ .

Common Mistakes

Using $\lambda = c/f$ without connecting $f$ to energy.
Forgetting to convert $\text{keV}$ to joules before using $h$ .
Using $E = hc/\lambda$ but rearranging incorrectly (e.g. $\lambda = E/hc$ ).

Things to Be Careful About

Use consistent SI units when using $h$ .
Quote wavelength in metres, typically in standard form for X-rays (here $\sim 10^{-11}\ \text{m}$ ).

Techniques used

use the relation E = hc/λconvert photon energy to joulesrearrange an equation to find wavelength

(d)

Explain why X-rays can be used to produce images of internal body structures that have good contrast.

3M Similar

DifficultyMedium

Worked solution

Answer

X-rays are attenuated (absorbed/scattered) as they pass through the body.
The attenuation depends on the material (e.g. density / thickness / atomic number), so different tissues absorb different amounts (bone absorbs more than soft tissue).
Hence different intensities reach the detector, producing light/dark regions and good contrast in the image.

Final answer

Different tissues attenuate X-rays by different amounts (depending on density/thickness/atomic number), so different transmitted intensities reach the detector, giving light/dark regions and good image contrast.

Detailed explanation

Background Concept

X-ray imaging works because X-rays are attenuated as they pass through matter. Attenuation means the intensity decreases due to absorption and scattering. If an incident intensity is $I_0$ , after passing through a material the transmitted intensity $I$ is smaller.

Crucially, attenuation depends on the properties of the material (and the X-ray energy):

denser material generally causes more interactions,
higher atomic number $Z$ generally increases absorption probability,
greater thickness increases the amount of attenuation.

Understanding the Question

The question asks why X-rays can produce images of internal body structures with good contrast (3 marks). “Good contrast” means different structures (e.g. bone vs soft tissue) show clearly different brightness on the detector.

Approach

Explain that different tissues absorb/attenuate X-rays by different amounts, so the transmitted intensity varies across the body. The detector converts these intensity variations into different shades, giving contrast.

Step-by-Step Reasoning

When X-rays pass into the body, some photons are removed from the beam by absorption and scattering, so intensity decreases (attenuation).
Different body structures are made of different materials and have different effective density/atomic number and thickness along the path of the beam.
- Bone (contains calcium; higher effective $Z$ and density) attenuates strongly.
- Soft tissue attenuates less.
- Air-filled regions attenuate very little.
Therefore, the number of photons reaching the detector varies with position.
The detector/film records high intensity as one shade and low intensity as another shade, so boundaries between tissues show up clearly: this is the contrast.

Key Takeaways

Contrast in an X-ray image comes from differential attenuation.
Bone absorbs more X-rays than soft tissue, so it appears differently on the image.

Common Mistakes

Saying only “X-rays penetrate the body” (penetration alone does not explain contrast).
Confusing contrast with resolution (resolution is about sharpness; contrast is about brightness difference).
Stating “X-rays reflect” (X-ray imaging is mainly transmission with attenuation, not reflection).

Things to Be Careful About

Use the correct term “attenuation” (absorption/scattering) rather than vague “blocked”.
Mention a reason for different attenuation (density / thickness / atomic number) to secure the explanation marks.

Techniques used

link image contrast to differential attenuation of X-raysrelate absorption to material properties such as density and atomic numberexplain how transmitted intensity differences form an image

Q9MediumNuclear Physics

(a)

Define half-life of a radioactive isotope.

1M Similar

DifficultyEasy

Worked solution

Answer

Half-life is the time taken for the number of undecayed nuclei (or the activity) of a radioactive isotope to fall to half its initial value.

Final answer

Time for the number of undecayed nuclei (or activity) to fall to half its initial value.

Detailed explanation

Background Concept

Radioactive decay is a random process in which unstable nuclei transform into other nuclei. For a large sample, the number of undecayed nuclei $N$ falls exponentially with time:

N = N_0 e^{-\lambda t}

where $N_0$ is the initial number of undecayed nuclei and $\lambda$ is the decay constant.

The half-life $t_{1/2}$ is a convenient time scale for exponential decay: it is the time for the number of undecayed nuclei (or equivalently the activity) to halve.

Understanding the Question

You are asked to define half-life. No calculation is required; you just need the standard statement in terms of either number of nuclei remaining or activity.

Approach

State the definition precisely:

what halves (number of undecayed nuclei or activity), and
relative to what (the initial value).

Step-by-Step Reasoning

In radioactive decay, $N$ decreases with time.
After one half-life, the sample has half the number of undecayed nuclei it started with.
Because activity $A$ is proportional to $N$ (since $A = \lambda N$ ), the activity also halves in the same time.

Key Takeaways

Half-life is a time.
It can be defined using either number of undecayed nuclei or activity.

Common Mistakes

Defining it as “time for all nuclei to decay”.
Saying “time for the mass to halve” without making clear it is the undecayed isotope (this is true but imprecise).
Confusing half-life with decay constant $\lambda$ .

Things to Be Careful About

Use “undecayed nuclei” (or “activity”) explicitly.
It is half the initial value, not half the value at some later time.

Techniques used

recall the definition of half-lifestate the definition in terms of number of undecayed nuclei or activity

(b)

Radioactive isotope X decays to isotope Y.

A sample contains only nuclei of X at time $t = 0$ . Fig. 9.1 shows the variation with $t$ of the numbers of nuclei of X and of Y as the sample decays.

Fig. 9.1

4M

(i)

State the name of the quantity represented by the magnitude of the gradient of line X in Fig. 9.1.

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The magnitude of the gradient of line X is the activity (rate of decay), $A = -\dfrac{\mathrm{d}N_X}{\mathrm{d}t}$ .

Final answer

Activity (rate of decay), A = −dNX/dt.

Detailed explanation

Background Concept

Activity $A$ is defined as the number of decays per second. If $N$ is the number of undecayed nuclei, then activity is the rate at which $N$ decreases:

A = -\frac{\mathrm{d}N}{\mathrm{d}t}

The minus sign is needed because $N$ decreases with time, so $\mathrm{d}N/\mathrm{d}t$ is negative. The magnitude of the gradient of an $N$ – $t$ graph is therefore the activity.

Understanding the Question

Curve X on Fig. 9.1 is “number of nuclei of X” plotted against time. The gradient of this curve at any point represents how fast $N_X$ is changing. The question asks specifically for the name of the quantity represented by the magnitude of this gradient.

Approach

Recognise that gradient on an $N$ vs $t$ graph is $\mathrm{d}N/\mathrm{d}t$ .
For radioactive decay, activity is proportional to how fast $N$ falls: $A = -\mathrm{d}N/\mathrm{d}t$ .
Therefore the magnitude of the gradient corresponds to $A$ .

Step-by-Step Reasoning

The gradient of curve X is $\mathrm{d}N_X/\mathrm{d}t$ .
Because $N_X$ decreases, this gradient is negative.
Activity is defined as a positive quantity (decays per second):

A = -\frac{\mathrm{d}N_X}{\mathrm{d}t}

Hence the magnitude of the gradient of X equals the activity.

Key Takeaways

Gradient of $N$ vs $t$ gives the rate of change of $N$ .
For decay, the magnitude of that rate is the activity.

Common Mistakes

Answering “decay constant $\lambda$ ” (this is related, but not equal to the gradient).
Forgetting that the gradient is negative and confusing the sign.

Things to Be Careful About

The question says magnitude of the gradient, so the intended quantity is positive.
Activity has units of $\text{s}^{-1}$ (or Bq), consistent with “nuclei per second”.

Techniques used

interpret the gradient of a number-of-nuclei vs time graphrelate the rate of change of number of nuclei to activity

(ii)

State three conclusions about X or Y that may be drawn from Fig. 9.1. The conclusions may be qualitative or quantitative. Use the space below for any working that you need.

3M Similar

DifficultyMedium

Worked solution

Answer

Any three conclusions from the graph, e.g.

$N_X$ decreases exponentially with time and $N_Y$ increases, approaching $4.0 \times 10^{22}$ .
At $t = 14\ \text{s}$ , $N_X = N_Y = 2.0 \times 10^{22}$ , so the half-life of X is $14\ \text{s}$ .
$N_X + N_Y$ is constant at $4.0 \times 10^{22}$ , so each nucleus of X decays into one nucleus of Y (closed system, no loss of nuclei).

Final answer

Examples: half-life of X is 14 s; NX+NY constant at 4.0×10^22 so X→Y one-to-one; NX decays exponentially while NY rises to 4.0×10^22.

Detailed explanation

Background Concept

In a simple decay chain where a parent isotope $X$ decays to a daughter $Y$ (and $Y$ is stable on the timescale considered):

$N_X$ decreases exponentially.
$N_Y$ increases because it is being produced by the decay of $X$ .
If every decay of $X$ produces exactly one $Y$ nucleus and none escape, then the total number of nuclei is conserved:

N_X + N_Y = \text{constant}

Half-life $t_{1/2}$ is the time for $N_X$ to fall to $\tfrac{1}{2}N_{X0}$ .

Understanding the Question

Fig. 9.1 shows two curves versus time $t$ :

Curve X: number of nuclei of $X$ .
Curve Y: number of nuclei of $Y$ .

You must state three valid conclusions that can be read from the curves. These can be numerical (e.g. half-life) or qualitative (e.g. exponential decrease).

Approach

Look for clear features of the curves:

Initial and final values.
Whether the shape is exponential/asymptotic.
Any special points like intersections.
Whether the sum appears constant.

Then translate those graphical observations into physics statements about decay.

Step-by-Step Reasoning

Initial values (at $t=0$ $t = 0$ ):
- The sample contains only $X$ , so $N_X$ starts at $4.0 \times 10^{22}$ and $N_Y$ starts at $0$ .
Intersection point:
- The curves cross at $t = 14\ \text{s}$ and $N_X = N_Y = 2.0 \times 10^{22}$ .
- Since $N_{X0} = 4.0 \times 10^{22}$ , at $t=14\ \text{s}$ the number of $X$ nuclei has fallen to half:

N_X = \frac{1}{2}N_{X0}

Therefore the half-life of $X$ is $14\ \text{s}$ .

Total number of nuclei:
- At $t=0$ , $N_X + N_Y = 4.0 \times 10^{22}$ .
- At the intersection, $2.0 \times 10^{22} + 2.0 \times 10^{22} = 4.0 \times 10^{22}$ .
- At large $t$ , $N_X$ tends to $0$ and $N_Y$ tends to $4.0 \times 10^{22}$ .
- So the total remains $4.0 \times 10^{22}$ : this supports “one $X$ nucleus produces one $Y$ nucleus” with no other losses.
Qualitative shapes:
- $N_X$ falls with a decreasing slope in magnitude: characteristic of exponential decay.
- $N_Y$ rises and levels off: it approaches a maximum as production slows when little $X$ remains.

Any three of these points are valid conclusions.

Key Takeaways

From a decay graph you can read half-life directly by finding when $N_X$ halves.
In a simple $X \to Y$ decay, $N_Y$ increases while $N_X$ decreases.
If the curves sum to a constant, it implies a one-to-one transformation with no leakage.

Common Mistakes

Saying “the half-life is when $N_X = N_Y$ ” without checking that $N_X$ at that time is half of the initial value.
Mixing up which curve corresponds to $X$ and which to $Y$ .
Claiming $Y$ decays as well (the graph here shows $N_Y$ tending to a constant, not falling).

Things to Be Careful About

Read values with the correct axis scale: the vertical axis is “number of nuclei / $10^{22}$ ”.
If you quote half-life, include the unit ( $\text{s}$ ).
“Exponential” is supported by the curve shape: steep initially then flattening, not a straight line.

Techniques used

read quantitative values from an exponential decay graphuse the intersection point of two curves to infer equal numbersdeduce half-life from the halving of the parent nucleiidentify conservation of total number in a decay-to-daughter system

(c)

The mass of radioactive isotope X in the sample in (b) is $7.3 \times 10^{-4}\ \text{kg}$ at time $t = 0$ .

Determine the nucleon number of isotope X.

nucleon number = ______

3M Similar

DifficultyMedium

Worked solution

Working

From Fig. 9.1 at $t=0$ :

N_X = 4.0 \times 10^{22}

Mass of sample at $t=0$ is $m = 7.3 \times 10^{-4}\ \text{kg}$ .

Assuming mass per nucleus $\approx A m_u$ where $m_u = 1.66 \times 10^{-27}\ \text{kg}$ :

m = N_X A m_u \;\Rightarrow\; A = \frac{m}{N_X m_u}

A = \frac{7.3 \times 10^{-4}}{\left(4.0 \times 10^{22}\right)\left(1.66 \times 10^{-27}\right)} = 11

Answer

nucleon number $= 11$

Final answer

11

Detailed explanation

Background Concept

The nucleon number (mass number) $A$ is the total number of protons + neutrons in a nucleus.

The mass of one nucleus is approximately

m_{\text{nucleus}} \approx A m_u

where $m_u = 1.66 \times 10^{-27}\ \text{kg}$ is the atomic mass unit. (The true nucleus mass differs slightly because of binding energy, but this approximation is what is expected here.)

For a sample containing $N$ nuclei, the total mass is

m = N \times m_{\text{nucleus}} \approx N A m_u

Understanding the Question

You are told the mass of isotope $X$ in the sample at $t=0$ is $7.3 \times 10^{-4}\ \text{kg}$ . From Fig. 9.1, at $t=0$ the sample contains only $X$ , and the graph gives the number of $X$ nuclei present initially.

You must find the nucleon number $A$ of isotope $X$ .

Approach

Read $N_X$ at $t=0$ from the graph.
Use $m \approx N_X A m_u$ .
Rearrange to $A = m/(N_X m_u)$ and substitute.

Step-by-Step Reasoning

From the graph at $t=0$ :

N_X = 4.0 \times 10^{22}

Total mass of these nuclei is

m = 7.3 \times 10^{-4}\ \text{kg}

Use the mass model:

m \approx N_X A m_u

Rearrange for $A$ :

A = \frac{m}{N_X m_u}

Substitute values:

A = \frac{7.3 \times 10^{-4}}{\left(4.0 \times 10^{22}\right)\left(1.66 \times 10^{-27}\right)}

Compute the denominator:

(4.0 \times 10^{22})(1.66 \times 10^{-27}) = 6.64 \times 10^{-5}

So

A = \frac{7.3 \times 10^{-4}}{6.64 \times 10^{-5}} \approx 11.0 \approx 11

The nucleon number must be an integer, so $A = 11$ .

Key Takeaways

Use the graph to get $N$ and use $m \approx N A m_u$ .
Keep track of powers of ten carefully.
Mass number is an integer.

Common Mistakes

Using $N_Y$ or the intersection value instead of $N_X$ at $t=0$ .
Forgetting the factor of $10^{22}$ on the vertical axis.
Using Avogadro’s constant unnecessarily (not needed when $N$ is already given).
Rounding $A$ to a non-integer.

Things to Be Careful About

The question says the mass of isotope $X$ at $t=0$ , so you must use $N_X$ at $t=0$ .
Ensure $m_u$ is in $\text{kg}$ .
Small differences due to binding energy are ignored at this level; the nearest integer is expected.

Techniques used

extract the initial number of nuclei from the graphuse mass per nucleus as nucleon number times atomic mass unitrearrange an expression to solve for nucleon numbercheck significant figures and units

Q10Medium-EasyAstronomy and Cosmology

(a)

5M

(i)

State what is meant by the luminosity of a star.

2M Similar

DifficultyEasy

Worked solution

Answer

Luminosity is the total energy emitted by a star per unit time (i.e. the total power output), over all wavelengths.

Final answer

Total power (energy per unit time) emitted by the star over all wavelengths.

Detailed explanation

Background Concept

Luminosity, usually denoted $L$ , is an intrinsic property of a star: it tells you how much energy the star produces and emits each second.

Because it is a power, its SI unit is the watt ( $\text{W}$ ), where $1\ \text{W} = 1\ \text{J s}^{-1}$ .

Luminosity is different from the radiant flux intensity (sometimes just called flux) $F$ that an observer measures, because $F$ depends on how far away the star is.

Understanding the Question

The question asks for the meaning of “luminosity of a star”. It is asking for a definition, not an equation or a calculation.

Approach

State clearly that luminosity is:

energy emitted per unit time, and
the total power output (not what is received at Earth), and
typically taken over all wavelengths.

Step-by-Step Reasoning

“Energy emitted per unit time” is the definition of power.
A star emits electromagnetic radiation across a wide range of wavelengths; luminosity refers to the total emission, not just visible light.
Since it is a property of the star itself, it does not change with distance.

Key Takeaways

Luminosity $L$ = total power output of the star.
Unit: watt ( $\text{W}$ ).
Do not confuse luminosity with flux $F$ measured by an observer.

Common Mistakes

Defining luminosity as “brightness as seen from Earth” (that is related to flux/apparent brightness).
Forgetting “per unit time” (energy alone is not luminosity).
Saying “light emitted” without making clear it is total power (and not just visible wavelengths).

Things to Be Careful About

Use wording that indicates total emission and per second.
If you mention “brightness”, qualify it carefully; examiners usually want “power output” explicitly.

Techniques used

recall the definition of luminositydistinguish between power output and observed flux

(ii)

Explain how a standard candle in a distant galaxy can be used to determine the distance of the galaxy from an observer.

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

A standard candle has known luminosity $L$ .
Measure its radiant flux intensity $F$ at Earth.
Use

F = \frac{L}{4\pi d^2}

and rearrange to find

d = \sqrt{\frac{L}{4\pi F}}.

Final answer

Use known luminosity of the standard candle and measured flux with F = L/(4πd²), so d = √(L/(4πF)).

Detailed explanation

Background Concept

As light (or any radiation) spreads out from a source, it distributes its power over the surface of a sphere.

Luminosity $L$ is the total power emitted by the source.
At distance $d$ , that power is spread over area $4\pi d^2$ .
The radiant flux intensity (power per unit area) measured by an observer is

F = \frac{L}{4\pi d^2}.

This is an inverse-square law: doubling $d$ makes $F$ four times smaller.

A standard candle is an astronomical object whose luminosity is known (e.g. certain types of supernovae or Cepheid variables once calibrated).

Understanding the Question

You are asked to explain how observing a standard candle in a distant galaxy lets you find the galaxy’s distance.

Given/available:

the standard candle’s known luminosity $L$ (intrinsic)
the measured flux $F$ at the observer (from telescope measurements)

Unknown:

distance $d$ to the galaxy.

Approach

Use the fact that a standard candle provides a known $L$ .
Measure $F$ (apparent brightness) from Earth.
Substitute into $F = L/(4\pi d^2)$ and rearrange to obtain $d$ .

Step-by-Step Reasoning

Because the luminosity $L$ is known, the only unknown in

F = \frac{L}{4\pi d^2}

is $d$ .

Rearranging:

4\pi d^2 = \frac{L}{F}

d^2 = \frac{L}{4\pi F}

d = \sqrt{\frac{L}{4\pi F}}.

So the distance is determined from one measurement of $F$ , provided $L$ is reliably known.

Key Takeaways

Standard candle: known $L$ .
Measure $F$ at Earth.
Use the inverse-square law to get $d$ .

Common Mistakes

Using $F = L/(4\pi d)$ (missing the square).
Saying “measure intensity then distance is found” without stating the relationship used.
Confusing luminosity $L$ with flux $F$ .

Things to Be Careful About

$F$ must be in $\text{W m}^{-2}$ if you want $d$ in metres.
In real astronomy there can be absorption/extinction by dust; unless asked, you typically ignore it at this level, but be aware it would make $F$ smaller and the inferred distance too large if uncorrected.

Techniques used

use the inverse-square relationship between flux and distancerearrange F = L / (4 pi d^2) to solve for distancelink 'standard candle' to known luminosity

(b)

The Sun has a radius of $6.96 \times 10^8\ \text{m}$ and a surface temperature of $5780\ \text{K}$ .

Light from the Sun is observed to have a peak intensity at a wavelength of $501\ \text{nm}$ .

4M

(i)

Calculate the luminosity of the Sun. Give a unit with your answer.

luminosity = ______ unit ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using Stefan–Boltzmann law:

L = 4\pi r^2 \sigma T^4

r = 6.96 \times 10^8\ \text{m},\quad T = 5780\ \text{K},\quad \sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}

L = 4\pi (6.96 \times 10^8)^2 (5.67 \times 10^{-8})(5780)^4

L \approx 3.86 \times 10^{26}\ \text{W}

Answer

$3.86 \times 10^{26}\ \text{W}$

Final answer

3.86 × 10^26 W

Detailed explanation

Background Concept

A star’s luminosity can be found if we approximate the star as a black body.

For a black body, the power radiated per unit area is:

\frac{P}{A} = \sigma T^4,

where:

$\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$ is the Stefan–Boltzmann constant,
$T$ is the absolute temperature in kelvin.

For a spherical star of radius $r$ , the surface area is $A = 4\pi r^2$ , so the total power (luminosity) is:

L = 4\pi r^2 \sigma T^4.

Understanding the Question

You are given the Sun’s:

radius $r = 6.96 \times 10^8\ \text{m}$
surface temperature $T = 5780\ \text{K}$

and asked to calculate its luminosity, including a unit.

(The peak wavelength information is not needed for this part; it is relevant to Wien’s law in part (b)(ii).)

Approach

Use Stefan–Boltzmann for a spherical black body:

Calculate $4\pi r^2$ .
Calculate $\sigma T^4$ .
Multiply to get $L$ .

Step-by-Step Reasoning

Start with:

L = 4\pi r^2 \sigma T^4.

Square the radius:

r^2 = (6.96 \times 10^8)^2 = 4.84 \times 10^{17}\ \text{m}^2 \ \text{(approximately)}

Surface area:

4\pi r^2 \approx 4\pi (4.84 \times 10^{17}) \approx 6.08 \times 10^{18}\ \text{m}^2.

Compute $T^4$ :

T^4 = (5780)^4 \approx 1.12 \times 10^{15}\ \text{K}^4.

Multiply by $\sigma$ :

\sigma T^4 \approx (5.67 \times 10^{-8})(1.12 \times 10^{15}) \approx 6.35 \times 10^{7}\ \text{W m}^{-2}.

Multiply by area to get luminosity:

L \approx (6.08 \times 10^{18})(6.35 \times 10^{7}) \approx 3.86 \times 10^{26}\ \text{W}.

The unit is watts because luminosity is power.

Key Takeaways

Use $L = 4\pi r^2 \sigma T^4$ .
Temperature must be in kelvin.
Radius must be in metres to get $L$ in watts.

Common Mistakes

Forgetting the $4\pi$ factor (using $\pi r^2$ or $4r^2$ ).
Using $T$ in degrees Celsius instead of kelvin.
Not giving a unit (must be $\text{W}$ ).
Using Wien’s law here unnecessarily.

Things to Be Careful About

Powers: $T^4$ is very large; keep track of standard form.
Significant figures: typically 3 s.f. is appropriate given the data.
Ensure $\sigma$ unit is $\text{W m}^{-2}\ \text{K}^{-4}$ so the final unit becomes $\text{W}$ after multiplying by $\text{m}^2$ and $\text{K}^4$ .

Techniques used

apply the Stefan-Boltzmann law for a spherical radiatorsubstitute values with correct SI unitsuse standard form and appropriate significant figures

(ii)

Another star emits radiation that has a peak intensity at a wavelength of $624\ \text{nm}$ .

Determine the surface temperature of this star.

surface temperature = ______ $\text{K}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using Wien’s displacement law:

\lambda_{\max} T = 2.90 \times 10^{-3}\ \text{m K}

\lambda_{\max} = 624\ \text{nm} = 6.24 \times 10^{-7}\ \text{m}

T = \frac{2.90 \times 10^{-3}}{6.24 \times 10^{-7}} \approx 4.65 \times 10^{3}\ \text{K}

Answer

$4.65 \times 10^{3}\ \text{K}$

Final answer

4.65 × 10^3 K

Detailed explanation

Background Concept

A (near) black-body radiator has an emission spectrum with a peak at wavelength $\lambda_{\max}$ . Wien’s displacement law relates this peak wavelength to the absolute temperature:

\lambda_{\max} T = b,

where $b \approx 2.90 \times 10^{-3}\ \text{m K}$ .

Hotter objects have smaller $\lambda_{\max}$ (peak shifts to shorter wavelengths).

Understanding the Question

A star has peak intensity at wavelength $624\ \text{nm}$ . You must find its surface temperature.

Given:

$\lambda_{\max} = 624\ \text{nm}$

Unknown:

$T$ in kelvin.

Approach

Convert $624\ \text{nm}$ into metres.
Use Wien’s law $T = b/\lambda_{\max}$ .
Quote the temperature to a sensible number of significant figures.

Step-by-Step Reasoning

Convert the wavelength:

624\ \text{nm} = 624 \times 10^{-9}\ \text{m} = 6.24 \times 10^{-7}\ \text{m}.

Apply Wien’s law:

T = \frac{2.90 \times 10^{-3}}{6.24 \times 10^{-7}}.

Divide the numbers and subtract powers of ten:

T \approx 0.465 \times 10^{4}\ \text{K} = 4.65 \times 10^{3}\ \text{K}.

Key Takeaways

Wien’s law connects peak wavelength and temperature.
Always convert nm to m before substituting into $b = 2.90 \times 10^{-3}\ \text{m K}$ .
Larger peak wavelength implies a cooler star.

Common Mistakes

Forgetting to convert nm to m (gives a temperature smaller by a factor of $10^9$ ).
Using $\lambda_{\max} T = 2.90 \times 10^{3}$ (wrong power of ten).
Giving temperature in $^\circ\text{C}$ instead of kelvin.

Things to Be Careful About

Keep track of powers of ten carefully: $10^{-3}/10^{-7} = 10^{4}$ .
Quote the answer to 3 s.f. to match typical data precision.
The constant $b$ can vary slightly depending on rounding; small differences in the final digits are usually acceptable.

Techniques used

apply Wien's displacement lawconvert wavelength units to metresrearrange an equation to solve for temperature

Physics 9702/43 — May/June 2024

Answer

Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

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Working

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Understanding the Question

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Step-by-Step Reasoning

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Working

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Understanding the Question

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Step-by-Step Reasoning

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning

Key Takeaways

Common Mistakes

Things to Be Careful About

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Background Concept

Understanding the Question

Approach

Step-by-Step Reasoning