(a) Define electric field. (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. (i) On Fig. 5.1, draw four field lines to represent the electric field between the plates. (ii) Determine the strength $E$ of the electric field between the plates. $E$ = \_\_\_\_\_\_ $\text{N C}^{-1}$ (iii) An electron travels at a speed of $2.6…

(a) Electric field strength is force per unit positive charge at a point (E = F/q). · (b)(i) Uniform field: straight parallel equally spaced lines from +430 V plate to 0 V plate. · (b)(ii) 6.4 × 10^3 N C^-1 · (b)(iii) Curves upward between plates, then straight-line continuation at exit. · (c)(i) Into the page. · (c)(ii) Electric and magnetic forces are equal and opposite, so resultant force is zero and no deflection. · (c)(iii) 2.5 × 10^-4 T

A circuit contains a power supply that provides a sinusoidal alternating input voltage $V_{\text{IN}}$. There is an output voltage $V_{\text{OUT}}$ 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 $V_{\text{OUT}}$ 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…

(a) To rectify the a.c. input, giving a unidirectional (pulsating d.c.) output across R (full-wave rectification). · (b)(i) 0.22 W · (b)(ii) See sketch: P ∝ V_OUT^2, always positive, peaks at 0.22 W with the same period as V_OUT. · (b)(iii) 0.11 W · (c) The same.

(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 halve. · (b)(i) Activity (decay rate) of X. · (b)(ii) Example conclusions: N_X(0)=4.0×10^22, N_Y(0)=0; half-life of X = 14 s; eventually N_Y→4.0×10^22 and N_X→0 (Y stable on this timescale). · (c) 11

9702/41 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 the point (with no change in kinetic energy).

Final answer

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

Detailed explanation

Background Concept

Gravitational potential $V$ at a point is defined using the idea of work and energy in a gravitational field. We choose the gravitational potential to be zero at infinity. Then, for a small test mass $m$ , the gravitational potential energy is $U = mV$ .

A key definition is:

$V$ at a point = work done per unit mass (by an external agent) to bring a test mass from infinity to that point, slowly, so that kinetic energy does not change.

Understanding the Question

You are asked to give the definition of gravitational potential at a point. No numbers are needed; you must include the reference point (infinity) and the “per unit mass” idea.

Approach

State the standard definition used in Cambridge A Level Physics:

from infinity, 2) work done (or energy required), 3) per unit mass, 4) no change in kinetic energy.

Step-by-Step Reasoning

“Work done” connects the field concept to energy.
“Per unit mass” distinguishes potential $V$ from potential energy $U$ .
“From infinity” sets the zero level of potential.
“No change in kinetic energy” means the work done corresponds only to a change in gravitational potential energy.

A fully correct definition is therefore: work done per unit mass by an external agent to bring a small test mass from infinity to the point.

Key Takeaways

Gravitational potential $V$ is energy per unit mass.
The reference level is $V=0$ at infinity.
Potential energy and potential are related by $U=mV$ .

Common Mistakes

Defining potential energy instead of potential (missing “per unit mass”).
Not mentioning that the mass is brought from infinity.
Saying “work done by the field” without clarity (the standard definition uses work done by an external agent).

Things to Be Careful About

Use the words “per unit mass” explicitly.
Include “from infinity”; otherwise the definition is incomplete.

Techniques used

state a definition in terms of work done per unit massreference the zero of potential at infinitylink gravitational potential to potential energy per unit mass

(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 gravity is attractive, a mass at X is in a bound state and work must be done to move it to infinity, so its potential (relative to infinity) is negative.

Final answer

Because potential is zero at infinity and gravity is attractive, a mass at X must gain energy (external work is required) to reach infinity, so the potential at X is negative.

Detailed explanation

Background Concept

Gravitational potential $V$ is defined with $V(\infty)=0$ . For a spherically symmetric mass, outside the mass we have

V = -\frac{GM}{r}

which is negative for any finite $r$ .

The negative sign comes from the fact that gravity is an attractive interaction: you must supply energy (do external work) to take a mass from a point in the field out to infinity.

Understanding the Question

You are told the gravitational potential at X (a finite distance from the planet) is $-\Phi$ . You must explain why it is negative, i.e. why the sign is “minus”.

Approach

Use the definition $V(\infty)=0$ and describe what happens energetically:

A mass at finite distance is attracted inward.
To move it out to infinity, an external agent must do positive work.
Therefore the potential energy at finite distance is lower than at infinity, meaning the potential is negative.

Step-by-Step Reasoning

Zero of potential is chosen at infinity: $V(\infty)=0$ .
At a point nearer the planet, a test mass is attracted towards the planet.
If you try to take the test mass from X to infinity, you must pull against the attractive force, so you must do positive work.
That means the test mass has lower potential energy at X than at infinity.
Therefore $V(X)$ (potential energy per unit mass) is less than zero, so it is negative.

Key Takeaways

The negative sign is a consequence of the choice $V(\infty)=0$ and the attractive nature of gravity.
Finite-distance gravitational states are “bound” and have negative potential (and potential energy).

Common Mistakes

Saying “because gravity is negative” without referencing the zero at infinity.
Mixing up potential and field strength: field strength $g$ is positive in magnitude but potential can be negative.

Things to Be Careful About

Always connect the sign to the reference level at infinity.
Use energy/work language clearly: work must be done to move away from the planet.

Techniques used

use the convention that potential is zero at infinityexplain sign using attractive force and work/energy argumentsrelate negative potential to bound states

(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 $r$ outside the planet,

V = -\frac{GM}{r}

At X, $r=4R$ and $V_X=-\Phi$ :

-\Phi = -\frac{GM}{4R} \Rightarrow \frac{GM}{R} = 4\Phi

At Y, $r=R$ :

V_Y = -\frac{GM}{R} = -4\Phi

Answer

$-4\Phi$

Final answer

-4\Phi

Detailed explanation

Background Concept

Outside a uniform spherical planet, the gravitational potential is the same as if all the planet’s mass were concentrated at its centre:

V = -\frac{GM}{r}

So $V \propto -\frac{1}{r}$ for points outside the planet.

Understanding the Question

Satellite X is at distance $4R$ from the planet’s centre and has potential $V_X=-\Phi$ . Satellite Y is at distance $R$ . You are asked for the potential at Y due to the planet, in terms of $\Phi$ .

Approach

Use the proportionality $V \propto -\frac{1}{r}$ :

If the radius becomes 4 times smaller (from $4R$ to $R$ ), the magnitude of the potential becomes 4 times larger.
Keep the negative sign because the potential remains negative at any finite radius.

Step-by-Step Reasoning

Write potential at radius $r$ :

V = -\frac{GM}{r}

At X:

V_X = -\frac{GM}{4R} = -\Phi

So

\frac{GM}{4R} = \Phi \Rightarrow \frac{GM}{R} = 4\Phi

At Y (radius $R$ ):

V_Y = -\frac{GM}{R} = -4\Phi

Key Takeaways

For a spherical planet, outside it: $V=-GM/r$ .
Potential scales as $1/r$ (not $1/r^2$ ).
The sign stays negative for finite $r$ when $V(\infty)=0$ .

Common Mistakes

Using $1/r^2$ (that is for field strength $g$ , not potential).
Forgetting the negative sign and writing $+4\Phi$ .

Things to Be Careful About

X is at $4R$ from the centre, not $4$ radii above the surface.
$\Phi$ here represents the magnitude of the potential at X (since the potential is given as $-\Phi$ ).

Techniques used

use the inverse-radius dependence of gravitational potential outside a spherecompare potentials at two radii using proportionalitysubstitute the given potential to find a new potential

(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), $V_X=-\Phi$ at $r=4R$ :

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

Field strength $g=GM_p/r^2$ .

Satellite X ( $r=4R$ ):

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

Satellite Y ( $r=R$ ):

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

Potential energy $U = mV$ .

Satellite X ( $m=M$ , $V=-\Phi$ ):

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

Satellite Y ( $m=2M$ , $V=-4\Phi$ ):

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

Answer

Field strength: X $= \dfrac{\Phi}{4R}$ , Y $= \dfrac{4\Phi}{R}$
Potential energy: X $= -M\Phi$ , Y $= -8M\Phi$

Final answer

X: g = \Phi/(4R), U = -M\Phi; Y: g = 4\Phi/R, U = -8M\Phi

Detailed explanation

Background Concept

For a spherically symmetric planet, at distances outside the planet:

V = -\frac{GM_p}{r}

and gravitational field strength (magnitude) is

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

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

U = mV

So once you know $V$ at a point, $U$ follows immediately by multiplying by the satellite’s mass.

Understanding the Question

You have two satellites at different orbital radii:

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

You must fill a table with expressions 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 or all of $M$ , $R$ , and $\Phi$ .

Approach

Use the given potential at X to eliminate the planet parameter $GM_p$ .
Use $g=GM_p/r^2$ to get $g_X$ and $g_Y$ .
Use $U=mV$ (with the potential values at X and Y) to get potential energies.

Step-by-Step Reasoning

1) Find $GM_p$ in terms of $\Phi$ and $R$ .
At X, $r=4R$ and

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

Cancel the minus signs:

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

So

GM_p = 4R\Phi

2) Field strengths
Use $g=GM_p/r^2$ .

At X, $r=4R$ :

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

At Y, $r=R$ :

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

These are magnitudes; the field direction is towards the planet.

3) Potential energies
Use $U=mV$ .

For X: $m=M$ and $V_X=-\Phi$ :

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

For Y: first use the potential at Y (from part (ii)), $V_Y=-4\Phi$ , and $m=2M$ :

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

Key Takeaways

Outside a spherical planet: $V \propto -1/r$ and $g \propto 1/r^2$ .
You can use a given potential at one radius to determine $GM_p$ and then find other quantities.
Potential energy is simply $U=mV$ .

Common Mistakes

Using $g \propto 1/r$ (confusing with potential).
Forgetting that satellite Y has mass $2M$ when calculating $U_Y$ .
Missing the negative sign in potential energy: bound gravitational potential energy is negative when zero is at infinity.

Things to Be Careful About

Distinguish carefully between:
- $V$ (potential, per unit mass),
- $U$ (potential energy),
- $g$ (field strength).
Use $r=4R$ and $r=R$ as distances from the centre (as stated), not height above the surface.
Final expressions should be algebraic in $M$ , $R$ , $\Phi$ with correct factors (4, 16, etc.).

Techniques used

use V = -GM/r to express GM in terms of a given potentialapply g = GM/r^2 at two different radiiuse U = mV to calculate gravitational potential energy

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 measures temperature via expansion of the liquid (and the glass), which is not directly proportional to thermodynamic temperature and depends on the materials. It must be calibrated using fixed points, so it does not measure thermodynamic temperature directly.

Final answer

It uses liquid expansion (material-dependent, not directly proportional), so it needs calibration and is not a direct thermodynamic temperature measurement.

Detailed explanation

Background Concept

A practical thermometer uses a thermometric property: a physical property that changes with temperature (e.g. length of a liquid column, electrical resistance, emf of a thermocouple). To read a temperature from that property, we must assume a relationship between the property and temperature.

Thermodynamic temperature is the “true” temperature defined by thermodynamics (independent of any particular substance). A practical thermometer generally cannot produce thermodynamic temperature directly; it must be calibrated so that its readings correspond to thermodynamic temperature.

Understanding the Question

You are asked why a laboratory liquid-in-glass thermometer (which uses the length/volume of the liquid column) does not directly measure thermodynamic temperature.

Approach

State what it measures (expansion of liquid) and why that is not the same as thermodynamic temperature (material dependence, non-linearity), so calibration is needed.

Step-by-Step Reasoning

In a liquid-in-glass thermometer, the thermometric property is the volume (or length) of the liquid column.
The expansion of the liquid is not perfectly proportional to thermodynamic temperature across all temperatures. Also, the glass bulb and capillary expand too, affecting the reading.
Because the relationship is not fundamental and depends on the liquid and glass, the thermometer must be calibrated using fixed points (e.g. ice point and steam point) to assign temperatures.
Therefore, the instrument’s reading is a practical temperature scale, not a direct measurement of thermodynamic temperature.

Key Takeaways

Practical thermometers measure a thermometric property, not temperature itself.
Thermodynamic temperature is substance-independent; liquid expansion depends on the liquid and the container.

Common Mistakes

Saying only “it is inaccurate” (too vague; the key idea is material dependence/calibration).
Claiming it “cannot measure kelvin” (it can be calibrated to read in K, but it is still not a direct thermodynamic measurement).

Things to Be Careful About

Make clear it’s not the presence of a scale marking that matters; it’s the assumed relationship between expansion and temperature.
Mentioning calibration or non-linearity/material dependence is usually essential for the mark.

Techniques used

identify the thermometric property used by a liquid-in-glass thermometerexplain the need for calibration against thermodynamic temperaturestate why the scale is not fundamental

(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 $\rho$ increases linearly with $T$ (straight line), so resistance (hence $\rho$ ) has a predictable, repeatable change with temperature and can be calibrated easily. The non-zero gradient means a change in $T$ gives a measurable change in $\rho$ (good sensitivity).

Final answer

ρ varies linearly with T with a clear gradient, giving a predictable, calibratable and sensitive change in resistance.

Detailed explanation

Background Concept

A resistance thermometer works because the electrical resistance of a metal changes with temperature.

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

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

If $L$ and $A$ are fixed, then measuring $R$ is effectively measuring resistivity $\rho$ . A good thermometric property should:

change significantly with temperature (sensitivity),
change in a predictable way (often close to linear),
be repeatable and stable.

Understanding the Question

You are given that the thermometer’s platinum wire is in the environment, and the resistance between terminals $X$ and $Y$ is measured. Fig. 2.2 plots platinum’s resistivity $\rho$ against thermodynamic temperature $T$ and shows a straight-line increase.

The question asks how that graph indicates platinum is suitable for a resistance thermometer.

Approach

Use the graph features that matter: straight line (linear relation) and clear positive gradient (measurable change per kelvin). Connect these to calibration and sensitivity.

Step-by-Step Reasoning

Since the graph of $\rho$ against $T$ is a straight line, $\rho$ is directly proportional to $T$ over the range shown.
Because $R \propto \rho$ (for constant geometry), the resistance also changes linearly with temperature.
A linear relationship makes calibration straightforward: equal temperature intervals correspond to equal resistance intervals.
The positive gradient means that a small change in $T$ produces a change in $\rho$ (and thus $R$ ) that can be detected; this indicates good sensitivity.

Key Takeaways

In resistance thermometers, $R$ is used as the thermometric property.
A linear, significant change of $R$ with $T$ makes a material suitable.

Common Mistakes

Just stating “it is suitable because it is a metal” (doesn’t use the graph evidence).
Talking about the intercept value at $T=0$ without linking it to thermometer performance.

Things to Be Careful About

The question refers to resistivity $\rho$ , but the instrument measures resistance $R$ ; you must link them using $R = \rho L/A$ .
Don’t claim the line must pass through the origin; for metals, $\rho$ at $0\ \text{K}$ need not be zero due to residual resistivity.

Techniques used

interpret a graph of thermometric property against temperaturelink linearity to ease of calibrationcomment on sensitivity using gradient

(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 slow response (large thermal mass / needs time to reach thermal equilibrium with the surroundings), so it cannot follow rapidly changing temperature.

Final answer

Slow response time due to thermal mass/time to reach thermal equilibrium.

Detailed explanation

Background Concept

A thermometer must exchange thermal energy with its surroundings until it reaches thermal equilibrium; only then does its thermometric property correspond to the surroundings’ temperature.

The time taken to respond depends on factors such as:

thermal mass (heat capacity) of the sensing element,
surface area and thermal contact,
thermal conductivity of materials,
insulation and protective casing.

A large thermal mass or poor heat transfer causes a time lag.

Understanding the Question

The platinum resistance thermometer shown is housed in a glass tube and contains a plastic strip with platinum wire wound around it. This construction likely increases the thermal mass and reduces how quickly the platinum wire’s temperature matches the environment.

You must suggest one reason it is not suitable when temperature changes rapidly.

Approach

State that it does not respond instantly because it needs time to come into equilibrium; link this to the physical construction (glass/plastic) giving thermal inertia.

Step-by-Step Reasoning

When the environment temperature changes, the platinum wire must gain/lose thermal energy to change its temperature.
The glass tube and plastic support add thermal mass and can reduce the rate of heat transfer to the wire.
Therefore the wire temperature (and hence resistance) lags behind the true environment temperature.
For rapid changes, the reading is always “behind”, making it unsuitable.

Key Takeaways

Rapidly changing temperature measurements require sensors with small thermal mass and fast heat transfer.
Platinum resistance thermometers can be accurate but may be slow.

Common Mistakes

Saying “resistance changes too slowly” without linking to thermal equilibrium/time lag.
Claiming the resistance change is non-linear (the given graph shows it is linear).

Things to Be Careful About

Don’t confuse electrical response time with thermal response time; the limitation is mainly thermal.
Keep the answer focused on one clear reason for the 1 mark (time lag / slow response).

Techniques used

relate thermometer response to thermal mass and heat transfer rateuse the idea of time lag/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

DifficultyEasy

Worked solution

Answer

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

Final answer

Thermistor resistance decreases with temperature (negative coefficient), unlike platinum which increases (approximately linearly).

Detailed explanation

Background Concept

In resistance thermometers, resistance depends on temperature, but different materials behave differently.

Metals (e.g. platinum): as temperature increases, lattice vibrations increase, electron scattering increases, so resistance typically increases approximately linearly over a wide range.
Thermistors (semiconductors, often metal oxides): as temperature increases, more charge carriers become available, so resistance typically decreases; the variation is usually strongly non-linear (often close to exponential over some range).

Understanding the Question

You are asked for one difference between how the resistance of a negative temperature coefficient (NTC) thermistor varies with temperature compared with a platinum wire.

Approach

Give one clear contrasting statement: either the direction of change (negative vs positive coefficient) or the shape (non-linear vs linear). One is enough.

Step-by-Step Reasoning

For an NTC thermistor, increasing temperature increases the number of free carriers in the semiconductor.
Increased carrier number reduces resistance.
For platinum, increasing temperature increases resistance (approximately linear).

Key Takeaways

Thermistor: resistance decreases with temperature (NTC), usually non-linear.
Platinum wire: resistance increases with temperature, often close to linear.

Common Mistakes

Saying both increase (confuses with metals).
Only saying “they are different” without stating how.

Things to Be Careful About

The question specifies negative temperature coefficient thermistor, so the sign of the change is the most direct, mark-winning difference.
If you mention non-linearity, make sure it is clearly contrasted with platinum’s approximately linear behaviour.

Techniques used

compare the temperature dependence of resistance for two sensorsstate the sign of the temperature coefficientdescribe linear versus non-linear behaviour

Q3MediumIdeal 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 the equation of state

pV = NkT \; (\text{or } pV = nRT)

at all temperatures and pressures, because its molecules occupy negligible volume and exert negligible intermolecular forces (except during elastic collisions).

Final answer

An ideal gas obeys pV = NkT (or pV = nRT) for all temperatures and pressures; molecules have negligible volume and no intermolecular forces (collisions elastic).

Detailed explanation

Background Concept

An ideal gas is a model used to simplify real gases. It is based on kinetic theory, which treats a gas as a very large number of small particles moving randomly.

The key mathematical description is the ideal gas equation of state:

pV = nRT \quad \text{or} \quad pV = NkT

where $p$ is pressure, $V$ is volume, $T$ is thermodynamic temperature, $n$ is number of moles, $N$ is number of molecules, $R$ is the molar gas constant and $k$ is the Boltzmann constant.

Understanding the Question

The question asks for what is meant by an ideal gas (2 marks). Typically, marks are awarded for (1) stating that it obeys the ideal gas equation and (2) stating the physical assumptions behind why it does so (negligible molecular volume and negligible intermolecular forces / elastic collisions).

Approach

Give the defining behaviour (obeys $pV = nRT$ or $pV = NkT$ for all conditions), then add one or more kinetic-theory assumptions that distinguish an ideal gas model from a real gas.

Step-by-Step Reasoning

State the equation of state: $pV = nRT$ (or $pV = NkT$ ).
Add the modelling assumptions:
- molecules are effectively point particles (negligible volume compared with the container),
- no intermolecular forces except during collisions,
- collisions are perfectly elastic.

Any two of these points usually secure full credit.

Key Takeaways

“Ideal gas” means it follows $pV \propto T$ exactly.
The reason is the simplifying kinetic-theory assumptions (no forces, negligible size, elastic collisions).

Common Mistakes

Saying only “it obeys $pV=nRT$ ” without indicating the “all temperatures and pressures” aspect.
Confusing ideal gas with “real gas at low pressure” (that is a condition where real gases approximate ideal behaviour, not the definition).

Things to Be Careful About

Temperature must be on the kelvin scale in the equation of state.
Do not introduce extra conditions like “low pressure only” as part of the definition (that describes when real gases behave ideally).

Techniques used

state the defining property of an ideal gaslink macroscopic gas behaviour 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, a basic kinetic-theory assumption is that intermolecular forces are negligible (except during collisions). Hence there is no significant intermolecular potential energy, so the potential energy associated with the random motion is constant and can be taken as zero.

Final answer

Intermolecular forces are negligible, so intermolecular potential energy is negligible/constant (taken as zero).

Detailed explanation

Background Concept

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

random translational kinetic energy,
rotational/vibrational kinetic energy (for non-monatomic gases),
intermolecular potential energy (due to forces between molecules).

A defining kinetic-theory assumption for an ideal gas is that molecules exert negligible forces on each other except during brief collisions. If there are no forces, there is no energy stored in separation (i.e. no intermolecular potential energy to consider).

Understanding the Question

You are asked to use one basic kinetic-theory assumption to deduce what can be said about the potential energy associated with random molecular motion in an ideal gas.

The intended assumption is: “no intermolecular forces (except during collisions)”. From that, we deduce that intermolecular potential energy is negligible (effectively zero or constant).

Approach

Choose the assumption: negligible intermolecular forces.
Translate “negligible force” into “negligible potential energy change / negligible potential energy associated with separation”.
State the consequence clearly.

Step-by-Step Reasoning

Potential energy between molecules exists only if there are intermolecular forces.
In an ideal gas model, intermolecular forces are ignored.
Therefore, molecules do not store energy in their relative positions; the intermolecular potential energy does not change and is taken as zero.

This is why (for an ideal gas) internal energy is treated as coming from kinetic energy only.

Key Takeaways

No intermolecular forces $\Rightarrow$ no intermolecular potential energy contribution.
For an ideal gas, internal energy is not increased by “separating” molecules (because no energy is stored in that separation).

Common Mistakes

Saying the potential energy is “maximum” or “increases with temperature” (that would require intermolecular forces).
Mixing up gravitational potential energy (macroscopic) with intermolecular potential energy (microscopic).

Things to Be Careful About

The question is about potential energy linked to random motion (i.e. intermolecular potential energy), not about kinetic energy.
You must explicitly connect the assumption (no forces) to the conclusion (no potential energy / constant potential energy).

Techniques used

use kinetic-theory assumptions to infer energy contributionsdistinguish kinetic and potential energy contributions to internal energy

(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)} = 1.3 \times 10^{25}

Answer

N = 1.3 \times 10^{25}

Final answer

1.3 × 10^25

Detailed explanation

Background Concept

For an ideal gas, the macroscopic variables are related by the equation of state:

pV = NkT

where $N$ is the number of molecules and $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ is the Boltzmann constant. This form is useful when the question asks for number of molecules directly.

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 $N$ , the number of molecules.

Approach

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

N = \frac{pV}{kT}

Then substitute the values (all already in SI units).

Step-by-Step Reasoning

Start from the ideal gas equation:

pV = NkT

Rearrange for $N$ :

N = \frac{pV}{kT}

Substitute:

$pV = (2.0 \times 10^5)(0.26) = 5.2 \times 10^4\ \text{Pa m}^3 = 5.2 \times 10^4\ \text{J}$
$kT = (1.38 \times 10^{-23})(290) = 4.00 \times 10^{-21}\ \text{J}$

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 asked for number of molecules.
$\text{Pa m}^3$ is equivalent to joules, which helps check consistency.

Common Mistakes

Using $R$ instead of $k$ without switching to moles ( $n$ ).
Using temperature in °C instead of K.
Dropping powers of ten in $k = 10^{-23}$ .

Things to Be Careful About

Significant figures: data are given to about 2 s.f., so $N$ should be quoted as $1.3 \times 10^{25}$ .
Ensure $V$ is in $\text{m}^3$ (not litres or $\text{cm}^3$ ).

Techniques used

apply the ideal gas equation pV = NkTrearrange an equation to make the required quantity the subjectsubstitute values consistently in SI units

(ii)

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

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Average translational kinetic energy per molecule:

E_K = \frac{3}{2}kT

E_K = \frac{3}{2}(1.38 \times 10^{-23})(290) = 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 random motion of molecules. For an ideal gas, the mean translational kinetic energy per molecule is

\langle E_{\text{trans}} \rangle = \frac{3}{2}kT

This is a key result: temperature measures the average translational kinetic energy of molecules.

Understanding the Question

You are asked to calculate the average translational kinetic energy of one molecule at $T = 290\ \text{K}$ . No other values are needed.

Approach

Use

E_K = \frac{3}{2}kT

and substitute $k = 1.38 \times 10^{-23}\ \text{J K}^{-1}$ and $T = 290\ \text{K}$ .

Step-by-Step Reasoning

Write the relation:

E_K = \frac{3}{2}kT

Substitute values:

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

Multiply:

$(1.38 \times 10^{-23}) \times 290 = 4.00 \times 10^{-21}\ \text{J}$ (approximately)
$1.5 \times 4.00 \times 10^{-21} = 6.0 \times 10^{-21}\ \text{J}$

Key Takeaways

For an ideal gas, mean translational kinetic energy is proportional to absolute temperature.
The factor $3/2$ comes from three translational degrees of freedom.

Common Mistakes

Using $E = kT$ instead of $\frac{3}{2}kT$ .
Forgetting that $T$ must be in kelvin.

Things to Be Careful About

This is energy per molecule, so use $k$ , not $R$ .
Quote the answer to appropriate significant figures and include the unit joule.

Techniques used

use the kinetic-theory result for mean translational kinetic energysubstitute numerical values with correct units

(iii)

the internal energy of the gas. Explain your reasoning.

internal energy = ______ $\text{J}$

2M Similar

DifficultyMedium

Worked solution

Working

For an ideal gas, intermolecular forces are negligible, so internal energy is due to random kinetic energy only:

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

Since $NkT = pV$ ,

U = \frac{3}{2}pV = \frac{3}{2}(2.0 \times 10^5)(0.26) = 7.8 \times 10^4\ \text{J}

Answer

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

Final answer

7.8 × 10^4 J

Detailed explanation

Background Concept

Internal energy $U$ is the total microscopic energy of a system. For a gas, it is usually described as:

the total random kinetic energy of the molecules, plus
the total intermolecular potential energy.

For an ideal gas, a key assumption is that intermolecular forces are negligible. That means the intermolecular potential energy is negligible/constant, so changes in internal energy come only from changes in molecular kinetic energy.

For a (monatomic) ideal gas, the total random translational kinetic energy is

U = \frac{3}{2}NkT

Using $pV = NkT$ , this can also be written as

U = \frac{3}{2}pV

Understanding the Question

You have the same gas sample as in (b): $p = 2.0 \times 10^5\ \text{Pa}$ and $V = 0.26\ \text{m}^3$ and $T = 290\ \text{K}$ . You must find the internal energy and explain why that expression is valid for an ideal gas.

Approach

State the physical reasoning: ideal gas has negligible intermolecular forces, so no potential energy contribution.
Use $U = \frac{3}{2}NkT$ .
Replace $NkT$ with $pV$ for a quicker calculation.

Step-by-Step Reasoning

Explanation part:
- With negligible intermolecular forces, there is no significant potential energy stored between molecules.
- Therefore internal energy is just the total random kinetic energy.
Use the kinetic-theory expression:

U = \frac{3}{2}NkT

Use the ideal gas equation $NkT = pV$ :

U = \frac{3}{2}pV

Substitute $p$ and $V$ :

pV = (2.0 \times 10^5)(0.26) = 5.2 \times 10^4\ \text{J}

Multiply by $\frac{3}{2}$ :

U = 1.5 \times 5.2 \times 10^4 = 7.8 \times 10^4\ \text{J}

Key Takeaways

Ideal gas: potential energy contribution is negligible, so $U$ depends only on $T$ .
Useful identity: for a monatomic ideal gas, $U = \frac{3}{2}pV$ .

Common Mistakes

Saying $U = pV$ (missing the factor $3/2$ ).
Claiming internal energy depends on volume directly for an ideal gas without specifying what is held constant.
Forgetting to explain why potential energy can be ignored.

Things to Be Careful About

The expression $U = \frac{3}{2}NkT$ is for translational kinetic energy (monatomic ideal gas). In A-level ideal-gas questions, this is the intended model when $\frac{3}{2}kT$ has been used.
Keep $p$ in pascals and $V$ in $\text{m}^3$ so that $pV$ comes out in joules.

Techniques used

relate internal energy to total random kinetic energy for an ideal gasuse U = (3/2)NkT and simplify using pV = NkTjustify neglect of potential energy using model assumptions

(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 pressure, $pV = NkT \Rightarrow T \propto V$ (for fixed $N$ ). For an ideal gas, $U \propto T$ , so $U \propto V$ .

Hence $U$ increases linearly with $V$ : a straight line through the origin with positive gradient.

diagram-c-UV

Final answer

Straight line through origin with positive gradient (U ∝ V at constant p).

Detailed explanation

Background Concept

For an ideal gas, internal energy depends only on temperature because intermolecular potential energy is negligible. Using the kinetic theory result (as in part b), for the same gas sample:

U = \frac{3}{2}NkT

So, for fixed $N$ , we have $U \propto T$ .

The ideal gas equation also tells us:

pV = NkT

If $p$ is kept constant and $N$ is fixed (same sample), then $T \propto V$ .

Understanding the Question

You now change the volume $V$ while keeping pressure constant. You must sketch how internal energy $U$ varies with $V$ on axes labelled $U$ (vertical) against $V$ (horizontal).

So the task is to decide whether $U$ is constant, increases linearly, curves, etc., and draw the correct qualitative graph.

Approach

Use $pV = NkT$ with $p$ constant to find how $T$ depends on $V$ .
Use the fact that for an ideal gas $U \propto T$ .
Combine these to get $U$ as a function of $V$ and sketch.

Step-by-Step Reasoning

Start with the ideal gas equation:

pV = NkT

Pressure is constant, and it is the same gas sample, so $N$ is constant. Rearranging:

T = \frac{pV}{Nk}

So:

T \propto V

For an ideal gas (using the same model as part (b)):

U = \frac{3}{2}NkT \Rightarrow U \propto T

Combine $U \propto T$ with $T \propto V$ :

U \propto V

Therefore, as $V$ increases, $U$ increases in direct proportion: the $U$ – $V$ graph is a straight line with positive slope passing through the origin (because $V=0$ would imply $T=0$ and hence $U=0$ in this idealised model).

diagram-c-UV

Key Takeaways

For an ideal gas, $U$ depends only on $T$ .
At constant $p$ , $T \propto V$ , so $U \propto V$ .
Sketching graphs often comes from chaining proportionalities.

Common Mistakes

Drawing $U$ constant with $V$ (that would be true if temperature were held constant, not pressure).
Drawing an inverse curve (confusing constant pressure with constant temperature or constant $pV$ ).
Drawing a straight line not through the origin without justification.

Things to Be Careful About

“Keeping its pressure constant” means $p$ fixed, not $pV$ fixed.
This is the same sample of gas: $N$ is constant.
A sketch mark is usually for the correct shape (straight line) and correct intercept behaviour (through the origin).

Techniques used

use pV = NkT to relate temperature to volume at constant pressureuse U \propto T for an ideal gassketch a straight-line graph with correct qualitative features

Q4MediumOscillations

(a)

State what is meant by resonance.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

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

Final answer

Resonance occurs when the driving frequency equals the natural frequency, giving maximum energy transfer and maximum amplitude.

Detailed explanation

Background Concept

Resonance is a feature of forced oscillations. A system that can oscillate (mass–spring, pendulum, vibrating string, etc.) has a natural frequency: the frequency at which it oscillates if displaced and released.

If the system is instead driven by an external periodic force (a “driver”) of frequency $f$ , the system responds by oscillating at (approximately) the driver’s frequency. The amplitude of that response depends strongly on how close $f$ is to the natural frequency.

Understanding the Question

The question asks for the meaning of resonance (2 marks). That normally requires:

the condition: driving frequency equals natural frequency
the effect: maximum amplitude (equivalently, maximum power/energy transfer from the driver to the oscillator)

Approach

Give a definition that includes both:

matching of driving and natural frequencies
resulting large/maximum amplitude due to maximum energy transfer.

Step-by-Step Reasoning

A driver supplies energy each cycle.
When $f$ equals the natural frequency, the driver’s force is in the correct timing (phase relationship) to add energy to the oscillator most effectively each cycle.
Therefore the rate of energy transfer is greatest, so the amplitude reaches a maximum value (limited in practice by damping).

Key Takeaways

Resonance happens in forced oscillations.
Condition: $f_{\text{drive}} = f_0$ .
Consequence: maximum amplitude / maximum energy transfer.

Common Mistakes

Saying only “large amplitude” without stating why/when (missing the frequency condition).
Saying “frequency becomes maximum” (frequency is set by the driver; it’s the amplitude that peaks).

Things to Be Careful About

In real systems, damping prevents the amplitude from becoming infinite; it is just a maximum at resonance.

Techniques used

state the condition for resonance in a driven oscillatorlink 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 / internal friction in the string) do work on the ball, so mechanical energy is transferred to the surroundings (as heat/sound). Hence the energy of oscillation decreases and the amplitude falls with time.

Final answer

Resistive forces dissipate energy to the surroundings, so the oscillation energy decreases and the amplitude falls.

Detailed explanation

Background Concept

For an oscillator, the total mechanical energy is the sum of kinetic and potential energy. For ideal SHM with no resistive forces, this energy stays constant and so does the amplitude.

With damping, a resistive force (often approximately opposite to velocity) removes energy:

the resistive force does negative work on the oscillator
energy is transferred to the surroundings, usually as thermal energy (and sometimes sound)

As energy decreases, the maximum displacement achievable (the amplitude) must also decrease.

Understanding the Question

The ball on a stretched string oscillates vertically after being released. The graph shows that the peak displacements get smaller as time goes on.

You are asked to explain why the amplitude decreases with time (2 marks): you need to mention energy loss (cause) and dissipation to surroundings (effect).

Approach

Explain damping in energy terms:

identify resistive forces present
state these forces remove mechanical energy each cycle
conclude less energy means smaller amplitude.

Step-by-Step Reasoning

As the ball moves, it experiences forces such as air resistance and internal friction/energy loss in the stretched string and at any attachment points.
These forces oppose the motion, so they do negative work on the ball.
Therefore the mechanical energy of the oscillation decreases each cycle.
The lost energy is transferred to the surroundings (mainly thermal energy, possibly some sound).
With less total energy, the oscillator cannot reach as large a maximum displacement, so the amplitude decays with time.

Key Takeaways

Damping = resistive forces remove energy from an oscillator.
Decreasing energy of oscillation implies decreasing amplitude.

Common Mistakes

Saying only “there is friction” without linking to energy transfer and amplitude.
Claiming the frequency decreases significantly; in light damping the frequency changes only slightly, but amplitude falls noticeably.

Things to Be Careful About

Use the idea of energy dissipation (to heat/sound) for clear, mark-scheme-style explanation.
Don’t confuse decreasing amplitude with decreasing period.

Techniques used

link decreasing amplitude to energy loss per cycleidentify resistive forces and the form of energy dissipation

(iii)

Determine the frequency of the oscillations of the ball.

frequency = ______ $\text{Hz}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 4.2, time between peaks:

T \approx 0.37 - 0.12 = 0.25\ \text{s}

(also $0.62-0.37=0.25\ \text{s}$ ).

f = \frac{1}{T} = \frac{1}{0.25} = 4.0\ \text{Hz}

Answer

$4.0\ \text{Hz}$

Final answer

4.0 Hz

Detailed explanation

Background Concept

For periodic oscillations:

the period $T$ is the time for one complete cycle
the frequency $f$ is the number of cycles per second

They are related by:

f = \frac{1}{T}

On a displacement–time graph, $T$ can be found by measuring the time between identical points on successive cycles (e.g. peak to peak).

Understanding the Question

The graph shows a damped oscillation, but it is still periodic. You need the frequency of the oscillation.

The given peaks are at approximately $t=0.12\ \text{s}$ , $0.37\ \text{s}$ and $0.62\ \text{s}$ , which allows you to find $T$ .

Approach

Find the period $T$ from the time separation between successive maxima.
Calculate $f=1/T$ .
Use a consistent value (or average) to reduce reading uncertainty.

Step-by-Step Reasoning

First period estimate:

T_1 = 0.37 - 0.12 = 0.25\ \text{s}

Second period estimate:

T_2 = 0.62 - 0.37 = 0.25\ \text{s}

These agree, so take $T \approx 0.25\ \text{s}$ .
Convert to frequency:

f = \frac{1}{0.25} = 4.0\ \text{Hz}

Key Takeaways

Read $T$ from repeating features (peaks/troughs) on a time graph.
Then use $f=1/T$ .

Common Mistakes

Using the time from $t=0$ to the first peak as the period (that is typically $T/4$ for a sine-like motion).
Reading peak positions inconsistently (e.g. peak to trough is $T/2$ ).

Things to Be Careful About

Damping reduces amplitude, but the period usually stays approximately constant (especially for light damping).
Use more than one cycle when possible to improve reliability (here, two intervals match well).

Techniques used

read the period from successive peaks on a displacement-time graphuse $ f = 1/T $ to find frequency

(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

Worked solution

Answer

Sketch a resonance curve: small amplitude at low $f$ , rising to a maximum at the natural frequency ( $f \approx 4\ \text{Hz}$ ), then decreasing at higher $f$ (peak is finite due to damping).

diagram-4c

Final answer

Resonance curve with a maximum at about 4 Hz and lower amplitudes at frequencies below and above this.

Detailed explanation

Background Concept

In forced oscillations, an oscillator is driven by an external periodic force of frequency $f$ .

The steady (long-term) amplitude depends on $f$ :

far below or far above the natural frequency, the response amplitude is small
near the natural frequency $f_0$ , energy transfer from the driver is most effective, so the amplitude is largest (resonance)
with damping, the resonance peak is finite and broader (not infinitely tall)

Understanding the Question

You switch on the vibration generator and slowly increase its frequency from $0$ to $10\ \text{Hz}$ . You must sketch how the ball’s oscillation amplitude varies with $f$ .

From part (b)(iii), the ball’s natural frequency is about $4\ \text{Hz}$ , so the maximum on the graph should occur near $4\ \text{Hz}$ .

Approach

Use the standard amplitude–frequency response for a driven, damped oscillator.
Place the resonance peak at $f \approx 4\ \text{Hz}$ .
Ensure the curve is low at very small and very large $f$ .

Step-by-Step Reasoning

At $f \to 0$ , the driver changes very slowly, so the system does not build up large oscillations → small amplitude.
As $f$ increases towards $f_0 \approx 4\ \text{Hz}$ , the driver supplies energy in step with the motion → amplitude increases.
At $f = f_0$ , resonance occurs → maximum amplitude.
For $f > f_0$ , the motion becomes increasingly out of phase with the driving force and energy transfer is less effective → amplitude decreases.
Because damping is present (seen in part b), the resonance peak should not be infinitely sharp; it has a rounded top.

diagram-4c

Key Takeaways

Resonance curve: amplitude peaks at the natural frequency.
Damping limits the maximum amplitude and broadens the peak.

Common Mistakes

Drawing the maximum at $10\ \text{Hz}$ (must be near the measured natural frequency $\approx 4\ \text{Hz}$ ).
Drawing a straight line increase with $f$ (the response must fall after resonance).
Drawing an infinitely tall spike (real systems are damped).

Things to Be Careful About

The sketch is qualitative, but the peak position should be consistent with part (b)(iii).
Ensure axes are used correctly: amplitude on vertical axis, frequency $f$ on horizontal axis.

Techniques used

relate response amplitude to driving frequency in forced oscillationsidentify the natural frequency from earlier datasketch a resonance curve with finite peak due to damping

Q5MediumElectric FieldsMagnetic Fields

(a)

Define electric field.

2M Similar

(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

DifficultyMedium-Easy

Worked solution

diagram-b-i

Answer

Four straight, parallel, equally spaced field lines perpendicular to the plates, with arrows from the $+430\ \text{V}$ plate to the $0\ \text{V}$ plate.

Final answer

Uniform field: straight parallel equally spaced lines from +430 V plate to 0 V plate.

Detailed explanation

Background Concept

Electric field lines are a visual way to represent an electric field:

The direction of the field line shows the direction of force on a positive test charge.
The spacing indicates relative field strength (closer = stronger).
Between large parallel plates (away from edges) the field is approximately uniform, so the lines are straight, parallel, and equally spaced.

Field lines go from positive to negative, and also from higher potential to lower potential.

Understanding the Question

The diagram shows two parallel plates, top at $+430\ \text{V}$ and bottom at $0\ \text{V}$ . You must draw four field lines between them.

Approach

Since the field between parallel plates is uniform:

draw straight lines,
make them perpendicular to the plates,
include arrowheads showing direction from the positive plate to the lower potential plate.

Step-by-Step Reasoning

Identify which plate is at higher potential: the top plate is $+430\ \text{V}$ .
Electric field direction is from higher potential to lower potential (same as from positive to negative): top to bottom.
Because it is uniform between parallel plates, draw several straight, parallel lines, equally spaced.

diagram-b-i

Key Takeaways

Uniform field between parallel plates: straight, parallel, equally spaced lines.
Direction: from $+$ plate (higher $V$ ) to lower $V$ .

Common Mistakes

Drawing curved lines between plates (suggests non-uniform field).
Forgetting arrowheads.
Drawing arrows from the $0\ \text{V}$ plate to the $+430\ \text{V}$ plate.

Things to Be Careful About

Keep lines perpendicular to plates.
Do not bunch lines together in the middle (would imply stronger field there).

Techniques used

use field-line conventions to show directionrepresent a uniform field with parallel, equally spaced linesshow field direction from higher potential to lower 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

For uniform field,

E = \frac{V}{d} = \frac{430}{6.7 \times 10^{-2}} = 6.42 \times 10^{3}

Answer

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

Final answer

6.4 × 10^3 N C^-1

Detailed explanation

Background Concept

Between parallel plates (ignoring edge effects), the electric field is uniform. The field strength is related to potential difference by

E = \frac{V}{d}

where:

$V$ is the potential difference across the plates,
$d$ is the separation between plates.

Units: $E$ is in $\text{N C}^{-1}$ (equivalently $\text{V m}^{-1}$ ).

Understanding the Question

You are given:

$V = 430\ \text{V}$
separation $d = 6.7\ \text{cm} = 6.7 \times 10^{-2}\ \text{m}$

You must find $E$ between the plates.

Approach

Use the uniform-field relation $E=V/d$ , ensuring $d$ is in metres.

Step-by-Step Reasoning

Convert distance:

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

Substitute into $E=V/d$ :

E = \frac{430}{6.7 \times 10^{-2}} = 6.42 \times 10^{3}\ \text{V m}^{-1}

Quote with appropriate significant figures (limited by 2 s.f. in $6.7$ ):

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

Key Takeaways

For parallel plates: $E=V/d$ .
Always convert $\text{cm}$ to $\text{m}$ .

Common Mistakes

Using $d=6.7$ instead of $0.067$ (gives an answer 100 times too small).
Omitting units.

Things to Be Careful About

$\text{N C}^{-1}$ and $\text{V m}^{-1}$ are equivalent for electric field strength.
Significant figures should usually follow the least precise given data.

Techniques used

apply $E = V/d$ for a uniform electric fieldconvert centimetres to metrescalculate and round to appropriate significant figures

(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 upwards between the plates (towards the $+430\ \text{V}$ plate), then continues beyond the plates in a straight line along the tangent to the curve at exit.

Final answer

Curves upward between plates, then straight-line continuation at exit.

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 an electron ( $q=-e$ ), the force is opposite to $E$ .

In a uniform field, the force (and hence acceleration) is constant in direction and magnitude, so the path while inside the plates is a parabola (like projectile motion) if the initial velocity has a component perpendicular to the acceleration.

Understanding the Question

The electron enters between the plates moving horizontally. The field between the plates is downward (from $+430\ \text{V}$ to $0\ \text{V}$ ). You must sketch the electron’s path between and beyond the plates.

Approach

Use the field direction to find the force direction on an electron.
Constant sideways (horizontal) velocity + constant vertical acceleration gives a curved (parabolic) path.
Once outside the plates, the field is (approximately) zero, so it travels in a straight line at constant velocity along the exit direction.

Step-by-Step Reasoning

Field lines point downward from the top plate to the bottom plate.
Electron has charge $-e$ , so electric force is upward (opposite to $E$ ), i.e. toward the positive top plate.
Therefore, as it moves to the right, it accelerates upward, producing a curve upward.
After it leaves the region between the plates, there is no longer a significant electric force, so it continues in a straight line tangent to the path at the exit.

diagram-b-iii

Key Takeaways

Direction of force depends on the sign of charge.
Uniform field gives constant acceleration, so trajectories are parabolic in the field region.

Common Mistakes

Curving the path downward (forgetting the electron is negative).
Drawing a curved path even after the electron has left the plates (there should be no force then).

Things to Be Careful About

Keep the sketch realistic: gentle curvature (not required to be to scale) and a straight-line continuation after exit.
The electron is attracted toward the positive plate.

Techniques used

determine force direction on a negative charge relative to field directioninfer acceleration direction from electric forcesketch trajectory: curved in field, straight beyond

(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

Worked solution

diagram-c-i

Answer

Magnetic field is into the page (so that the magnetic force on the electron is downward, opposing the upward electric force).

Final answer

Into the page.

Detailed explanation

Background Concept

A moving charge in a magnetic field experiences a magnetic force

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

Key points:

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

Understanding the Question

In part (b), the electron was deflected by the electric field. Now a uniform magnetic field is applied so that the electron travels undeviated through the region. You need the direction of $\vec{B}$ .

The electron velocity is to the right. The electric field between the plates points downward, so the electric force on an electron is upward. Therefore the magnetic force must be downward to cancel it.

Approach

Determine the direction of electric force on the electron.
Require magnetic force to be opposite.
Use $\vec{F}_B = q(\vec{v} \times \vec{B})$ with $q=-e$ to choose $\vec{B}$ direction (into/out of page).

Step-by-Step Reasoning

Electric field $\vec{E}$ is downward (from +430 V to 0 V).
Electric force on electron:

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

Since $q=-e$ , $\vec{F}_E$ is upward.
3. For undeviated motion, need $\vec{F}_B$ downward.
4. With $\vec{v}$ to the right, choose $\vec{B}$ so that $\vec{v} \times \vec{B}$ points upward (then the electron’s force, opposite to that, is downward). This occurs when $\vec{B}$ is into the page.

diagram-c-i

Key Takeaways

Always include the sign of the charge when using $\vec{F}_B = q(\vec{v} \times \vec{B})$ .
For straight-line undeflected motion: magnetic and electric forces must be equal and opposite.

Common Mistakes

Forgetting the electron is negative and giving the opposite $\vec{B}$ direction.
Mixing up “into the page” and “out of the page”.

Things to Be Careful About

State direction clearly using standard symbols: crosses ( $\times$ ) for into the page, dots ( $\bullet$ ) for out of the page.
Ensure the magnetic force opposes the electric force, not the field direction itself.

Techniques used

use Fleming’s left-hand rule / cross product for $\vec{F} = q\vec{v} \times \vec{B}$account for the negative charge of the electronchoose magnetic field direction to oppose the electric force

(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

The electric field exerts a force on the electron upward ( $F_E = eE$ ). The magnetic field exerts a force downward ( $F_B = evB$ ).

Since these forces are equal in magnitude and opposite in direction, resultant force is zero, so there is no acceleration and the electron is undeviated.

Final answer

Electric and magnetic forces are equal and opposite, so resultant force is zero and no deflection.

Detailed explanation

Background Concept

An electron in crossed electric and magnetic fields can experience two forces:

Electric force:

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

Magnetic force (if moving):

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

If these forces are equal and opposite, the net force is zero, so acceleration is zero and the particle continues in a straight line at constant velocity.

Understanding the Question

The electron previously curved due to the electric field. Now a magnetic field is applied such that it travels straight through the region. You must explain this by referring to the forces from each field.

Approach

State:

direction of the electric force on the electron,
direction of the magnetic force on the electron,
that they balance (equal magnitude, opposite direction), so the resultant force is zero and no deflection occurs.

Step-by-Step Reasoning

The electric field between the plates is downward. For an electron ( $q=-e$ ), the electric force is upward.
With the chosen magnetic field direction, the magnetic force on the moving electron acts downward.
When adjusted so that

F_E = F_B

the resultant force is

\vec{F}_{\text{resultant}} = \vec{F}_E + \vec{F}_B = 0

Zero resultant force means zero acceleration (Newton’s 2nd law), so the electron’s velocity direction does not change: it is undeviated.

Key Takeaways

Undeflected motion in crossed fields requires zero net force.
Balance condition is $F_E = F_B$ .

Common Mistakes

Saying “fields cancel” instead of “forces cancel” (the fields do not cancel; they both exist).
Forgetting that magnetic force depends on velocity.

Things to Be Careful About

Make clear you are talking about forces on the electron.
Mention both magnitude (equal) and direction (opposite).

Techniques used

identify electric and magnetic forces on a moving chargecompare directions of forces from the two fieldsapply the condition for undeviated motion: resultant force zero

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

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

Answer

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

Final answer

2.5 × 10^-4 T

Detailed explanation

Background Concept

In crossed fields, if $\vec{E}$ is perpendicular to $\vec{v}$ and $\vec{B}$ is perpendicular to $\vec{v}$ , the magnitudes of the forces are:

F_E = eE

and (when $\vec{v} \perp \vec{B}$ ):

F_B = evB

For no deflection, these must balance:

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

Understanding the Question

You already found the electric field strength between the plates in (b)(ii), and you are given the electron speed $v = 2.6 \times 10^{7}\ \text{m s}^{-1}$ . You must calculate the magnetic flux density $B$ that makes the electron travel straight.

Approach

Use the balance condition $eE = evB$ (forces equal and opposite), cancel $e$ , and substitute values into $B=E/v$ .

Step-by-Step Reasoning

Write the force balance for undeflected motion:

eE = evB

Cancel the common factor $e$ :

B = \frac{E}{v}

Substitute $E \approx 6.4 \times 10^{3}\ \text{N C}^{-1}$ and $v = 2.6 \times 10^{7}\ \text{m s}^{-1}$ :

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

Quote to 2 s.f.:

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

Key Takeaways

For a velocity selector (no deflection): $B=E/v$ .
Unit of magnetic flux density is the tesla (T).

Common Mistakes

Using $B = v/E$ instead of $E/v$ .
Forgetting that $F_B = qvB$ only when $v \perp B$ .
Missing the unit T.

Things to Be Careful About

Use the speed in $\text{m s}^{-1}$ .
Keep consistent significant figures with $E$ and $v$ .

Techniques used

set electric and magnetic forces equal for undeviated motionuse $B = E/v$ for perpendicular fieldssubstitute values and include 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 p.d. across the resistor is initially $12\ \text{V}$ so the current is initially maximum.

As the capacitor discharges, its charge and p.d. fall, so the p.d. across the resistor falls and hence $I = V/R$ decreases.

Since the capacitor discharge is exponential, $I$ decreases exponentially towards zero (rate of decrease becomes smaller with time).

Final answer

Current decays exponentially to zero because the capacitor voltage (and hence p.d. across R) decreases exponentially during discharge.

Detailed explanation

Background Concept

For a capacitor discharging through a resistor:

The capacitor voltage $V_C$ is related to charge by $Q = CV_C$ .
The current in the circuit is related to how fast charge leaves the capacitor:

I = -\frac{\mathrm{d}Q}{\mathrm{d}t}

The discharge obeys an exponential law:

Q = Q_0 e^{-t/RC}

So the voltage also falls exponentially:

V_C = V_0 e^{-t/RC}

Because the resistor is the only resistance, the current is

I = \frac{V_R}{R}

and in this simple series discharge, $V_R = V_C$ , so

I = \frac{V_0}{R} e^{-t/RC} = I_0 e^{-t/RC}.

Understanding the Question

Initially, the capacitor has $12\ \text{V}$ across it. When the switch is closed at $t=0$ , the capacitor starts to discharge through $R$ , producing a current in the resistor. The graph shows that current starts at some value and then falls with time.

You are asked to explain why the graph has that shape (why it is not a straight line, and why it approaches zero).

Approach

Connect the graph shape to how the capacitor’s p.d. changes during discharge:

Use the fact that the resistor current is set by the p.d. across it: $I = V/R$ .
Recognise that the p.d. across the resistor comes from the capacitor, and that $V_C$ decreases as the capacitor discharges.
Recall that the discharge of a capacitor through a resistor is exponential, so $I$ must also fall exponentially.

Step-by-Step Reasoning

At the instant the switch is closed, the capacitor still has the full $12\ \text{V}$ across it. That means the p.d. driving current through the resistor is largest at $t=0$ , so the current is largest at $t=0$ .
As current flows, charge leaves the capacitor plates. Since $Q = CV$ , the charge decrease implies the capacitor voltage $V_C$ decreases.
The current at any time is $I = V_R/R$ . But the resistor’s p.d. is supplied by the capacitor during discharge, so $V_R = V_C$ .
Therefore, as $V_C$ decreases, $I$ decreases.
The key point for the shape is that $V_C$ (and $Q$ ) does not decrease linearly: it decreases exponentially with time constant $\tau = RC$ . Hence current follows

I = I_0 e^{-t/RC}

so the graph is a steep drop at first and then it flattens off, approaching zero but never going negative.

Key Takeaways

In a discharging $RC$ circuit, $Q$ , $V_C$ , and $I$ all decay exponentially with the same time constant $\tau = RC$ .
The current is largest initially because the capacitor p.d. is initially largest.
The current tends to zero because eventually the capacitor becomes nearly uncharged, so there is almost no p.d. to drive current.

Common Mistakes

Saying “current decreases because resistance increases” (the resistance is constant).
Claiming the current becomes zero at a fixed finite time; exponential decay approaches zero asymptotically.
Forgetting to link the decreasing current to the decreasing capacitor p.d. via $I = V/R$ .

Things to Be Careful About

Distinguish between why current falls (voltage falls as capacitor discharges) and why it is exponential (the governing differential equation leads to an exponential form).
Avoid describing it as a straight-line decrease; the curve’s changing gradient is essential.

Techniques used

relate current to potential difference using Ohm's lawuse the exponential discharge relationship for a capacitorlink the decreasing capacitor charge/voltage to 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, at $t=0$ , $I_0 = 0.13\ \text{mA} = 1.3 \times 10^{-4}\ \text{A}$ .

Initially $V = 12\ \text{V}$ , so

R = \frac{V}{I_0} = \frac{12}{1.3 \times 10^{-4}} = 9.23 \times 10^{4}\ \Omega.

Answer

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

Final answer

9.2 × 10^4 Ω

Detailed explanation

Background Concept

At any instant in an $RC$ discharge, the resistor current is set by Ohm’s law:

I = \frac{V_R}{R}.

At $t=0$ (just after closing the switch), the capacitor still holds its initial p.d., so the resistor initially has the full driving p.d. across it.

Understanding the Question

You are told the capacitor initially has $12\ \text{V}$ across it. The graph provides the current at $t=0$ (the initial current). Using those two pieces of information, you can find $R$ from $R = V/I$ .

Approach

Read $I_0$ from the graph at $t=0$ .
Convert mA to A.
Use $R = V/I_0$ with $V = 12\ \text{V}$ .

Step-by-Step Reasoning

From the graph, the y-intercept at $t=0$ is $I_0 = 0.13\ \text{mA}$ .
Convert to amperes:

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{V}{I_0} = \frac{12}{1.3 \times 10^{-4}} = 9.23 \times 10^{4}\ \Omega.

Quote to an appropriate s.f. based on the graph reading: $9.2 \times 10^{4}\ \Omega$ .

Key Takeaways

The initial current in a discharge is $I_0 = V_0/R$ .
Always convert mA to A before using SI equations.

Common Mistakes

Using $I$ at a later time instead of $I_0$ .
Forgetting the $10^{-3}$ when converting mA to A.
Using $R = I/V$ instead of $R = V/I$ .

Things to Be Careful About

Graph readings usually justify only 2 s.f.
Ensure you use the initial capacitor p.d. ( $12\ \text{V}$ ) at $t=0$ , not a later smaller value.

Techniques used

read the initial value from a graphapply Ohm's law to relate potential difference, current and resistanceconvert between mA and A

(ii)

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

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

3M Similar

DifficultyMedium

Worked solution

Working

Initial current $I_0 = 0.13\ \text{mA}$ .

At $t = \tau$ ,

I = \frac{I_0}{e} = \frac{0.13}{e} = 0.0478\ \text{mA} \approx 0.048\ \text{mA}.

From Fig. 6.2, the time when $I \approx 0.048\ \text{mA}$ is $t \approx 3.0\ \text{s}$ .

Answer

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

Final answer

3.0 s

Detailed explanation

Background Concept

For exponential decay of current in an $RC$ discharge:

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

The time constant is

\tau = RC.

A very useful definition is: at $t = \tau$ , the quantity has fallen to $1/e$ of its initial value:

I(\tau) = \frac{I_0}{e} \approx 0.368 I_0.

Understanding the Question

The graph gives $I$ against $t$ for a discharging capacitor. You must extract the circuit time constant $\tau$ from that curve.

Approach

Read the initial current $I_0$ from the graph.
Calculate $I_0/e$ (or $0.37 I_0$ ).
On the graph, find the time at which the current equals this value; that time is $\tau$ .

Step-by-Step Reasoning

From the y-intercept at $t=0$ , $I_0 = 0.13\ \text{mA}$ .
Compute the current at one time constant:

I(\tau) = \frac{I_0}{e} = \frac{0.13}{2.718} = 0.0478\ \text{mA} \approx 0.048\ \text{mA}.

Locate $I = 0.048\ \text{mA}$ on the vertical axis, draw a horizontal line to the decay curve, then drop vertically to the time axis to read $t$ .
From the provided curve, this occurs at about $t = 3.0\ \text{s}$ , so

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

(Any close value consistent with a careful graph reading would typically be credited.)

Key Takeaways

For any exponential decay $x = x_0 e^{-t/\tau}$ , the time constant is the time to fall to $x_0/e$ .
Reading $\tau$ from a graph usually means using the $1/e$ (or 0.37) rule.

Common Mistakes

Using $I_0/2$ instead of $I_0/e$ .
Mixing up $e$ with $10$ (i.e. treating the decay as base-10).
Reading the time where $I$ is almost zero rather than applying the defined $1/e$ criterion.

Things to Be Careful About

Keep units consistent (here mA throughout is fine, since you are only scaling by $1/e$ ).
Don’t round $I_0/e$ too aggressively before locating it on the graph; it can shift the time reading noticeably.
Ensure you start from the correct $I_0$ at exactly $t=0$ .

Techniques used

use the definition of time constant from an exponential decaycalculate the 1/e fraction of an initial valueread the corresponding time from a graph

(c)

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

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

C = \frac{3.0}{9.2 \times 10^{4}} = 3.26 \times 10^{-5}\ \text{F}.

Answer

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

Final answer

3.3 × 10^-5 F

Detailed explanation

Background Concept

The time constant $\tau$ of an $RC$ circuit is defined as

\tau = RC.

It sets the timescale for the exponential discharge:

I = I_0 e^{-t/\tau}.

Once you know $R$ and $\tau$ , you can find the capacitance using algebra.

Understanding the Question

You have already determined (from part (b)):

the resistance $R$ using $R = V/I_0$
the time constant $\tau$ using the $1/e$ point on the decay graph.

Now you must combine them to obtain $C$ .

Approach

Rearrange $\tau = RC$ to make $C$ the subject:

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

Then substitute the numerical values with units.

Step-by-Step Reasoning

Start with

\tau = RC.

Rearrange for $C$ :

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

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

C = \frac{3.0}{9.2 \times 10^{4}}\ \text{F} = 3.26 \times 10^{-5}\ \text{F}.

Round appropriately (limited by graph readings):

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

Key Takeaways

The link between the graph (time constant) and circuit component values is $\tau = RC$ .
Capacitance can be found once $R$ and $\tau$ are known.

Common Mistakes

Using $C = R/\tau$ instead of $C = \tau/R$ .
Forgetting that $\Omega\,\text{F} = \text{s}$ , so the units must come out as farads.
Using an inconsistent $R$ or $\tau$ (e.g. mixing a later current value with $V=12\ \text{V}$ ).

Things to Be Careful About

Significant figures: $R$ and $\tau$ are obtained from a graph, so 2 s.f. is usually appropriate.
Ensure $R$ is in ohms (not k\Omega) when used in the calculation unless you explicitly convert.

Techniques used

use the time constant relation tau = RCrearrange an equation to make capacitance the subjectsubstitute numerical values with correct powers of ten

Q7MediumAlternating Currents

A circuit contains a power supply that provides a sinusoidal alternating input voltage $V_{\text{IN}}$ . There is an output voltage $V_{\text{OUT}}$ 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

DifficultyEasy

Worked solution

Answer

To rectify the a.c. input, giving a unidirectional (pulsating d.c.) output across $R$ (full-wave rectification).

Final answer

To rectify the a.c. input, giving a unidirectional (pulsating d.c.) output across R (full-wave rectification).

Detailed explanation

Background Concept

A bridge rectifier uses four diodes arranged so that, for both halves of an a.c. input, the current through the load resistor flows in the same direction. The output is therefore not a steady d.c. value, but a pulsating d.c. waveform (the negative half-cycles are flipped to be positive).

Understanding the Question

You are shown a bridge rectifier circuit with an a.c. input voltage $V_{\text{IN}}$ and an output voltage $V_{\text{OUT}}$ across a load resistor $R$ . The question asks what the circuit is for (i.e. what it does to the input).

Approach

Recognise the diode bridge arrangement and state its standard function: converting a.c. to a one-direction (unidirectional) output across the load.

Step-by-Step Reasoning

In one half-cycle of the a.c. input, two diodes conduct and route current through $R$ in (say) the left-to-right direction.
In the opposite half-cycle, the other two diodes conduct, but the circuit routes current through $R$ in the same left-to-right direction again.
Therefore, the output p.d. across $R$ keeps the same polarity: this is full-wave rectification.

Key Takeaways

A bridge rectifier produces a full-wave rectified (pulsating d.c.) output.
“Unidirectional current through the load” is the key phrase.

Common Mistakes

Saying it “smooths” the signal (that needs a capacitor/reservoir, not shown here).
Saying it “increases voltage” (it does not; ideal diodes only redirect current).

Things to Be Careful About

Use correct language: rectifies a.c. to d.c. (more precisely, pulsating d.c.).
If asked for “full-wave” explicitly, include it.

Techniques used

identify the function of a bridge rectifier circuitrelate rectification to the sign of the output potential difference

(b)

Fig. 7.2 shows the variation of $V_{\text{OUT}}$ 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, $V_{\text{OUT,max}} = 9.0\ \text{V}$ .

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

P_{\text{max}} = \frac{(9.0)^2}{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, the instantaneous power converted to thermal energy is

P = IV

Using Ohm's law $V = IR$ , we can write power in forms that use either $I$ or $V$ :

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

When the voltage varies with time, the instantaneous power varies with time too, and the maximum power occurs when the magnitude of the voltage is at its maximum.

Understanding the Question

From the full-wave rectified output graph (Fig. 7.2), you are given a peak output voltage across the resistor $R = 370\ \Omega$ . You must show the maximum power dissipated is $0.22\ \text{W}$ .

Approach

Read the peak output voltage from the graph.
Use the resistor power relation $P = V^2/R$ with $V = V_{\text{max}}$ .
Evaluate and round appropriately.

Step-by-Step Reasoning

The graph indicates a peak output voltage of $9.0\ \text{V}$ . This is the maximum instantaneous voltage across the resistor.
For a resistor,

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

Substitute:

P_{\text{max}} = \frac{(9.0\ \text{V})^2}{370\ \Omega} = \frac{81}{370}\ \text{W} = 0.219\ \text{W}

To 2 s.f., $0.219\ \text{W} \approx 0.22\ \text{W}$ .

Key Takeaways

Maximum power in a resistor corresponds to maximum instantaneous voltage: $P_{\text{max}} = V_{\text{max}}^2/R$ .
Always include the unit (watts).

Common Mistakes

Using $P = VI$ but not finding $I$ correctly.
Using $R/V^2$ instead of $V^2/R$ .
Reading the wrong vertical value (e.g. using an average instead of the peak).

Things to Be Careful About

Ensure you use the peak value from the graph, not an rms value.
Quote the final power to sensible significant figures consistent with $9.0\ \text{V}$ .

Techniques used

use the resistor power equation in terms of potential differenceread a peak value from a waveform graphsubstitute into an equation and evaluate with correct unit

(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_{\text{OUT}}^2}{R}

So $P$ is always positive, is zero when $V_{\text{OUT}}=0$ , and has maximum value $0.22\ \text{W}$ when $V_{\text{OUT}}=9.0\ \text{V}$ .

diagram-7bii

Answer

Sketch as shown: repeating $\sin^2$ -shaped pulses with peak $0.22\ \text{W}$ and the same zeros/timing as $V_{\text{OUT}}$ .

Final answer

See sketch: P ∝ V_OUT^2, always positive, peaks at 0.22 W with the same period as V_OUT.

Detailed explanation

Background Concept

For a resistor,

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

Squaring has two crucial effects:

Negative voltages produce the same power as positive voltages of the same magnitude.
The shape becomes more “peaked” because squaring makes values near the peak relatively larger compared with values near zero.

If $V(t)$ is a (rectified) sine wave, then $P(t)$ is proportional to $\sin^2$ .

Understanding the Question

You are given $V_{\text{OUT}}$ versus time for a full-wave rectified output. You must sketch the power dissipated in the load resistor $R$ versus time on the provided axes.

Key information inherited from the stem and earlier part:

$V_{\text{OUT}}$ is a sequence of positive half-sine pulses (full-wave rectified).
Peak $V_{\text{OUT}} = 9.0\ \text{V}$ .
$R = 370\ \Omega$ .
From (b)(i), $P_{\text{max}} = 0.22\ \text{W}$ .

Approach

Use $P(t)=V_{\text{OUT}}(t)^2/R$ .
Mark the key times when $V_{\text{OUT}}=0$ ; at those times $P=0$ .
At the times of voltage peaks, set power peaks to $0.22\ \text{W}$ .
Draw smooth curves between these points with a $\sin^2$ -like shape.

Step-by-Step Reasoning

Start from the relationship:

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

Identify zeros: whenever the rectified voltage waveform touches the time axis (at the ends of each half-cycle pulse), $V_{\text{OUT}}=0$ so

P=0

Identify maxima: at each peak of $V_{\text{OUT}}$ , the power is

P_{\text{max}} = \frac{(9.0)^2}{370} \approx 0.22\ \text{W}

So every pulse in the power graph reaches $0.22\ \text{W}$ .

Shape: between 0 and the peak, $V_{\text{OUT}}$ is sinusoidal in shape; squaring gives a curve that rises more gently near zero and is more rounded near the top (a $\sin^2$ pulse). The power never goes negative.

diagram-7bii

Key Takeaways

Instantaneous power in a resistor: $P(t)=V(t)^2/R$ .
Squaring means power is always positive and peaks when voltage peaks.
For rectified sine, the power waveform is a series of $\sin^2$ pulses.

Common Mistakes

Drawing negative power during parts of the cycle (power in a resistor cannot be negative).
Keeping the same shape as $V_{\text{OUT}}$ instead of squaring (power should be more “peaked”).
Using the wrong peak power (must match $0.22\ \text{W}$ from part (i)).

Things to Be Careful About

Put peaks at the correct times (same times as voltage peaks).
Ensure the graph touches $P=0$ at the same times $V_{\text{OUT}}=0$ .
Use the given axis scale up to $0.4\ \text{W}$ and label the peak at about $0.22\ \text{W}$ .

Techniques used

relate instantaneous power to instantaneous potential differencetransform a waveform using a square relationshipsketch a periodic graph with correct peak and zeros

(iii)

Calculate the mean power dissipated in $R$ .

mean power = ______ $\text{W}$

1M Similar

DifficultyMedium-Easy

Worked solution

Working

For a sinusoidal voltage across a resistor,

P_{\text{mean}} = \frac{1}{2}P_{\text{max}}

P_{\text{mean}} = \frac{1}{2} \times 0.22 = 0.11\ \text{W}

Answer

$0.11\ \text{W}$

Final answer

0.11 W

Detailed explanation

Background Concept

Instantaneous power in a resistor is

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

For a sinusoidal voltage $V(t)=V_0\sin(\omega t)$ (or its full-wave rectified form), the square involves $\sin^2$ :

P(t)=\frac{V_0^2}{R}\sin^2(\omega t)

The key average is

\langle \sin^2(\omega t) \rangle = \frac{1}{2}

So the mean power is half the maximum power.

Understanding the Question

You already found the maximum power in (b)(i) as $0.22\ \text{W}$ . You are now asked for the mean power dissipated in $R$ for the waveform shown (full-wave rectified sine).

Approach

Use the standard result that the time average of $\sin^2$ over a whole number of cycles is $1/2$ , hence $P_{\text{mean}} = P_{\text{max}}/2$ .

Step-by-Step Reasoning

The output voltage is sinusoidal in magnitude; power depends on $V^2$ , so it is proportional to $\sin^2$ .
Over one complete pulse period, the average value of $\sin^2$ is $1/2$ .
Therefore,

P_{\text{mean}} = \frac{1}{2}P_{\text{max}} = \frac{1}{2}\times 0.22 = 0.11\ \text{W}

(Using more digits from (b)(i) would give $0.1095\ \text{W}$ , which rounds to $0.11\ \text{W}$ .)

Key Takeaways

For a sinusoidal voltage across a resistor: $P_{\text{mean}} = \frac{V_0^2}{2R}$ .
Equivalently: $P_{\text{mean}} = \frac{1}{2}P_{\text{max}}$ .

Common Mistakes

Averaging the voltage instead of averaging $V^2$ .
Thinking rectification changes the mean power for the same peak voltage (it does not, because power depends on $V^2$ ).

Things to Be Careful About

Make sure you average over a whole number of cycles.
Quote the answer with unit and sensible significant figures.

Techniques used

use the mean value of a squared sinusoidrelate mean power to peak power for a sinusoidal waveform

(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-Easy

Worked solution

Answer

The mean power is the same.

Power in a resistor is $P \propto V^2$ , so changing the negative half-cycles to negative values (instead of rectifying them) does not change $V^2$ and hence does not change the mean power.

Final answer

The same.

Detailed explanation

Background Concept

For a resistor,

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

The crucial point is that power depends on the square of the voltage, so the sign of $V$ does not matter:

(-V)^2 = V^2

Therefore, for the same peak voltage, a sine wave $V_0\sin(\omega t)$ and a full-wave rectified wave $|V_0\sin(\omega t)|$ produce the same instantaneous power waveform $\propto \sin^2(\omega t)$ , and hence the same mean power.

Understanding the Question

You remove the bridge rectifier and connect the resistor $R$ directly to the a.c. supply. You must compare the new mean power with the mean power found in (b)(iii), with no calculations.

Approach

Think about what changes when the bridge rectifier is removed:

The voltage across $R$ will now be positive for half the time and negative for half the time.
But power uses $V^2$ , so negative voltages still give positive power.
Hence the mean power is unchanged (assuming ideal diodes and the same peak voltage across $R$ ).

Step-by-Step Reasoning

With the bridge: $V_{\text{OUT}}(t)$ is always positive, but varies sinusoidally in magnitude.
Without the bridge: $V(t)$ alternates positive and negative (a standard sine wave).
In both cases, the power is

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

Squaring removes the sign, so the power variation is the same $\sin^2$ form in both situations.
Therefore the time average (mean) power is the same as in (b)(iii).

(Practical note: in a real bridge rectifier, there are diode voltage drops so the rectified peak may be slightly smaller; then removing the diodes would slightly increase the mean power. Unless explicitly stated, exam questions usually treat diodes as ideal.)

Key Takeaways

Mean power in a resistor depends on $\langle V^2 \rangle$ , not on the mean value of $V$ .
Rectification changes the sign of voltage/current but not the power for the same voltage magnitude.

Common Mistakes

Saying the mean power becomes zero because the mean voltage is zero (mean voltage is not what determines mean power).
Claiming the mean power halves because “half the cycle is negative” (negative voltage does not mean negative power in a resistor).

Things to Be Careful About

The comparison assumes the same peak magnitude across $R$ ; if the question mentioned diode drops explicitly, you would consider that effect.
State the key physics link: $P \propto V^2$ (or $P=I^2R$ ) to justify the comparison.

Techniques used

compare mean power using the dependence of power on squared voltagereason qualitatively about the effect of sign reversal on 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:

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 potential energy lost by one electron becomes kinetic energy gained:

E = eV

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

The maximum energy photon is produced if one electron gives (approximately) all its kinetic energy to a single photon in the target.

Also:

$1\ \text{eV} = 1.60 \times 10^{-19}\ \text{J}$
$1\ \text{keV} = 10^3\ \text{eV}$
$1\ \text{MeV} = 10^6\ \text{eV}$

Understanding the Question

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

Approach

Use $E_{\max} = eV$ .
Recognise that a voltage in kV directly corresponds to energy in keV for a single electron.
Convert keV to MeV.

Step-by-Step Reasoning

An electron accelerated through $32\ \text{kV}$ gains energy $32\ \text{keV}$ .
Therefore the maximum photon energy is:

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

Convert to MeV:

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

Key Takeaways

Accelerating a charge $e$ through $V$ gives energy $eV$ .
$V$ in kV corresponds to electron energy in keV.

Common Mistakes

Writing $32\ \text{kV} = 32\ \text{MeV}$ (wrong by a factor of $10^3$ ).
Converting to joules unnecessarily and then making a power-of-ten error.

Things to Be Careful About

Use “maximum”: this assumes one photon takes essentially all the electron energy.
Keep track of prefixes: $\text{k} = 10^3$ , $\text{M} = 10^6$ .

Techniques used

use energy transfer from accelerating potential difference via E = eVconvert between electronvolts and megaelectronvolts

(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} = eV = 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

$1.71\times 10^{-23}\ \text{N s}$

Final answer

1.71 × 10^-23 N s

Detailed explanation

Background Concept

For a photon:

p = \frac{E}{c}

where $p$ is momentum, $E$ is energy, and $c$ is the speed of light.

To use this equation numerically with SI units, $E$ must be in joules.

Understanding the Question

You already found (or can find) the maximum photon energy from the $32\ \text{kV}$ accelerating voltage. You must calculate the corresponding maximum photon momentum.

Approach

Find $E_{\max}$ in joules using $E = eV$ .
Substitute into $p = E/c$ .

Step-by-Step Reasoning

Convert the accelerating voltage energy to joules:

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

Use the photon momentum relation:

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

Units: $\text{J} / (\text{m s}^{-1}) = (\text{N m})/(\text{m s}^{-1}) = \text{N s}$ .

Key Takeaways

Photon momentum is proportional to photon energy: $p = E/c$ .
Convert eV to joules when using SI constants.

Common Mistakes

Using $p = mc$ (photons have zero rest mass).
Forgetting to convert $32\ \text{keV}$ to joules.

Things to Be Careful About

Use $c = 3.00\times 10^8\ \text{m s}^{-1}$ .
Ensure the final unit is $\text{N s}$ (equivalently $\text{kg m s}^{-1}$ ).

Techniques used

convert photon energy from electronvolts to joulesuse 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\times 10^3 \times 1.60\times 10^{-19} = 5.12\times 10^{-15}\ \text{J}

\lambda_{\min} = \frac{hc}{E_{\max}} = \frac{6.63\times 10^{-34}\times 3.00\times 10^8}{5.12\times 10^{-15}} = 3.89\times 10^{-11}\ \text{m}

Answer

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

Final answer

3.9 × 10^-11 m

Detailed explanation

Background Concept

Photon energy is related to frequency by $E = hf$ and wave speed by $c = f\lambda$ . Eliminating $f$ gives the key link between photon energy and wavelength:

E = \frac{hc}{\lambda}

So:

\lambda = \frac{hc}{E}

In an X-ray tube, the minimum wavelength corresponds to the maximum photon energy (because higher energy means shorter wavelength).

Understanding the Question

With accelerating voltage $32\ \text{kV}$ , electrons have a maximum kinetic energy $eV$ . If that becomes photon energy, we get $E_{\max}$ , and then the minimum wavelength is found from $\lambda_{\min} = hc/E_{\max}$ .

Approach

Calculate $E_{\max}$ from the voltage (in joules).
Use $\lambda_{\min} = hc/E_{\max}$ .

Step-by-Step Reasoning

Maximum photon energy:

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

Substitute into the wavelength relation:

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

Compute $hc$ :

hc = (6.63\times 10^{-34})(3.00\times 10^8) = 1.99\times 10^{-25}\ \text{J m}

Then:

\lambda_{\min} = \frac{1.99\times 10^{-25}}{5.12\times 10^{-15}} = 3.89\times 10^{-11}\ \text{m}

So $\lambda_{\min} \approx 3.9\times 10^{-11}\ \text{m}$ .

Key Takeaways

Minimum wavelength corresponds to maximum photon energy.
Use $\lambda = hc/E$ with SI units.

Common Mistakes

Using $E = h/\lambda$ (missing the factor $c$ ).
Using $E$ in eV while using $h$ and $c$ in SI without conversion.

Things to Be Careful About

The question asks for minimum wavelength (so use maximum energy).
Powers of ten: keV to joules is often where errors occur.

Techniques used

use the relationship between photon energy and wavelength via E = hc/\lambdasubstitute maximum photon energy to find minimum wavelengthcarry out unit conversion consistently in SI units

(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

Different body tissues absorb (attenuate) X-rays by different amounts.

Bone (higher density / higher effective atomic number) absorbs more strongly than soft tissue, so fewer X-rays reach the detector through bone.

The detector records different transmitted intensities, producing light and dark regions and hence good contrast between structures.

Final answer

X-rays are attenuated by different amounts in different tissues (bone absorbs much more than soft tissue), so the transmitted intensity varies across the body and the detector records this variation as contrasting light/dark regions.

Detailed explanation

Background Concept

X-ray imaging is based on attenuation (reduction in intensity) of X-rays as they pass through matter. If $I_0$ is the incident intensity and $I$ is the transmitted intensity, then thicker or more strongly absorbing material gives a smaller transmitted intensity. Different materials attenuate X-rays differently.

In general, attenuation is greater for:

larger thickness,
higher density,
higher (effective) atomic number.

An image is formed because a detector (film or digital sensor) measures the transmitted intensity pattern across many rays.

Understanding the Question

The question asks why X-rays can produce images of internal body structures with good contrast. “Good contrast” means different structures show up as noticeably different brightness/greyscale on the final image.

Approach

Explain, in linked steps:

different tissues attenuate X-rays differently,
this leads to different transmitted intensities,
the detector converts these intensity differences into light/dark regions (contrast).

Step-by-Step Reasoning

X-rays passing through the body are partly absorbed/scattered; the intensity decreases.
Bone contains calcium and is denser with higher effective atomic number than soft tissue, so it attenuates X-rays much more.
Therefore, through bone, the transmitted intensity at the detector is much lower than through surrounding soft tissue.
A detector (or photographic film) responds differently to different X-ray intensities, so regions behind bone and regions behind soft tissue appear with different brightness. This difference produces a high-contrast image of internal structures.

Key Takeaways

Contrast arises from differences in attenuation between tissues.
Bone attenuates X-rays much more strongly than soft tissue, so it is easily distinguished.

Common Mistakes

Saying only “X-rays pass through the body” (penetration alone does not explain contrast).
Not mentioning that different tissues absorb different amounts.

Things to Be Careful About

Use the correct idea: contrast comes from transmitted intensity differences, not from reflection.
Mention at least one physical reason for different attenuation (density / atomic number / thickness).

Techniques used

relate image contrast to differential attenuation of radiationlink absorption to material properties such as density and atomic numberdescribe how detector intensity variations form an image

Q9MediumNuclear Physics

(a)

Define half-life of a radioactive isotope.

1M Similar

(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

Magnitude of gradient of line X represents the activity (decay rate) of isotope X, $A = \left|\dfrac{\mathrm{d}N}{\mathrm{d}t}\right|$ .

Final answer

Activity (decay rate) of X.

Detailed explanation

Background Concept

For a radioactive sample, the activity $A$ is the number of decays per second.

If $N$ is the number of undecayed nuclei, then activity is defined by

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

The negative sign is because $N$ decreases with time. The magnitude of the gradient of an $N$ against $t$ graph is therefore the activity.

Understanding the Question

In Fig. 9.1, line X is a graph of number of nuclei of X, $N_X$ , against time $t$ .

The question asks what physical quantity is represented by the magnitude of the gradient of that curve.

Approach

Recognise this is a “rate of change” question: gradient means $\mathrm{d}N/\mathrm{d}t$ .
For radioactive decay, $-\mathrm{d}N/\mathrm{d}t$ is activity.
Since it asks for the magnitude, we ignore the negative sign.

Step-by-Step Reasoning

The gradient of X on the $N$ – $t$ graph is $\dfrac{\mathrm{d}N_X}{\mathrm{d}t}$ .
Because X is decaying, $\dfrac{\mathrm{d}N_X}{\mathrm{d}t}$ is negative.
Activity is

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

Therefore the magnitude of the gradient is

\left|\frac{\mathrm{d}N_X}{\mathrm{d}t}\right| = A

Key Takeaways

Gradient on an $N$ – $t$ graph is a rate.
For radioactive decay, that rate corresponds to activity.

Common Mistakes

Saying “decay constant” (the decay constant relates to the fractional rate $\lambda = -\tfrac{1}{N}\tfrac{\mathrm{d}N}{\mathrm{d}t}$ , not the gradient itself).
Forgetting the question asks for magnitude and writing a negative quantity.

Things to Be Careful About

Activity has unit $\text{s}^{-1}$ (or Bq), whereas the gradient is “nuclei per second” which is numerically the same thing.
The gradient changes with time (curve is not a straight line), so activity decreases as time increases.

Techniques used

interpret the gradient of a graph physicallyrelate 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-Easy

Worked solution

Answer

Any three conclusions, e.g.

At $t=0$ , $N_X = 4.0 \times 10^{22}$ and $N_Y = 0$ .
The curves intersect at $t = 14\ \text{s}$ with $N_X = N_Y = 2.0 \times 10^{22}$ , so the half-life of X is $14\ \text{s}$ .
As $t$ increases, $N_Y$ tends to $4.0 \times 10^{22}$ while $N_X \to 0$ , so essentially all X eventually becomes Y (Y is stable / does not decay significantly on this timescale).

Final answer

Example conclusions: N_X(0)=4.0×10^22, N_Y(0)=0; half-life of X = 14 s; eventually N_Y→4.0×10^22 and N_X→0 (Y stable on this timescale).

Detailed explanation

Background Concept

When a parent isotope X decays to a daughter isotope Y, the number of X nuclei $N_X$ decreases with time, typically exponentially. If Y is stable (or its half-life is very long compared with the time shown), the number of Y nuclei $N_Y$ increases and tends towards the initial number of X nuclei.

If every X decay produces one Y nucleus, then (ignoring any subsequent decay of Y):

N_X + N_Y = \text{constant}

Also, half-life can be read from a decay curve as the time for $N_X$ to halve.

Understanding the Question

You are given a graph of $N_X$ and $N_Y$ against time.

You must state three conclusions that are supported by features of the curves. These can be:

numerical (values at given times, half-life), or
qualitative (shape of curves, what happens at long times).

Approach

Pick three clear, unambiguous statements that can be directly read from the graph:

initial values, 2) half-life from the halving point, 3) long-time behaviour / stability of Y / conservation of nuclei.

Step-by-Step Reasoning

Initial values: read values at $t=0$ .
- The X curve starts at $4.0$ on the axis labelled “ $\times 10^{22}$ ”, so $N_X(0)=4.0\times 10^{22}$ .
- The Y curve starts at $0$ , so $N_Y(0)=0$ .
Half-life of X: find when $N_X$ is half of its initial value.
- Half of $4.0\times 10^{22}$ is $2.0\times 10^{22}$ .
- The graph shows the two curves intersect at $N_X=N_Y=2.0\times 10^{22}$ and $t=14\ \text{s}$ .
- At that time, $N_X$ has fallen to half its initial value, so

t_{1/2} = 14\ \text{s}

Long-time behaviour / stability of Y:
- As $t$ becomes large, the X curve approaches zero, meaning almost all original X has decayed.
- The Y curve approaches $4.0\times 10^{22}$ (the original number of X nuclei), implying each X nucleus becomes a Y nucleus and Y does not significantly decay during the time shown.
- This is consistent with $N_X + N_Y \approx 4.0\times 10^{22}$ throughout.

Other valid conclusions you could state include:

X decays exponentially (shape of X curve).
The activity of X decreases with time (slope magnitude of X decreases).

Key Takeaways

You can read half-life directly from where $N$ halves.
For a simple parent-daughter decay with stable daughter, $N_X + N_Y$ stays constant.
Exponential decay gives a curve whose slope becomes less steep with time.

Common Mistakes

Claiming the half-life is $28\ \text{s}$ (confusing the intersection time with “two half-lives” or misreading the axis).
Saying Y decays faster because it increases quickly at first (increase in Y is due to production, not its own decay).
Forgetting the axis scale factor of $10^{22}$ .

Things to Be Careful About

Quote values with the correct power of ten: “ $2.0 \times 10^{22}$ ”, not just “2.0”.
“Y is stable” is an inference from the curve tending to a constant; phrase it as “does not decay significantly on this timescale” to be safe.
Ensure your conclusions are actually supported by the graph, not general nuclear facts.

Techniques used

extract numerical values from intersections on a graphidentify half-life from a decay curveuse conservation of nucleon number to infer daughter-product behaviour

(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 one nucleus:

m_{\text{nucleus}} = \frac{7.3 \times 10^{-4}}{4.0 \times 10^{22}} = 1.825 \times 10^{-26}\ \text{kg}

Using nucleon mass $m_n \approx 1.67 \times 10^{-27}\ \text{kg}$ :

A = \frac{m_{\text{nucleus}}}{m_n} = \frac{1.825 \times 10^{-26}}{1.67 \times 10^{-27}} = 10.9 \approx 11

Answer

Nucleon number $= 11$

Final answer

11

Detailed explanation

Background Concept

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

A nucleus has mass approximately equal to $A$ times the nucleon mass (proton/neutron mass), because each nucleon has mass about

1.67 \times 10^{-27}\ \text{kg}

(The actual nucleus mass is slightly smaller due to binding energy, but for finding an integer nucleon number this approximation is sufficient.)

Understanding the Question

You are told the total mass of isotope X in the sample at $t=0$ is $7.3 \times 10^{-4}\ \text{kg}$ .

From part (b) / the graph, at $t=0$ the sample contains only X, so the number of X nuclei present initially is the initial value on the X curve:
$N_X(0) = 4.0 \times 10^{22}$ .

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

Approach

Read $N_X$ at $t=0$ from the graph.
Find mass per nucleus: $m_{\text{nucleus}} = \dfrac{m_{\text{total}}}{N_X}$ .
Divide by nucleon mass to estimate how many nucleons are in each nucleus:

A \approx \frac{m_{\text{nucleus}}}{1.67 \times 10^{-27}}

Then round to the nearest whole number (since nucleon number is an integer).

Step-by-Step Reasoning

Number of nuclei at $t=0$
From the graph, curve X starts at 4.0 on the scale of $\times 10^{22}$ :

N_X(0)=4.0 \times 10^{22}

Mass per nucleus
Total mass is $7.3 \times 10^{-4}\ \text{kg}$ , shared among $4.0 \times 10^{22}$ identical nuclei:

m_{\text{nucleus}} = \frac{7.3 \times 10^{-4}}{4.0 \times 10^{22}}

Do the powers of ten carefully:

m_{\text{nucleus}} = 1.825 \times 10^{-26}\ \text{kg}

Convert mass per nucleus to nucleon number
Divide by nucleon mass $m_n \approx 1.67 \times 10^{-27}\ \text{kg}$ :

A \approx \frac{1.825 \times 10^{-26}}{1.67 \times 10^{-27}} = 10.9

Since $A$ must be an integer:

A = 11

Key Takeaways

You can determine a nucleus’s mass from total mass divided by number of nuclei.
Nucleon number is found by comparing nucleus mass with the mass of one nucleon.
Always round to an integer for nucleon number.

Common Mistakes

Using $2.0 \times 10^{22}$ (the intersection value) instead of $4.0 \times 10^{22}$ at $t=0$ .
Dropping the $\times 10^{22}$ scale factor from the graph.
Rounding to $10$ instead of $11$ (must be the nearest integer).

Things to Be Careful About

Keep powers of ten consistent when dividing.
Use an appropriate nucleon mass (typically $1.67 \times 10^{-27}\ \text{kg}$ ).
The slight difference between nuclear mass and $A m_n$ (binding energy) is not large enough to change the nearest integer here.

Techniques used

extract the number of nuclei from a graph at a specified timecompute mass per nucleus from total mass and number of nucleiestimate nucleon number using the nucleon mass

Q10Medium-EasyAstronomy and Cosmology

(a)

5M

(i)

State what is meant by the luminosity of a star.

2M Similar

(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 the flux (apparent brightness) $F$ received by the observer.

Use

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

so

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

Final answer

Use known luminosity L of the standard candle and measured flux F, with F = L/(4πd^2) so d = √(L/(4πF)).

Detailed explanation

Background Concept

For an isotropic source (emitting equally in all directions), the emitted power $L$ spreads out over the surface of a sphere of radius $d$ .

Area of that sphere:

A = 4\pi d^2

Flux (sometimes called radiant flux intensity) $F$ is power received per unit area:

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

So flux decreases with the square of distance (inverse-square law).

A “standard candle” is an astronomical object whose luminosity $L$ is known (because its type has a known absolute luminosity).

Understanding the Question

The question asks how observing a standard candle in a distant galaxy lets you find the galaxy’s distance. You are expected to connect:

known intrinsic luminosity $L$ of the candle,
measured flux $F$ at Earth,
inverse-square law to solve for distance $d$ .

Approach

Identify that the standard candle provides a known $L$ .
State that we measure $F$ (apparent brightness) at the observer.
Use $F = L/(4\pi d^2)$ and rearrange for $d$ .

Step-by-Step Reasoning

Because the candle’s luminosity $L$ is known, the only unknown in $F = \frac{L}{4\pi d^2}$ is $d$ once $F$ is measured.
Rearranging: $d^2 = \frac{L}{4\pi F}$ $d = \sqrt{\frac{L}{4\pi F}}.$
The distance to the standard candle is then the distance to the galaxy containing it (to a good approximation).

Key Takeaways

Standard candle gives known $L$ .
Measure $F$ and apply inverse-square law.
Distance follows from a single rearrangement.

Common Mistakes

Saying “use redshift” (not what a standard candle method is).
Confusing luminosity $L$ with flux $F$ .
Missing the square root when solving for $d$ .

Things to Be Careful About

$F$ is power per unit area at the observer.
The factor $4\pi$ must be included for isotropic emission.
The method assumes negligible absorption/extinction or that it has been corrected for.

Techniques used

use the inverse-square law relating flux and luminosityrearrange an equation to solve for distancelink measured flux to intrinsic luminosity of a standard candle

(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

Stefan-Boltzmann law:

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

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

L = 3.85 \times 10^{26}

Answer

$L = 3.85 \times 10^{26}\ \text{W}$

Final answer

3.85 × 10^26 W

Detailed explanation

Background Concept

A star’s luminosity can be related to its surface temperature if it behaves approximately like a black body.

For a black body, the power emitted per unit area (the intensity at the surface) is given by the Stefan-Boltzmann law:

I = \sigma T^4

where:

$I$ is power per unit area ( $\text{W m}^{-2}$ )
$\sigma = 5.67 \times 10^{-8}\ \text{W m}^{-2}\ \text{K}^{-4}$
$T$ is absolute temperature in kelvin.

The star’s total power output (luminosity) is then

L = (\text{surface area}) \times I = 4\pi r^2 \sigma T^4.

Understanding the Question

You are given the Sun’s radius $r = 6.96 \times 10^8\ \text{m}$ and its surface temperature $T = 5780\ \text{K}$ .

You must calculate the Sun’s luminosity $L$ and state its unit. (The wavelength information is not needed for this particular calculation.)

Approach

Use the Stefan-Boltzmann expression:

Calculate the surface area $4\pi r^2$ .
Calculate $T^4$ .
Multiply by $\sigma$ .
Quote $L$ in watts.

Step-by-Step Reasoning

Start with

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

Substitute $r = 6.96 \times 10^8\ \text{m}$ :

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

Surface area:

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

Now evaluate $T^4$ :

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

Multiply $\sigma T^4$ to get emitted power per unit area:

\sigma T^4 = (5.67 \times 10^{-8})(1.12 \times 10^{15}) \approx 6.33 \times 10^7\ \text{W m}^{-2}.

Finally multiply by area:

L = (6.08 \times 10^{18})(6.33 \times 10^7) \approx 3.85 \times 10^{26}\ \text{W}.

This matches the known order of magnitude for the Sun’s luminosity ( $\approx 10^{26}\ \text{W}$ ), which is a useful sanity check.

Key Takeaways

Use $L = 4\pi r^2 \sigma T^4$ for a (black-body) star.
Keep SI units: $r$ in metres, $T$ in kelvin.
Final unit of luminosity is $\text{W}$ .

Common Mistakes

Using $L = \sigma T^4$ without multiplying by surface area.
Using diameter instead of radius.
Using $T$ in degrees Celsius instead of kelvin.
Arithmetic slips with the power $T^4$ and indices.

Things to Be Careful About

Quote $\sigma$ with correct powers and units.
When squaring $r$ and raising $T$ to the fourth power, keep track of standard form.
Give the final answer to a sensible number of significant figures (typically 3 s.f. here).

Techniques used

apply the Stefan-Boltzmann lawcalculate surface area of a spheresubstitute values with consistent SI units

(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

Wien’s law:

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

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

T = \frac{2.90 \times 10^{-3}}{6.24 \times 10^{-7}} = 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 star that approximately behaves as a black body emits a continuous spectrum with a peak at some wavelength $\lambda_{\text{max}}$ .

Wien’s displacement law relates the peak wavelength to the absolute temperature:

\lambda_{\text{max}} T = b

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

So a hotter star has a smaller (bluer) peak wavelength, and a cooler star has a larger (redder) peak wavelength.

Understanding the Question

You are told that another star’s radiation has peak intensity at $\lambda_{\text{max}} = 624\ \text{nm}$ . You must find its surface temperature.

The key is to:

convert $\text{nm}$ to $\text{m}$ ,
apply Wien’s law and rearrange for $T$ .

Approach

Write Wien’s law $\lambda_{\text{max}} T = 2.90 \times 10^{-3}$ .
Convert $624\ \text{nm}$ into metres.
Rearrange: $T = b/\lambda_{\text{max}}$ .

Step-by-Step Reasoning

Convert wavelength:

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

Use Wien’s law:

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

Compute powers of ten:

\frac{10^{-3}}{10^{-7}} = 10^{4}.

Compute the numerical factor:

\frac{2.90}{6.24} \approx 0.465.

So

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

This is cooler than the Sun (which peaks at a shorter wavelength), which is consistent with the longer peak wavelength of $624\ \text{nm}$ .

Key Takeaways

Wien’s law: $\lambda_{\text{max}} T = 2.90 \times 10^{-3}\ \text{m K}$ .
Always convert $\text{nm}$ to $\text{m}$ before substituting.
Longer peak wavelength means lower temperature.

Common Mistakes

Forgetting to convert $\text{nm}$ to $\text{m}$ , giving a temperature too small by a factor of $10^9$ .
Using $b = 2.90 \times 10^{-3}$ with the wrong unit (it must be $\text{m K}$ ).
Mixing up the rearrangement and calculating $\lambda_{\text{max}}/b$ .

Things to Be Careful About

Keep standard form and check the power of ten carefully.
Quote the final temperature in kelvin, not degrees Celsius.
Use an appropriate number of significant figures (typically 3 s.f. to match the given wavelength).

Techniques used

apply Wien's displacement lawconvert wavelength units to metresrearrange to solve for temperature

Physics 9702/41 — 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

Common Mistakes

Things to Be Careful About

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Understanding the Question

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Step-by-Step Reasoning

Key Takeaways

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Working

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Background Concept

Understanding the Question

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Step-by-Step Reasoning

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Common Mistakes

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Background Concept

Understanding the Question

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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

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

Common Mistakes

Things to Be Careful About

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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

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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