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

(a) For two point charges separated by distance r, F = (1/4πϵ0) Q1Q2 /r^2, along the line joining them; like repel, unlike attract. · (b)(i) +2e · (b)(ii) 1.6 × 10^{-8} N · (c)(i) 1.7 × 10^6 m s^-1 · (c)(ii) 6.2 × 10^-16 s · (d)(i) 0.125 · (d)(ii) Resultant inward force is reduced (other electron opposes nucleus), so orbit is less tightly bound: lower speed for same radius / larger radius for same speed.

9702/41 May/June 2025 — 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 in bringing a small test mass from infinity to the point (with no change in kinetic energy).

Final answer

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

Detailed explanation

Background Concept

Gravitational potential $\phi$ at a point is a measure of how much energy per unit mass is associated with position in a gravitational field.

It is defined by the work done per unit mass by an external agent in bringing a small test mass from infinity to that point, with its speed unchanged (so the work goes into gravitational potential energy, not kinetic energy).

Mathematically, gravitational potential energy is related by

U = m\phi

and for a spherical mass $M$ ,

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

where $r$ is distance from the centre.

Understanding the Question

You are asked to give the definition of gravitational potential at a point. This is not gravitational field strength ( $g$ ), and it is not potential energy (which depends on the satellite mass). The key phrases needed are “per unit mass” and “from infinity”.

Approach

Write the standard definition: work done (or potential energy) per unit mass, for a small test mass, moved from infinity to the point without changing kinetic energy.

Step-by-Step Reasoning

“Potential” is energy per unit mass, so the definition must include “per unit mass”.
The reference level for gravitational potential is taken at infinity ( $\phi = 0$ at $r \to \infty$ ), so the definition must include “from infinity”.
Mentioning “no change in kinetic energy” clarifies that the work done corresponds to a change in gravitational potential energy only.

Key Takeaways

Gravitational potential $\phi$ is energy (or work done) per unit mass.
The zero of gravitational potential is defined at infinity.

Common Mistakes

Defining $g$ instead (force per unit mass).
Saying “work done to move a mass from the point to infinity” without making it per unit mass or without a clear sign/reference.
Giving potential energy instead of potential (forgetting “per unit mass”).

Things to Be Careful About

Include “per unit mass” explicitly.
Include the reference “from infinity” explicitly.
Do not confuse potential (scalar, $\text{J kg}^{-1}$ ) with field strength (vector, $\text{N kg}^{-1}$ ).

Techniques used

state the definition of gravitational potentiallink gravitational potential to work done per unit mass from infinity

(b)

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

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

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

5M

(i)

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

3M Similar

DifficultyMedium

Worked solution

Working

For a satellite of mass $m$ at radius $r$ :

U=-\frac{GMm}{r}

So the increase in GPE when moving from $r_1$ to $r_2$ is

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

r_1 = 3.4\times 10^6 + 1.7\times 10^6 = 5.1\times 10^6\ \text{m}

r_2 = 3.4\times 10^6 + 6.8\times 10^6 = 1.02\times 10^7\ \text{m}

M=\frac{\Delta U}{Gm\left(\frac{1}{r_1}-\frac{1}{r_2}\right)} =\frac{5.1\times 10^8}{(6.67\times 10^{-11})(122)\left(\frac{1}{5.1\times 10^6}-\frac{1}{1.02\times 10^7}\right)} \approx 6.4\times 10^{23}\ \text{kg}

Answer

$M \approx 6.4\times 10^{23}\ \text{kg}$

Final answer

6.4 × 10^23 kg

Detailed explanation

Background Concept

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

U=-\frac{GMm}{r}

The negative sign appears because we define $U=0$ at infinity; at finite $r$ the satellite is bound to the planet, so the energy is negative.

When you move a satellite to a higher orbit (larger $r$ ), $U$ becomes less negative, so $U$ increases (a positive $\Delta U$ ).

Understanding the Question

You are told:

Mars radius $R = 3.4\times 10^6\ \text{m}$
Satellite mass $m = 122\ \text{kg}$
Initial height above surface $h_1 = 1.7\times 10^6\ \text{m}$
Final height above surface $h_2 = 6.8\times 10^6\ \text{m}$
Increase in gravitational potential energy $\Delta U = 5.1\times 10^8\ \text{J}$

You must use this change in GPE to find the unknown mass of Mars, $M$ , and show it is $6.4\times 10^{23}\ \text{kg}$ .

Approach

Convert heights above the surface into orbital radii from the centre: $r = R+h$ .
Write $U_1=-GMm/r_1$ and $U_2=-GMm/r_2$ .
Form $\Delta U = U_2-U_1$ and simplify to an expression containing $M$ .
Rearrange to solve for $M$ and substitute the numbers.

Step-by-Step Reasoning

Convert to radii from the centre:

r_1 = R+h_1 = 3.4\times 10^6 + 1.7\times 10^6 = 5.1\times 10^6\ \text{m}

r_2 = R+h_2 = 3.4\times 10^6 + 6.8\times 10^6 = 1.02\times 10^7\ \text{m}

Write the potential energies:

U_1=-\frac{GMm}{r_1}, \qquad U_2=-\frac{GMm}{r_2}

Take the difference:

\Delta U = U_2-U_1 = -\frac{GMm}{r_2}+\frac{GMm}{r_1}

Factorise:

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

This bracket is positive because $r_2>r_1$ , matching the given “increases by $5.1\times 10^8\ \text{J}$ ”.

Rearrange for $M$ :

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

Substitute:

M\approx 6.4\times 10^{23}\ \text{kg}

which is the required value.

Key Takeaways

Always use distance from the centre ( $r=R+h$ ), not height above the surface, in $-GMm/r$ .
Moving outward increases $U$ (it becomes less negative).
Changes in GPE in a gravitational field are often easier than using forces or orbital speed.

Common Mistakes

Using $h$ instead of $r$ in $U=-GMm/r$ .
Getting the sign wrong for $\Delta U$ (forgetting that $U$ is negative).
Arithmetic slip in evaluating $\left(\frac{1}{r_1}-\frac{1}{r_2}\right)$ .

Things to Be Careful About

Keep powers of ten consistent when calculating reciprocals of $r$ .
Use $G=6.67\times 10^{-11}\ \text{N m}^2\text{kg}^{-2}$ .
Quote $M$ to appropriate significant figures (here $6.4\times 10^{23}$ matches the given data).

Techniques used

use the expression for gravitational potential energy of a mass in a spherical fieldcompute change in potential energy between two radiirearrange to solve for the central mass

(ii)

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

$\phi$ = ______ unit ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

At the surface, $r=R$ and

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

\phi=-\frac{(6.67\times 10^{-11})(6.4\times 10^{23})}{3.4\times 10^6} \approx -1.3\times 10^7

Answer

$\phi \approx -1.3\times 10^7\ \text{J kg}^{-1}$

Final answer

−1.3 × 10^7 J kg−1

Detailed explanation

Background Concept

Gravitational potential $\phi$ is gravitational potential energy per unit mass:

\phi = \frac{U}{m}

For a uniform sphere (or any spherically symmetric mass), outside the sphere it behaves like a point mass at its centre:

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

The unit of $\phi$ is $\text{J kg}^{-1}$ (equivalently $\text{m}^2\text{s}^{-2}$ ).

Understanding the Question

You are asked for the gravitational potential at the surface of Mars. That means the distance from the centre is the radius $R = 3.4\times 10^6\ \text{m}$ . You use the Mars mass found in (b)(i), $M=6.4\times 10^{23}\ \text{kg}$ .

Approach

Use

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

Substitute values and give the unit.

Step-by-Step Reasoning

Identify $r$ at the surface:

r=R=3.4\times 10^6\ \text{m}

Substitute into the potential expression:

\phi=-\frac{(6.67\times 10^{-11})(6.4\times 10^{23})}{3.4\times 10^6}

The result is about $1.26\times 10^7$ in magnitude, so

\phi\approx -1.3\times 10^7\ \text{J kg}^{-1}

The negative sign is expected because $\phi=0$ at infinity and is negative at finite distance.

Key Takeaways

Gravitational potential depends only on the planet: $\phi=-GM/r$ .
The surface value uses $r=R$ .
Unit: $\text{J kg}^{-1}$ .

Common Mistakes

Forgetting the negative sign.
Using $r=R+h$ even though $h=0$ at the surface (or accidentally using the orbit height instead of surface radius).
Giving unit $\text{J}$ (that is for potential energy, not potential).

Things to Be Careful About

Use $R$ in metres.
Keep correct significant figures consistent with the given data.
State the unit explicitly (the question asks for it).

Techniques used

use gravitational potential of a spherical masssubstitute values with correct sign and units

(c)

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

2M

(i)

The orbit has a period of $25$ hours.

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

1M Similar

(ii)

State one other feature of this orbit.

1M Similar

DifficultyEasy

Worked solution

Answer

The orbit is in the equatorial plane of Mars (satellite above the equator).

Final answer

Orbit is in Mars’s equatorial plane (above the equator).

Detailed explanation

Background Concept

For a satellite to remain above the same point on a rotating planet (a stationary orbit), three key conditions apply:

The orbit must be circular.
The orbit must be in the planet’s equatorial plane.
The satellite must orbit in the same direction as the planet’s rotation.

The equatorial requirement matters because the “same point” on the surface traces a circle parallel to the equator as the planet rotates; only an equatorial orbit can stay directly above it.

Understanding the Question

You are asked to state one other feature of the orbit besides the period matching the planet’s rotation (already considered in part (i)). Any one correct feature earns the mark.

Approach

State one standard stationary-orbit condition: equatorial plane, circular orbit, or same direction as rotation.

Step-by-Step Reasoning

A valid statement is:

The orbit must lie in Mars’s equatorial plane (so the satellite remains above a fixed latitude and longitude).

(Alternatives that would also be acceptable for “one feature” include: the orbit is circular; the satellite moves west-to-east, i.e. same direction as Mars’s rotation.)

Key Takeaways

A stationary orbit is not just “period matches rotation”; its geometry must also match the rotating surface.

Common Mistakes

Saying “it must be close to Mars” (not necessarily; the radius is determined by the period).
Saying “it must pass over the poles” (polar orbits cannot stay over one point).

Things to Be Careful About

Only one feature is needed; do not contradict yourself by listing an incorrect one.
Keep the statement specific (e.g. “equatorial plane” rather than vague “around the middle”).

Techniques used

recall conditions for a stationary orbitstate an additional geometric property of the orbit

Q2MediumElectric FieldsMotion in a Circle

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

Fig. 2.1

(a)

State Coulomb’s law.

2M Similar

DifficultyEasy

Worked solution

Answer

For two point charges separated by distance $r$ , the force magnitude is

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

The force acts along the line joining the charges; like charges repel and unlike charges attract.

Final answer

For two point charges separated by distance r, F = (1/4πϵ0) |Q1Q2|/r^2, along the line joining them; like repel, unlike attract.

Detailed explanation

Background Concept

Coulomb’s law gives the electrostatic force between two point charges. The magnitude depends on:

the product of the charges, $Q_1Q_2$ ;
the inverse square of their separation, $1/r^2$ .

In SI units,

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

where $\varepsilon_0$ is the permittivity of free space. The direction of the force is always along the straight line joining the two charges.

Understanding the Question

You are asked to state the law, so the mark is for the correct proportionalities/equation and the correct directional/attraction-repulsion statements.

Approach

Write the standard form of Coulomb’s law (either in words or as the equation) and add the key directional statement (line of centres) plus whether it is attractive/repulsive.

Step-by-Step Reasoning

For point charges, the force magnitude is proportional to $|Q_1Q_2|$ .
It is inversely proportional to $r^2$ .
The constant of proportionality in SI is $1/(4\pi\varepsilon_0)$ .
State that the force acts along the line joining the charges, and that like charges repel and unlike charges attract.

Key Takeaways

Coulomb’s law is an inverse-square law.
Direction is along the line joining the charges.
Sign of charges determines attraction vs repulsion.

Common Mistakes

Forgetting the $r^2$ (writing $1/r$ ).
Omitting the direction (line joining charges).
Confusing attraction/repulsion for like/unlike charges.

Things to Be Careful About

Use “point charges” or “small compared with separation” to indicate applicability.
If writing the equation, include $4\pi\varepsilon_0$ (or equivalently $k = 1/(4\pi\varepsilon_0)$ ).

Techniques used

state Coulomb’s law in wordsinclude direction and attraction/repulsion conditions

(b)

2M

(i)

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

charge = ______ $e$

1M Similar

(ii)

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

Separation $r = 170\ \text{pm} = 1.70\times 10^{-10}\ \text{m}$ .

F = \frac{1}{4\pi\varepsilon_0}\frac{|Q_1Q_2|}{r^2} = (8.99\times 10^9)\frac{(2e)(e)}{(1.70\times 10^{-10})^2}

with $e = 1.60\times 10^{-19}\ \text{C}$ :

F = 8.99\times 10^9\times \frac{2(1.60\times 10^{-19})^2}{(1.70\times 10^{-10})^2} = 1.6\times 10^{-8}\ \text{N}.

Answer

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

Final answer

1.6 × 10^{-8} N

Detailed explanation

Background Concept

The electrostatic force between two point charges is given by Coulomb’s law:

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

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

Understanding the Question

You must show that the attractive force between the helium nucleus (charge $+2e$ ) and one electron (charge $-e$ ) at radius $170\ \text{pm}$ has magnitude $1.6\times 10^{-8}\ \text{N}$ . The word “show” means your method and substitution must be visible.

Approach

Convert $170\ \text{pm}$ into metres.
Use $Q_\text{nucleus} = +2e$ and $Q_\text{electron} = -e$ .
Substitute into Coulomb’s law using magnitudes.

Step-by-Step Reasoning

Unit conversion:

170\ \text{pm} = 170\times 10^{-12}\ \text{m} = 1.70\times 10^{-10}\ \text{m}.

Charge magnitudes:

|Q_1Q_2| = |(+2e)(-e)| = 2e^2.

Substitute into Coulomb’s law:

F = (8.99\times 10^9)\frac{2(1.60\times 10^{-19})^2}{(1.70\times 10^{-10})^2}.

Evaluate powers of ten carefully:

$(1.60\times 10^{-19})^2 = 2.56\times 10^{-38}$
$(1.70\times 10^{-10})^2 = 2.89\times 10^{-20}$

So

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

Key Takeaways

Always convert pm to m before substitution.
Use magnitudes for the force size; the sign tells direction (attraction here).

Common Mistakes

Using $r = 170\times 10^{-12}$ incorrectly as $170\times 10^{-10}$ .
Forgetting the factor 2 from the nuclear charge $+2e$ .
Squaring $r$ incorrectly.

Things to Be Careful About

The question wants the force between nucleus and one electron, not between the two electrons.
Significant figures: $1.6\times 10^{-8}\ \text{N}$ matches the given target value.

Techniques used

apply Coulomb’s law between two point chargesconvert picometres to metressubstitute constants and evaluate powers of ten

(c)

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

4M

(i)

Calculate the speed of the orbiting electrons.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Using $F = \dfrac{mv^2}{r}$ with $F = 1.6\times 10^{-8}\ \text{N}$ , $r = 170\ \text{pm} = 1.70\times 10^{-10}\ \text{m}$ , $m = 9.11\times 10^{-31}\ \text{kg}$ :

v = \sqrt{\frac{Fr}{m}} = \sqrt{\frac{(1.6\times 10^{-8})(1.70\times 10^{-10})}{9.11\times 10^{-31}}} = 1.7\times 10^{6}\ \text{m s}^{-1}.

Answer

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

Final answer

1.7 × 10^6 m s^-1

Detailed explanation

Background Concept

Uniform circular motion requires a centripetal (inward) acceleration

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

so the required inward (centripetal) force is

F = ma = m\frac{v^2}{r}.

If the only force is electrostatic attraction, that electrostatic force provides the centripetal force.

Understanding the Question

You are told to assume the force found in (b)(ii), $1.6\times 10^{-8}\ \text{N}$ , is the only force on an electron moving in a circle of radius $170\ \text{pm}$ . You must find the electron’s speed $v$ .

Approach

Set

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

and solve for $v$ .

Step-by-Step Reasoning

Convert the radius:

r = 170\ \text{pm} = 1.70\times 10^{-10}\ \text{m}.

Use electron mass $m = 9.11\times 10^{-31}\ \text{kg}$ .
Rearrange centripetal force formula:

F = \frac{mv^2}{r} \Rightarrow v^2 = \frac{Fr}{m} \Rightarrow v = \sqrt{\frac{Fr}{m}}.

Substitute:

\frac{Fr}{m} = \frac{(1.6\times 10^{-8})(1.70\times 10^{-10})}{9.11\times 10^{-31}} \approx 3.0\times 10^{12}.

Square root:

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

Key Takeaways

In circular motion problems, identify what provides the centripetal force.
Use $F = mv^2/r$ and solve for the unknown.

Common Mistakes

Forgetting to convert $\text{pm}$ to $\text{m}$ .
Using $F = m\omega^2 r$ without knowing $\omega$ (it’s fine, but you must connect it correctly).
Using proton mass instead of electron mass.

Things to Be Careful About

The force is radial; tangential forces would change speed, but none are assumed here.
Quote speed to an appropriate number of significant figures (typically 2–3).

Techniques used

equate centripetal force to the resultant radial forcerearrange F = mv^2/r to solve for vsubstitute values with consistent SI units

(ii)

Calculate the period of the orbit of the electrons.

period = ______ $\text{s}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

T = \frac{2\pi r}{v} = \frac{2\pi (1.70\times 10^{-10})}{1.7\times 10^{6}} = 6.2\times 10^{-16}\ \text{s}.

Answer

$6.2\times 10^{-16}\ \text{s}$

Final answer

6.2 × 10^-16 s

Detailed explanation

Background Concept

For uniform circular motion, the speed is constant and the distance for one complete orbit is the circumference $2\pi r$ . The period $T$ is time for one orbit:

T = \frac{\text{distance in one orbit}}{\text{speed}} = \frac{2\pi r}{v}.

Equivalently, $v = r\omega$ and $\omega = 2\pi/T$ lead to the same result.

Understanding the Question

You have already found the electron speed in (c)(i), and the orbital radius is $170\ \text{pm}$ . The question asks for the orbital period $T$ .

Approach

Use

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

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

Step-by-Step Reasoning

Convert radius:

r = 1.70\times 10^{-10}\ \text{m}.

Use the speed from (c)(i): $v \approx 1.7\times 10^{6}\ \text{m s}^{-1}$ .
Substitute:

T = \frac{2\pi (1.70\times 10^{-10})}{1.7\times 10^{6}}.

Estimate magnitude: numerator $\sim 10^{-9}$ , denominator $\sim 10^6$ , so $T \sim 10^{-15}$ to $10^{-16}\ \text{s}$ , which is consistent.
Calculation gives

T \approx 6.2\times 10^{-16}\ \text{s}.

Key Takeaways

Period in circular motion can be found from geometry: $T = 2\pi r / v$ .
Always keep units consistent.

Common Mistakes

Using diameter instead of radius in $2\pi r$ .
Forgetting $2\pi$ and using $T = r/v$ .
Using $r = 170\times 10^{-12}$ without converting to standard form (still okay if done consistently, but errors are common).

Things to Be Careful About

Use the speed from (c)(i) consistently (do not re-round too aggressively before substituting).
Period is extremely small; powers of ten are the main place students lose marks.

Techniques used

use circumference divided by speed to find periodsubstitute radius and speed in SI unitsapply T = 2 pi r / v

(d)

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

3M

(i)

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

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

ratio = ______

2M Similar

DifficultyMedium

Worked solution

Working

At one electron:

E_{\text{nucleus}} = k\frac{2e}{r^2}

and the other electron is at distance $2r$ so

E_{\text{electron}} = k\frac{e}{(2r)^2} = k\frac{e}{4r^2}.

Hence

\frac{E_{\text{electron}}}{E_{\text{nucleus}}} = \frac{\frac{ke}{4r^2}}{\frac{2ke}{r^2}} = \frac{1}{8} = 0.125.

Answer

$0.125$

Final answer

0.125

Detailed explanation

Background Concept

The electric field strength due to a point charge has magnitude

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

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

Understanding the Question

We consider one electron and compare:

the field at its position due to the nucleus (charge $+2e$ ) at distance $r$ ;
the field at its position due to the other electron (charge $-e$ ) at distance $2r$ (because the electrons are diametrically opposite).

The question asks for

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

Approach

Write $E = k|Q|/d^2$ for each source charge, using the correct separation $d$ .
Form the ratio so $k$ and $e$ cancel.

Step-by-Step Reasoning

Field due to nucleus (magnitude): charge $2e$ , distance $r$ :

E_{\text{nucleus}} = k\frac{2e}{r^2}.

Field due to other electron (magnitude): charge $e$ , distance is the diameter $2r$ :

E_{\text{electron}} = k\frac{e}{(2r)^2} = k\frac{e}{4r^2}.

Ratio:

\frac{E_{\text{electron}}}{E_{\text{nucleus}}} = \frac{k\frac{e}{4r^2}}{k\frac{2e}{r^2}}.

Cancel $k$ , cancel $e$ , cancel $r^2$ :

\frac{E_{\text{electron}}}{E_{\text{nucleus}}} = \frac{1/4}{2} = \frac{1}{8} = 0.125.

Key Takeaways

Electric field from a point charge obeys an inverse-square dependence.
Ratios often simplify dramatically because constants cancel.

Common Mistakes

Using distance $r$ instead of $2r$ for the electron–electron separation.
Forgetting the nucleus has charge $+2e$ .
Mixing up force and field (field is force per unit positive charge).

Things to Be Careful About

The ratio is of field strengths (magnitudes). Direction matters in the next part, but here the question is about the ratio of strengths.

Techniques used

use E = kQ/r^2 for point chargescompare field strengths by forming a ratiouse relative distances and charges to simplify algebra

(ii)

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

1M Similar

DifficultyMedium-Easy

Worked solution

Answer

The other electron produces a field (and hence force) opposite to that due to the nucleus at the electron’s position, so the resultant inward (centripetal) force is reduced (by about $12.5\%$ ). Therefore the electron would orbit with a smaller speed for the same radius, or equivalently a larger radius for the same speed (less tightly bound).

Final answer

Resultant inward force is reduced (other electron opposes nucleus), so orbit is less tightly bound: lower speed for same radius / larger radius for same speed.

Detailed explanation

Background Concept

An electron in a circular orbit needs an inward centripetal force

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

Electric forces provide this in atomic models. Forces (or fields) are vectors, so direction matters: a field due to a positive charge points away from it, and due to a negative charge points towards it. The force on an electron is opposite to the direction of the electric field (because $q=-e$ ).

Understanding the Question

You found in (d)(i) that the other electron’s field strength at the position of the first electron is $1/8$ of the nucleus’s field strength. Now you must use this to describe how the orbit changes when you include the effect of the other electron.

Approach

Work out the directions of the forces on one electron due to:
- the nucleus (attractive, inward towards nucleus),
- the other electron (repulsive, away from the other electron).
Compare magnitudes using the ratio $1/8$ .
Conclude what happens to the required centripetal force and thus to $v$ or $r$ .

Step-by-Step Reasoning

diagram-dii-1

Force due to the nucleus on the chosen electron is attractive (nucleus is $+2e$ , electron is $-e$ ), so it acts towards the nucleus (inward).
The other electron is negative; like charges repel, so it pushes the chosen electron away from the other electron. Since the electrons are opposite ends of a diameter, “away from the other electron” is along the diameter towards the outside, i.e. opposite to the inward pull of the nucleus.
From (d)(i), the magnitude of the other electron’s field (and so the force it would produce on a given charge) is $1/8$ of the nucleus’s. So the outward effect is significant but smaller.
Therefore the net inward force is reduced from the value in part (b)(ii). With a smaller centripetal force available,

\frac{mv^2}{r} \text{ is smaller.}

So either:

for the same radius $r$ , the speed $v$ must be smaller; or
for the same speed $v$ , the orbit radius $r$ must increase.

In words: the electron is less strongly pulled towards the nucleus, so it is less tightly bound and its orbit would be “larger”/slower than predicted when ignoring the second electron.

Key Takeaways

Always consider direction when adding electric effects from multiple charges.
A repulsive electron–electron interaction reduces the net attractive (centripetal) effect of the nucleus.

Common Mistakes

Treating fields as scalars and simply adding magnitudes without directions.
Thinking the other electron increases attraction (it does the opposite: it repels).
Saying “orbit becomes faster” despite a reduced inward force.

Things to Be Careful About

The ratio $1/8$ applies to field strength magnitudes; you must still state the direction to decide whether it increases or reduces the net inward force.
The question asks you to “suggest and explain” qualitatively; a short, direction-based argument is what scores.

Techniques used

determine directions of electric forces from attraction/repulsioncombine forces/fields to find the resultant centripetal effectrelate reduced inward force to a change in orbital speed or radius

Q3MediumTemperature

(a)

Define specific latent heat.

2M Similar

DifficultyEasy

Worked solution

Answer

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

Final answer

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

Detailed explanation

Background Concept

When a substance changes state (e.g. solid to liquid, liquid to gas), energy is transferred but the temperature stays constant during the change.

The specific latent heat $L$ of a particular change of state is defined by

E = mL

where:

$E$ is the energy transferred,
$m$ is the mass undergoing the change of state,
$L$ is energy per unit mass (units typically $\text{J kg}^{-1}$ , or sometimes $\text{J g}^{-1}$ ).

Understanding the Question

You are asked to define specific latent heat, so you must state what it means physically and include the key condition that temperature does not change during the phase change.

Approach

Give a one-sentence definition: “energy per unit mass needed to change state at constant temperature”.

Step-by-Step Reasoning

“Specific” means per unit mass.
“Latent heat” refers to energy transferred during a change of state.
The crucial point is that this energy transfer occurs without a change in temperature.

Key Takeaways

Use $E = mL$ .
Specific latent heat is energy per unit mass for a phase change.
Temperature remains constant during the change of state.

Common Mistakes

Saying “heat to raise the temperature” (that is specific heat capacity, not latent heat).
Missing “per unit mass”.
Forgetting to mention “no change in temperature”.

Things to Be Careful About

The definition must apply to any phase change (fusion, vaporisation, etc.).
Units may be $\text{J kg}^{-1}$ or $\text{J g}^{-1}$ depending on the question, but the definition is the same.

Techniques used

state the definition of specific latent heat in terms of energy per unit massspecify that the change of state occurs at constant temperature

(b)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

In vaporisation, molecules become much more widely separated than in fusion, so more energy is needed to overcome intermolecular forces (greater increase in molecular potential energy).

Also, during vaporisation the substance expands greatly and does work against atmospheric pressure (extra energy needed), whereas fusion involves much less expansion.

Final answer

Vaporisation needs more energy because molecules separate much further (greater increase in potential energy to overcome intermolecular forces) and the liquid does significant expansion work against atmospheric pressure; fusion needs less separation and little work done.

Detailed explanation

Background Concept

Latent heat is associated mainly with changes in molecular potential energy due to intermolecular forces.

In a solid, molecules are in fixed positions and strongly bound.
In a liquid, molecules are still close together but can move around each other; many bonds/attractions are weakened but not completely overcome.
In a gas, molecules are far apart and intermolecular forces are negligible.

For a phase change at constant temperature, the energy supplied does not increase average kinetic energy (since temperature is constant). Instead it goes into:

Increasing molecular potential energy by overcoming intermolecular attractions.
(For liquid to gas) doing work as the substance expands against external pressure.

Understanding the Question

You must explain why the specific latent heat of vaporisation $L_v$ is usually larger than the specific latent heat of fusion $L_f$ . This is a comparison of what must happen to molecules in each change of state.

Approach

Compare the microscopic changes:

fusion: solid $\to$ liquid (molecules remain close)
vaporisation: liquid $\to$ gas (molecules separate greatly)
Then add the macroscopic energy requirement for vaporisation: expansion work $W = p\Delta V$ .

Step-by-Step Reasoning

Fusion (melting): the ordered solid structure is broken, but molecules in the liquid are still close together. Only some of the intermolecular bonds/attractions are overcome, so the increase in potential energy per kg is relatively smaller.
Vaporisation (boiling/evaporation): molecules must separate to large distances so that they can move freely as a gas. This requires overcoming intermolecular attractions much more completely, giving a much larger increase in potential energy.
Expansion work: going from liquid to gas involves a very large increase in volume. The system must do work on the surroundings:

W = p\Delta V

This extra energy contributes to $L_v$ but is much smaller (often negligible) in fusion because volume change is small.

Key Takeaways

$L_v > L_f$ because vaporisation requires much greater separation of molecules (greater increase in potential energy).
Vaporisation also includes significant work done against atmospheric pressure due to large expansion.

Common Mistakes

Saying “temperature increases more” (temperature does not change during the phase change).
Only mentioning “bonds are broken” without comparing extent of separation in gas versus liquid.
Forgetting expansion work against external pressure as an additional reason.

Things to Be Careful About

Use clear comparative language: “much greater separation” and “much larger volume increase”.
Do not imply intermolecular forces are completely absent in a liquid; they are reduced, not eliminated.

Techniques used

compare molecular separation and intermolecular forces in different phasesrelate energy transfer to changes in molecular potential energyinclude work done against external pressure during expansion

(c)

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

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

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

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

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

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

4M Similar

DifficultyMedium

Worked solution

Working

Heat lost by warm water:

Q_{\text{lost}} = mc\Delta\theta = 208 \times 4.18 \times (26.4 - 10.3)

Q_{\text{lost}} = 1.40 \times 10^{4}\ \text{J}

Heat gained by ice to melt and then warm from $0.0\ ^{\circ}\text{C}$ to $10.3\ ^{\circ}\text{C}$ :

Q_{\text{gain}} = mL_f + mc\Delta\theta = 37.0L_f + 37.0 \times 4.18 \times 10.3

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

1.40 \times 10^{4} = 37.0L_f + 1.59 \times 10^{3}

L_f = \frac{1.40 \times 10^{4} - 1.59 \times 10^{3}}{37.0} = 3.35 \times 10^{2}\ \text{J g}^{-1}

Answer

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

Final answer

335 J g^-1

Detailed explanation

Background Concept

In an insulated system with negligible heat capacity of the container, energy is conserved: thermal energy lost by the warmer object equals thermal energy gained by the colder object.

Two energy relations are used:

Sensible heating/cooling:

Q = mc\Delta\theta

where $m$ is mass, $c$ is specific heat capacity, and $\Delta\theta$ is temperature change.

Change of state (fusion here):

Q = mL_f

where $L_f$ is the specific latent heat of fusion.

Understanding the Question

Ice: $m_i = 37.0\ \text{g}$ at $0.0\ ^{\circ}\text{C}$ .
Water: $m_w = 208\ \text{g}$ at $26.4\ ^{\circ}\text{C}$ .
Final equilibrium temperature: $10.3\ ^{\circ}\text{C}$ .
$c$ for water: $4.18\ \text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$ .
No heat exchange with surroundings; beaker heat capacity negligible.

What happens physically:

The warm water cools from $26.4\ ^{\circ}\text{C}$ to $10.3\ ^{\circ}\text{C}$ , losing energy.
The ice melts at $0.0\ ^{\circ}\text{C}$ (requires latent heat).
The meltwater then warms from $0.0\ ^{\circ}\text{C}$ to $10.3\ ^{\circ}\text{C}$ .

We are asked to find $L_f$ .

Approach

Set up an energy balance:

\text{heat lost by warm water} = \text{heat to melt ice} + \text{heat to warm melted water}

Compute each sensible heat term using $Q = mc\Delta\theta$ , then solve for $L_f$ using $Q = mL_f$ .

Step-by-Step Reasoning

Heat lost by the initial warm water as it cools:

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

Q_{\text{lost}} = 208 \times 4.18 \times 16.1

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

Heat gained after the ice has melted: first, the melted ice (now liquid water) warms from $0.0\ ^{\circ}\text{C}$ to $10.3\ ^{\circ}\text{C}$ :

Q_{\text{warm}} = 37.0 \times 4.18 \times 10.3 = 1.593 \times 10^{3}\ \text{J}

Heat used to melt the ice at $0.0\ ^{\circ}\text{C}$ :

Q_{\text{melt}} = 37.0 L_f

Energy conservation (insulated system):

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

So

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

Rearrange:

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

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

To three significant figures:

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

Key Takeaways

In an insulated mixing problem: heat lost = heat gained.
Remember to include both:
- latent heat for melting,
- sensible heat to warm the melted ice up to the final temperature.
Keep units consistent: here $c$ is in $\text{J g}^{-1}\ ^{\circ}\text{C}^{-1}$ so masses must be in $\text{g}$ .

Common Mistakes

Forgetting to include the heating of the melted ice from $0$ to $10.3\ ^{\circ}\text{C}$ .
Using the total final mass of water for the warming term (only the melted ice mass is warmed from $0$ ).
Mixing $\text{kg}$ and $\text{g}$ with $c$ given per gram.
Using $\Delta\theta = 10.3 - 26.4$ without taking the magnitude for heat lost.

Things to Be Careful About

The ice starts at $0.0\ ^{\circ}\text{C}$ , so there is no term for warming ice up to $0.0\ ^{\circ}\text{C}$ before melting.
Quote $L_f$ to three significant figures as requested.
Ensure the beaker being “perfectly insulated” means no energy transfer to surroundings, so the only exchange is between the water and the ice.

Techniques used

apply conservation of energy in an isolated thermal systemcalculate heat transfer using Q = mc\Delta\thetaequate heat lost by hot water to heat gained in melting and warming the ice meltwaterrearrange E = mL to determine specific latent heat

Q4MediumThermodynamicsIdeal Gases

(a)

4M

(i)

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

2M Similar

DifficultyEasy

Worked solution

Answer

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

Final answer

Sum of random kinetic and potential energies of the molecules.

Detailed explanation

Background Concept

Internal energy $U$ is the energy a substance has because of the microscopic motion and interactions of its particles. It is not the macroscopic kinetic energy of the whole object moving, and it is not necessarily equal to “heat”.

For a material made of many particles, internal energy is the total of:

random kinetic energy of the particles (translational, and possibly rotational/vibrational), and
potential energy associated with forces between particles (e.g. attractive/repulsive intermolecular forces).

Understanding the Question

You are asked to state what is meant by internal energy. That means you should give a precise definition rather than examples or a long explanation.

Approach

Use the standard Cambridge definition: internal energy is a sum of microscopic kinetic and potential energies. Include both parts to gain full credit.

Step-by-Step Reasoning

The system contains many molecules.
Each molecule has random motion, so it has kinetic energy.
Molecules also interact via intermolecular forces, so there is potential energy associated with their separations.
Adding all these microscopic contributions gives the internal energy $U$ .

Key Takeaways

Internal energy refers to microscopic energy, not macroscopic motion.
For full marks, mention both random kinetic energy and intermolecular potential energy.

Common Mistakes

Saying “internal energy is heat energy” (heat is energy transferred due to temperature difference, not energy stored).
Mentioning only kinetic energy and forgetting potential energy (loses a mark in typical schemes).
Confusing internal energy with temperature (temperature relates to average kinetic energy, not total energy).

Things to Be Careful About

Use the word “random” (or “microscopic”) to distinguish from bulk kinetic energy.
State “sum/total” clearly: it is for all the molecules in the system.

Techniques used

state the definition of internal energyidentify the microscopic energy stores that contribute to internal energy

(ii)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

For an ideal gas there are no intermolecular forces, so the internal energy is the total random kinetic energy of the molecules.

The average kinetic energy per molecule is proportional to thermodynamic temperature, e.g.

\langle E_k \rangle = \frac{3}{2}kT.

Hence total internal energy $U = N\langle E_k \rangle \propto T$ .

Final answer

For an ideal gas, internal energy is molecular kinetic energy and (\langle E_k\rangle = \tfrac{3}{2}kT), so (U \propto T).

Detailed explanation

Background Concept

An ideal gas model assumes:

molecules are point particles with negligible volume,
no intermolecular forces except during perfectly elastic collisions,
therefore (importantly here) there is no intermolecular potential energy store.

So for an ideal gas, internal energy is purely the random kinetic energy of the molecules.

Kinetic theory gives a direct link between thermodynamic temperature $T$ and molecular motion. One key result is the average translational kinetic energy per molecule:

\langle E_k \rangle = \frac{3}{2}kT,

where $k$ is the Boltzmann constant.

Understanding the Question

You must explain why internal energy $U$ is directly proportional to $T$ for an ideal gas. So you need two ideas:

in an ideal gas, internal energy comes only from kinetic energy (no potential energy term), and
kinetic theory shows average kinetic energy is proportional to $T$ .

Approach

Start from the definition of internal energy as kinetic + potential energies.
Use the ideal-gas assumption “no intermolecular forces” to remove the potential part.
Use the kinetic theory result that average kinetic energy per molecule is proportional to $T$ .
Multiply by the number of molecules $N$ to get total internal energy proportional to $T$ .

Step-by-Step Reasoning

For any system,

internal energy $U$ is the total random kinetic energy + potential energy of molecules.

For an ideal gas,

there are no intermolecular forces (except during brief collisions), so there is no significant intermolecular potential energy stored.
Therefore $U$ is just the total random kinetic energy.

Kinetic theory gives

\langle E_k \rangle = \frac{3}{2}kT.

This shows that as $T$ increases, the average kinetic energy per molecule increases in direct proportion.

The total internal energy is the sum over all molecules:

U = N\langle E_k \rangle.

Substitute the kinetic theory relation:

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

Since $\frac{3}{2}Nk$ is a constant for a fixed amount of gas, $U \propto T$ .

Key Takeaways

Ideal gas: internal energy depends only on temperature because it is only kinetic energy.
The proportionality comes from $\langle E_k \rangle = \frac{3}{2}kT$ .

Common Mistakes

Forgetting to mention why potential energy is absent (must connect to “no intermolecular forces”).
Saying “ $U$ is proportional to $T$ because $pV = nRT$ ” without linking to internal energy (that equation of state alone does not define $U$ ).
Confusing average kinetic energy per molecule with total internal energy (must multiply by $N$ ).

Things to Be Careful About

The proportionality is to thermodynamic temperature in kelvin.
This statement is true for an ideal gas; for real gases, intermolecular potential energy can change so $U$ can depend on both $T$ and volume/pressure.

Techniques used

apply the kinetic theory model for an ideal gasuse the proportionality of average molecular kinetic energy to thermodynamic temperaturelink total internal energy to total molecular kinetic energy

(b)

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

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

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

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

Table 4.1

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

4M Similar

DifficultyMedium

Worked solution

Working

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

At $T$ : internal energy $= U$ .
At $3T$ : internal energy $= 3U$ .
At $2T$ : internal energy $= 2U$ .

Use first law (work done on gas):

\Delta U = q + W.

Compression: $\Delta U = 3U - U = 2U$ , and $W = +W$ .

q = \Delta U - W = 2U - W.

Cooling at constant volume: $W = 0$ .

\Delta U = 2U - 3U = -U \quad\Rightarrow\quad q = -U.

Answer

Compression: work $= +W$ , thermal energy supplied $= 2U - W$ , increase in internal energy $= +2U$ .

Cooling: work $= 0$ , thermal energy supplied $= -U$ , increase in internal energy $= -U$ .

Final answer

Compression: (+W, 2U − W, +2U). Cooling: (0, −U, −U).

Detailed explanation

Background Concept

Two key ideas are being tested:

Internal energy of an ideal gas depends only on temperature
For a fixed amount of ideal gas, internal energy is proportional to thermodynamic temperature:

U \propto T.

So if temperature is multiplied by 3, the internal energy is multiplied by 3.

First law of thermodynamics
Using the Cambridge sign convention stated in many mark schemes:

\Delta U = q + W,

where

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

Understanding the Question

You are told:

initially the gas is at temperature $T$ and has internal energy $U$ .
it is compressed, temperature rises to $3T$ , and work $+W$ is done on the gas.
then it is cooled at constant volume until the temperature falls to $2T$ .

You must fill the table entries (in terms of $W$ , $T$ , $U$ ) for each process:

work done on the gas,
thermal energy supplied to the gas ( $q$ ),
increase (change) in internal energy ( $\Delta U$ ).

Approach

Use $U \propto T$ to find the internal energy at $3T$ and $2T$ .
For each process, calculate $\Delta U$ from final minus initial internal energy.
Apply $\Delta U = q + W$ to solve for $q$ .
Remember: constant volume implies $\Delta V = 0$ , so $W = p\Delta V = 0$ for that stage.

Step-by-Step Reasoning

1) Write internal energies at each temperature

Given $U$ at $T$ and $U \propto T$ :

At $3T$ :

U_{3T} = 3U.

At $2T$ :

U_{2T} = 2U.

2) Compression stage ( $T \to 3T$ )

Change in internal energy:

\Delta U = U_{3T} - U_T = 3U - U = 2U.

Work done on gas is given as $+W$ .

Apply first law:

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

So the gas may still receive heat during compression if $2U - W$ is positive, or lose heat if it is negative. The expression itself is what is required.

3) Cooling stage at constant volume ( $3T \to 2T$ )

At constant volume, $\Delta V = 0$ so the mechanical work term is:

W = p\Delta V = 0.

Change in internal energy from $3U$ down to $2U$ :

\Delta U = U_{2T} - U_{3T} = 2U - 3U = -U.

Apply first law:

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

Negative $q$ correctly indicates that thermal energy is removed from the gas during cooling.

Key Takeaways

For an ideal gas, temperature ratios give internal energy ratios directly.
Use a consistent sign convention in $\Delta U = q + W$ .
Constant volume implies zero work done.

Common Mistakes

Using the wrong sign convention (e.g. $\Delta U = q - W$ while still taking compression work as positive).
Taking cooling change as $+U$ instead of $-U$ (final internal energy is smaller).
Forgetting that constant volume implies $W = 0$ .
Using $3T \to 2T$ but subtracting in the wrong order when finding $\Delta U$ .

Things to Be Careful About

“Increase in internal energy” in the table can be negative for cooling; you should write $-U$ (not just $U$ ).
Keep $q$ as “thermal energy supplied to the gas”: if energy leaves, it must be negative.
Do not assume compression is adiabatic; the question does not say that, so $q$ is not necessarily zero.

Techniques used

use the proportionality of internal energy to thermodynamic temperature for an ideal gasapply the first law of thermodynamics with a consistent sign conventiondeduce heat transfer from changes in internal energy and work done

Q5MediumOscillations

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

Fig. 5.1

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

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

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

Fig. 5.2

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

Fig. 5.3

(a)

2M

(i)

Determine the amplitude of the oscillations.

amplitude = ______ $\text{m}$

1M Similar

DifficultyEasy

Worked solution

Working

From Fig. 5.3, turning points at $h = 0.8\ \text{m}$ and $h = 2.0\ \text{m}$ .

Equilibrium at midpoint: $h_0 = \dfrac{0.8 + 2.0}{2} = 1.4\ \text{m}$ .

Amplitude $x_0 = 2.0 - 1.4 = 0.60\ \text{m}$ .

Answer

$0.60\ \text{m}$

Final answer

0.60 m

Detailed explanation

Background Concept

For an oscillator, the amplitude $x_0$ is the maximum displacement from the equilibrium position.

In a displacement-style graph, the turning points (where the object changes direction) are where the speed is zero. Since kinetic energy is

E_K = \frac{1}{2}mv^2,

turning points correspond to $E_K = 0$ .

Understanding the Question

The depth of the block’s base is $h$ . The block oscillates up and down, so $h$ varies between two extreme values.

Fig. 5.3 is a graph of $E_K$ against $h$ . We use the two values of $h$ where $E_K=0$ as the two turning points, and then find the amplitude as the distance from equilibrium to either turning point.

Approach

Read the two $h$ values where $E_K=0$ (turning points).
Find the equilibrium depth $h_0$ (midpoint between turning points).
Amplitude is $x_0 = |h_{\text{turn}} - h_0|$ .

Step-by-Step Reasoning

From Fig. 5.3, $E_K=0$ at $h=0.8\ \text{m}$ and $h=2.0\ \text{m}$ . These are the turning points.
The equilibrium position lies halfway between turning points:

h_0 = \frac{0.8 + 2.0}{2} = 1.4\ \text{m}.

The amplitude is the maximum displacement from equilibrium:

x_0 = 2.0 - 1.4 = 0.60\ \text{m}

(or equally $1.4-0.8=0.60\ \text{m}$ ).

Key Takeaways

Turning points occur where $v=0$ , so $E_K=0$ .
Equilibrium is midway between turning points for symmetric oscillations.
Amplitude is the distance from equilibrium to a turning point.

Common Mistakes

Using the full peak-to-peak range ( $2.0-0.8=1.2\ \text{m}$ ) as the amplitude (that is actually $2x_0$ ).
Taking the $h$ value at maximum $E_K$ as the amplitude (it is the equilibrium depth).

Things to Be Careful About

Read $h$ values carefully from the axis scale.
Quote the amplitude to a sensible precision consistent with the graph (here $0.60\ \text{m}$ or $0.6\ \text{m}$ would be acceptable depending on expected precision).

Techniques used

read turning points from an energy–position graphuse equilibrium position from the zero-acceleration pointcalculate amplitude as maximum displacement from equilibrium

(ii)

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

1M Similar

DifficultyEasy

Worked solution

Answer

The line shows $a$ is proportional to $-\,$ displacement from equilibrium (straight line with negative gradient), so the motion is simple harmonic.

Final answer

Acceleration is proportional to negative displacement from equilibrium, so the oscillations are SHM.

Detailed explanation

Background Concept

Simple harmonic motion (SHM) is defined by the condition

a = -\omega^2 x,

where:

$x$ is displacement from equilibrium,
$a$ is acceleration,
$\omega$ is angular frequency (constant).

This means acceleration is directly proportional to displacement, and always directed towards equilibrium (opposite sign to $x$ ).

Understanding the Question

Fig. 5.2 plots acceleration $a$ against depth $h$ . Depth $h$ plays the role of “position”. The equilibrium position is where the acceleration is zero. The question asks what the straight line implies about the nature of the oscillations.

Approach

Identify that displacement from equilibrium is $x = h - h_0$ .
Check whether the graph shows a straight-line relationship with negative gradient between $a$ and $h-h_0$ .
Conclude SHM if it matches $a \propto -x$ .

Step-by-Step Reasoning

The graph of $a$ against $h$ is a straight line with negative gradient.
It crosses $a=0$ at a particular depth $h_0$ (the equilibrium depth).
For $h > h_0$ , the acceleration is negative (upwards), pushing back toward $h_0$ .
For $h < h_0$ , the acceleration is positive (downwards), again pushing back toward $h_0$ .
This is exactly the SHM condition $a = -\omega^2(h-h_0)$ .

Key Takeaways

SHM can be identified from an $a$ –position graph: a straight line through equilibrium with negative slope.

Common Mistakes

Saying “acceleration is constant” (a straight line on an $a$ – $h$ graph is not constant acceleration; constant $a$ would be a horizontal line).
Not referencing equilibrium (SHM is about acceleration relative to equilibrium position).

Things to Be Careful About

The proportionality is to displacement from equilibrium $x=h-h_0$ , not to $h$ measured from the surface.

Techniques used

interpret a linear acceleration–displacement relationshipidentify simple harmonic motion from proportionality to negative displacement

(b)

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

3M Similar

DifficultyMedium

Worked solution

Working

From Fig. 5.2, $a=0$ at $h_0 = 1.4\ \text{m}$ (equilibrium depth).

Using points $(h,a)=(2.0,-0.6)$ and $(1.4,0)$ :

\text{gradient} = \frac{\Delta a}{\Delta h} = \frac{-0.6-0}{2.0-1.4} = -1.0\ \text{s}^{-2}

So $a = -\omega^2(h-h_0)$ gives $\omega^2 = 1.0\ \text{s}^{-2}$ and $\omega = 1.0\ \text{rad s}^{-1}$ .

From (a)(i), $x_0 = 0.60\ \text{m}$ so

v_{\max} = \omega x_0 = 1.0 \times 0.60 = 0.60\ \text{m s}^{-1}.

From Fig. 5.3, $E_{K\max} = 8.0\ \text{J}$ , so

8.0 = \frac{1}{2} m (0.60)^2 \Rightarrow m = \frac{16}{0.36} = 44\ \text{kg}.

Answer

Equilibrium depth $h_0 = 1.4\ \text{m}$ .

$\omega = 1.0\ \text{rad s}^{-1}$ (so $T = 2\pi\ \text{s}$ ).

Mass of block $m \approx 44\ \text{kg}$ .

Final answer

h0 = 1.4 m; ω = 1.0 rad s^-1 (T = 2π s); m ≈ 44 kg

Detailed explanation

Background Concept

For SHM, if $x$ is displacement from equilibrium,

a = -\omega^2 x.

So an acceleration–position graph is a straight line with gradient $-\omega^2$ .

Also, the maximum speed in SHM is

v_{\max} = \omega x_0,

and maximum kinetic energy occurs at equilibrium where the speed is maximum:

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

The turning points have $v=0$ so $E_K=0$ .

Understanding the Question

You are given two graphs:

Fig. 5.2: acceleration $a$ versus depth $h$ .
Fig. 5.3: kinetic energy $E_K$ versus depth $h$ .

You must state three quantitative conclusions about the block/oscillation (numerical values). Typical useful quantities are equilibrium depth, angular frequency/period, maximum speed/acceleration, maximum kinetic energy, mass, etc.

Approach

Use the graphs to extract three numbers:

Read $h_0$ from where $a=0$ (equilibrium).
Find $\omega$ from the gradient of the straight-line $a$ – $h$ graph.
Combine $\omega$ , amplitude $x_0$ (from turning points) and $E_{K\max}$ to find another quantity such as $v_{\max}$ or $m$ .

Step-by-Step Reasoning

Equilibrium depth

Equilibrium is where the restoring acceleration is zero.
From Fig. 5.2, $a=0$ at $h=1.4\ \text{m}$ .
So $h_0 = 1.4\ \text{m}$ .

Angular frequency

In SHM, $a = -\omega^2(h-h_0)$ .
This means a plot of $a$ against $h$ is a straight line with gradient $-\omega^2$ .
Using two clear points from the line, e.g. $(1.4,0)$ and $(2.0,-0.6)$ :

\frac{\Delta a}{\Delta h} = \frac{-0.6-0}{2.0-1.4} = -1.0\ \text{s}^{-2}.

Therefore $\omega^2 = 1.0\ \text{s}^{-2}$ and

\omega = 1.0\ \text{rad s}^{-1}.

You could also state the period:

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

Maximum speed and mass

From Fig. 5.3, $E_K=0$ at turning points $h=0.8\ \text{m}$ and $2.0\ \text{m}$ , so amplitude is $x_0=0.60\ \text{m}$ .
Then

v_{\max} = \omega x_0 = 1.0 \times 0.60 = 0.60\ \text{m s}^{-1}.

Maximum kinetic energy from Fig. 5.3 is $8.0\ \text{J}$ , and this equals $\tfrac{1}{2}mv_{\max}^2$ :

8.0 = \frac{1}{2}m(0.60)^2.

Solving:

m = \frac{2\times 8.0}{0.60^2} = \frac{16}{0.36} = 44.4\ \text{kg} \approx 44\ \text{kg}.

Any three of these numerical results would count as “three quantitative conclusions”, but they must be distinct and supported by the graphs.

Key Takeaways

$a$ –position straight line: gradient gives $-\omega^2$ .
$E_K=0$ identifies turning points; max $E_K$ occurs at equilibrium.
Combine $v_{\max}=\omega x_0$ with $E_{K\max}=\tfrac12 mv_{\max}^2$ to find the mass.

Common Mistakes

Using gradient as $-\omega$ instead of $-\omega^2$ .
Forgetting that displacement is from equilibrium: $x=h-h_0$ .
Using $E_{K\max}=\tfrac12 m (\omega^2 x_0^2)$ directly but mishandling brackets/powers.

Things to Be Careful About

Units: gradient of $a$ (m s $^{-2}$ ) versus $h$ (m) has units s $^{-2}$ , matching $\omega^2$ .
Read-off accuracy: choose two well-separated points on the straight line to reduce reading error.
Significant figures: values taken from a sketch/graph are usually given to 2 s.f. at best.

Techniques used

read equilibrium and extreme positions from graphsdetermine angular frequency from the gradient of an acceleration–displacement graphuse SHM relation between maximum speed, angular frequency and amplitudeuse maximum kinetic energy to calculate the mass

(c)

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

Fig. 5.4

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Total energy $E = E_K + E_P$ is constant.
From Fig. 5.3, $E_{K\max}=8.0\ \text{J}$ at $h=1.4\ \text{m}$ , so take $E=8.0\ \text{J}$ .

Hence $E_P = 8.0 - E_K$ :

$E_P=0$ at $h=1.4\ \text{m}$ ,
$E_P=8.0\ \text{J}$ at $h=0.8\ \text{m}$ and $h=2.0\ \text{m}$ ,
draw an upward-opening parabola symmetric about $h=1.4\ \text{m}$ through these points.

diagram-c-ep-vs-h

Final answer

Upward-opening parabola with minimum EP = 0 at h = 1.4 m and EP = 8 J at h = 0.8 m and 2.0 m.

Detailed explanation

Background Concept

For (undamped) SHM, the total mechanical energy is constant:

E = E_K + E_P.

At equilibrium: speed is maximum, so $E_K$ is maximum and $E_P$ is minimum.
At turning points: speed is zero, so $E_K=0$ and $E_P$ is maximum.

Energy graphs against position are typically parabolic:

$E_P \propto x^2$ (upward-opening parabola),
$E_K = E - E_P$ (downward-opening parabola).

Understanding the Question

You are given the kinetic energy variation with depth $h$ (Fig. 5.3) and asked to sketch the potential energy $E_P$ against $h$ on Fig. 5.4.

So we must draw a curve that, when added to $E_K(h)$ , gives a constant total energy.

Approach

Read the maximum value of $E_K$ (this equals total energy if we set $E_P=0$ at equilibrium).
Use $E_P = E - E_K$ .
Plot key anchor points: equilibrium and turning points.
Sketch the smooth curve through them with the correct symmetry.

Step-by-Step Reasoning

From Fig. 5.3, $E_K$ has a maximum of $8.0\ \text{J}$ at $h=1.4\ \text{m}$ .
At that position, the block passes through equilibrium, so $E_P$ is minimum. A standard convention is to take this minimum as zero for the oscillation energy.
Thus total energy $E = 8.0\ \text{J}$ .
At the turning points, $E_K=0$ (since $v=0$ ), which occur at $h=0.8\ \text{m}$ and $h=2.0\ \text{m}$ . Therefore

E_P = E - E_K = 8.0 - 0 = 8.0\ \text{J}

at both turning points.

So the $E_P$ $E_{P}$ graph must:
- have a minimum at $(1.4,0)$ ,
- pass through $(0.8,8.0)$ and $(2.0,8.0)$ ,
- be symmetric about $h=1.4\ \text{m}$ ,
- curve upwards (since $E_P$ increases with $|h-h_0|$ ).

diagram-c-ep-vs-h

Key Takeaways

For SHM without damping, $E_K + E_P$ is constant.
Turning points give $E_K=0$ and maximum $E_P$ .
The potential energy curve is the “complement” of the kinetic energy curve.

Common Mistakes

Drawing $E_P$ as a straight line or a downward-opening parabola.
Making $E_P=0$ at the turning points (it is $E_K$ that is zero there).
Putting the minimum $E_P$ at the wrong $h$ (it must be at the same $h$ where $E_K$ is maximum).

Things to Be Careful About

The sketch must pass through the correct numerical points read from the graph.
Keep symmetry about the equilibrium depth $h=1.4\ \text{m}$ .
Use a smooth curve (parabola-like), not sharp corners.

Techniques used

use conservation of mechanical energy to relate kinetic and potential energyinfer potential-energy graph shape from the kinetic-energy graphidentify key points (turning points and equilibrium) to anchor a sketch

Q6MediumAlternating CurrentsCapacitance

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

Fig. 6.1

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

(a)

2M

(i)

State what is meant by rectification.

1M Similar

(ii)

State the purpose of capacitor $C$ .

1M Similar

DifficultyEasy

Worked solution

Answer

To smooth $V_{OUT}$ (reduce the ripple): the capacitor charges to a peak value and then discharges through $R$ between peaks to maintain the output p.d.

Final answer

To smooth the output (reduce ripple) by charging at peaks and discharging through R between peaks.

Detailed explanation

Background Concept

A capacitor stores charge and energy. In a rectifier circuit, a capacitor connected in parallel with the load can act as a reservoir:

when the rectified output rises, the capacitor charges up,
when the rectified output falls, the capacitor discharges through the load, supplying current and keeping the output voltage from dropping too quickly.

This reduces the variation (ripple) in the output voltage.

Understanding the Question

Capacitor $C$ is in parallel with $R$ across the output terminals. The question asks what its purpose is in this position.

Approach

State the smoothing function and mention charging at peaks and discharging between peaks.

Step-by-Step Reasoning

During times when the rectified voltage is high, $C$ charges so that its p.d. becomes close to the peak output voltage.
When the rectified voltage starts to fall (e.g. between peaks), the capacitor cannot change its voltage instantly, so it discharges through $R$ .
This discharge provides current to the resistor and keeps $V_{OUT}$ larger than it would be without the capacitor.
The result is a more steady (smoothed) d.c. output with reduced ripple.

Key Takeaways

A parallel capacitor across the load is a smoothing capacitor.
It works by charge storage and discharge between peaks.

Common Mistakes

Saying the capacitor “increases the voltage”: it mainly reduces the drop between peaks; it does not create extra energy.
Saying it “filters out negative half-cycles”: that is the diode/rectifier’s job.

Things to Be Careful About

The smoothing effectiveness depends on the time constant $RC$ and the time between peaks of the rectified waveform.

Techniques used

describe the charge and discharge of a capacitor in a rectifier circuitlink capacitor action to reduction of ripple in the output voltage

(b)

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

Fig. 6.2

8M

(i)

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

V_{IN} = A \cos Bt

where $A$ and $B$ are constants.

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

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

From Fig. 6.2, peak value $= 12\ \text{V}$ so

A = 12\ \text{V}

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

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

Answer

$A = 12\ \text{V}$

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

Final answer

A = 12 V, B = 3.14 × 10^2 rad s^-1

Detailed explanation

Background Concept

A sinusoidal voltage can be written in different but equivalent forms:

V = V_0 \sin(\omega t) \quad \text{or} \quad V = V_0 \cos(\omega t)

where:

$V_0$ is the peak (amplitude),
$\omega$ is the angular frequency in $\text{rad s}^{-1}$ ,
$\omega$ and the period $T$ are related by

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

In the expression $V_{IN} = A \cos Bt$ , $A$ is the amplitude and $B$ is the angular frequency.

Understanding the Question

You are given the graph of $V_{IN}$ against time. You must read off the peak voltage and the period, then use these to find $A$ and $B$ in $V_{IN} = A \cos Bt$ .

Approach

Read amplitude (peak value) from the vertical axis to get $A$ .
Read period $T$ from the horizontal axis.
Compute $B$ using $B = 2\pi/T$ .

Step-by-Step Reasoning

From the graph, the sinusoid reaches a maximum of $+12\ \text{V}$ (and a minimum of $-12\ \text{V}$ ), so the amplitude is $12\ \text{V}$ . Hence $A = 12\ \text{V}$ .
One complete cycle takes $20\ \text{ms}$ , so $T = 20\ \text{ms} = 20 \times 10^{-3}\ \text{s} = 2.0 \times 10^{-2}\ \text{s}.$
Then $B = \omega = \frac{2\pi}{T} = \frac{2\pi}{2.0 \times 10^{-2}} \approx 3.14 \times 10^{2}\ \text{rad s}^{-1}.$

Key Takeaways

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

Common Mistakes

Using $B = 1/T$ instead of $2\pi/T$ (that would give frequency $f$ , not angular frequency).
Forgetting to convert milliseconds to seconds.
Using peak-to-peak value ( $24\ \text{V}$ ) as the amplitude.

Things to Be Careful About

The question already states the unit for $B$ as $\text{rad s}^{-1}$ ; ensure your value matches that.
$A$ must have a unit (volts).

Techniques used

read peak value and period from a sinusoidal graphuse angular frequency $\omega = 2\pi/T$match a sinusoid to the form $A\cos Bt$

(ii)

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

1M Similar

(iii)

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

Fig. 6.3

2M Similar

DifficultyMedium

Worked solution

diagram-6biii-bridge

Answer

Four diodes connected as a bridge rectifier between W, X (a.c. input) and Y, Z (d.c. output).

Final answer

Bridge rectifier (four diodes) drawn inside the box.

Detailed explanation

Background Concept

A bridge rectifier uses four diodes arranged so that, regardless of the polarity of the a.c. input, current through the load flows in the same direction. The a.c. input is connected to one pair of opposite corners of the bridge; the d.c. output is taken from the other pair.

In each half-cycle, two diodes conduct (forward-biased) and two block (reverse-biased), producing a unidirectional output.

Understanding the Question

You are given a box labelled “rectification circuit” with four terminals W, X (input) and Y, Z (output). You must draw what components are inside this box to produce the $V_{OUT}$ waveform identified as full-wave rectified.

Approach

Draw a standard bridge rectifier:

Put four diodes in a diamond.
Connect W and X to the left and right nodes (a.c. input).
Connect Y and Z to the top and bottom nodes (d.c. output), with diode directions chosen so the same output terminal is always positive.

Step-by-Step Reasoning

diagram-6biii-bridge

Arrange diodes so that when W is positive relative to X, current is guided to make Y positive relative to Z.
When X is positive relative to W (the opposite half-cycle), a different pair of diodes conducts, but current through the load still makes Y positive relative to Z.
This produces full-wave rectification, consistent with the output peaks occurring every $T/2$ .

Key Takeaways

Full-wave rectification in this context is most commonly achieved with a four-diode bridge.
Correct diode orientation is essential: the output polarity must not reverse.

Common Mistakes

Drawing only one diode (half-wave rectifier) or two diodes with a centre-tapped transformer (not shown here).
Reversing one diode so that the bridge does not give a consistent output polarity.
Connecting the input to adjacent corners instead of opposite corners.

Things to Be Careful About

Label which terminals are the a.c. input (W, X) and which are the d.c. output (Y, Z).
Make sure exactly two diodes would be forward-biased in each half-cycle.

Techniques used

draw a bridge rectifier using four diodesset diode orientations so the output polarity is the same for both half-cycles

(iv)

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

time constant = ______ $\text{s}$

3M Similar

DifficultyMedium

Worked solution

Working

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

For discharge,

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

So

\frac{8}{12} = e^{-\Delta t/\tau}

\tau = \frac{\Delta t}{\ln\left(\frac{12}{8}\right)} = \frac{1.0 \times 10^{-2}}{\ln(1.5)} = 2.47 \times 10^{-2}\ \text{s}

Answer

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

Final answer

2.5 × 10^-2 s

Detailed explanation

Background Concept

When a capacitor discharges through a resistor, the capacitor voltage decreases exponentially:

V = V_0 e^{-t/RC}

The product $RC$ is the time constant $\tau$ :

\tau = RC

After a time $\tau$ , the voltage has fallen to $1/e \approx 0.37$ of its initial value.

In a rectifier with smoothing capacitor, the capacitor charges rapidly near the peaks of the rectified waveform, then discharges through the load resistor between peaks. The shape between peaks is approximately an exponential decay, so we can estimate $\tau$ from the ripple drop.

Understanding the Question

You are asked to estimate the discharge time constant of $C$ through $R$ using the output ripple graph. The graph shows $V_{OUT}$ decaying from near the peak to a minimum before the next recharge.

Approach

Identify two voltages on the discharge curve: $V_0$ at the start (near a peak) and $V$ at the end (just before the next peak).
Identify the time interval $\Delta t$ for that drop (time between charging peaks).
Use

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

and rearrange for $\tau$ .

Step-by-Step Reasoning

From the graph, the output is topped up to about $12\ \text{V}$ after each peak, then decays to about $8\ \text{V}$ before the next topping up.
Because the rectification is full-wave, the capacitor is recharged twice per input cycle. With input period $20\ \text{ms}$ , the time between recharges is $\Delta t = \frac{T}{2} = \frac{20\ \text{ms}}{2} = 10\ \text{ms} = 1.0 \times 10^{-2}\ \text{s}.$
Apply exponential discharge: $\frac{V}{V_0} = e^{-\Delta t/\tau}.$ Using $V_0 = 12\ \text{V}$ , $V = 8\ \text{V}$ : $\frac{8}{12} = e^{-\Delta t/\tau}.$
Take natural logs: $\ln\left(\frac{8}{12}\right) = -\frac{\Delta t}{\tau}$ so $\tau = \frac{\Delta t}{\ln\left(\frac{12}{8}\right)}.$
Substitute $\Delta t = 1.0 \times 10^{-2}\ \text{s}$ : $\tau = \frac{1.0 \times 10^{-2}}{\ln(1.5)} \approx 2.47 \times 10^{-2}\ \text{s} \approx 2.5 \times 10^{-2}\ \text{s}.$

Key Takeaways

Discharge of a capacitor through a resistor is exponential.
In a smoothed rectifier, the time between peaks (recharging events) is crucial: full-wave gives $T/2$ .
Estimating $\tau$ from two points uses a log rearrangement.

Common Mistakes

Using $\Delta t = 20\ \text{ms}$ even though the output is recharged every $10\ \text{ms}$ for full-wave.
Treating the decay as linear and using $\Delta V/\Delta t$ .
Using $\log_{10}$ instead of $\ln$ without converting properly.

Things to Be Careful About

Read $V_0$ and $V$ consistently from the graph (peak and trough of the ripple).
Convert milliseconds to seconds.
The graph-based value is an estimate; quoting to 2 s.f. is appropriate.

Techniques used

model capacitor discharge using exponential decayread time interval between peaks from a graphrearrange an exponential to solve for the time constant

(c)

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

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

resistance = ______ $\Omega$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

$C = 570\ \mu\text{F} = 570 \times 10^{-6}\ \text{F}$ .
Using $\tau = 2.47 \times 10^{-2}\ \text{s}$ :

R = \frac{2.47 \times 10^{-2}}{570 \times 10^{-6}} = 4.33 \times 10^{1}\ \Omega

Answer

$R \approx 43\ \Omega$ .

Final answer

43 Ω

Detailed explanation

Background Concept

For a capacitor $C$ discharging through a resistor $R$ , the time constant is

\tau = RC.

This comes directly from the exponential decay law $V = V_0 e^{-t/RC}$ .

Understanding the Question

You are given the capacitance $C = 570\ \mu\text{F}$ and asked to find $R$ using the time constant you estimated in part (b)(iv).

Approach

Rearrange $\tau = RC$ to $R = \tau/C$ , convert $\mu\text{F}$ to $\text{F}$ , then substitute.

Step-by-Step Reasoning

Start with $\tau = RC \Rightarrow R = \frac{\tau}{C}.$
Convert capacitance: $570\ \mu\text{F} = 570 \times 10^{-6}\ \text{F}.$
Substitute $\tau \approx 2.47 \times 10^{-2}\ \text{s}$ : $R = \frac{2.47 \times 10^{-2}}{570 \times 10^{-6}} \approx 4.33 \times 10^{1}\ \Omega \approx 43\ \Omega.$

Key Takeaways

Time constant links circuit components: $\tau = RC$ .
Always convert prefixes correctly (micro $= 10^{-6}$ ).

Common Mistakes

Using $570 \times 10^{6}$ instead of $570 \times 10^{-6}$ .
Forgetting units, or giving $R$ in the wrong unit.
Using $R = C/\tau$ (inverting the relationship).

Things to Be Careful About

Use a consistent number of significant figures: $\tau$ from a graph is approximate, so 2 s.f. for $R$ is appropriate.
Ensure $\tau$ is in seconds and $C$ is in farads before dividing.

Techniques used

use the definition of time constant $\tau = RC$convert microfarads to faradsrearrange to solve for resistance

Q7MediumMagnetic FieldsElectric Fields

(a)

Define magnetic flux density.

2M Similar

DifficultyEasy

Worked solution

Answer

Magnetic flux density $B$ is defined as the force per unit current per unit length on a straight conductor placed at right angles to the magnetic field:

B = \frac{F}{IL}

Final answer

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

Detailed explanation

Background Concept

Magnetic flux density $B$ (unit: tesla, $\text{T}$ ) describes the strength of a magnetic field in terms of the force it can exert.

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

F = BIL\sin\theta

where $\theta$ is the angle between the current direction and the magnetic field. The maximum force occurs when the conductor is perpendicular to the field ( $\theta = 90^\circ$ ), so $\sin\theta = 1$ .

Understanding the Question

The question asks you to define “magnetic flux density”. For Cambridge A Level, the expected definition is the one based on force on a current-carrying conductor, including the condition that the conductor is at right angles to the field.

Approach

Use the maximum-force case ( $\theta = 90^\circ$ ) in $F = BIL\sin\theta$ to express $B$ as a ratio involving force, current, and length.

Step-by-Step Reasoning

When the conductor is perpendicular to the magnetic field,

F = BIL

Rearrange to make $B$ the subject:

B = \frac{F}{IL}

So, $B$ is force per unit current per unit length, for a conductor placed at right angles to the field.

Key Takeaways

Use $F = BIL\sin\theta$ for force on a current-carrying wire.
The definition of $B$ uses the perpendicular (maximum force) arrangement.

Common Mistakes

Forgetting to state “per unit length” or “perpendicular to the field”.
Defining $B$ using a moving charge ( $F = BQv$ ) instead of the standard conductor definition (often not credited for a “define” command word).

Things to Be Careful About

Include the condition “at right angles to the field” (or equivalent) to secure the full definition marks.
Quote the defining equation clearly as $B = F/(IL)$ (not $B = F/(I)$ ).

Techniques used

state a standard definition using force per unit current per unit lengthspecify the perpendicular condition for maximum force

(b)

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

Fig. 7.1

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

3M

(i)

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

$F$ = ______

1M Similar

(ii)

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

1M Similar

(iii)

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

1M Similar

DifficultyMedium-Easy

Worked solution

diagram-b-iii

Answer

A circular arc curving upwards through the field region.

Final answer

Circular arc curving upwards

Detailed explanation

Background Concept

In a uniform magnetic field, if a charged particle’s velocity is perpendicular to $\vec{B}$ , the magnetic force has constant magnitude $F = BQv$ and is always perpendicular to the instantaneous velocity.

A force that is always perpendicular to velocity changes the direction of motion but not the speed. This is exactly the condition for uniform circular motion: the force acts as a centripetal force.

Understanding the Question

You found in (b)(ii) that the magnetic force at entry is upwards. The question asks you to draw a possible path through the field region. Because the force is always perpendicular to the motion, the path must curve.

Approach

Start with velocity to the right at Y.
Force is upwards at that instant, so the path initially curves upwards.
In a uniform field, the trajectory is a circular arc.

Step-by-Step Reasoning

At Y:

$\vec{v}$ is to the right.
$\vec{F}$ is upwards.
So the particle accelerates upwards, making its path bend upwards.

As soon as the velocity turns slightly upward, the magnetic force also turns so that it remains perpendicular to the new velocity, continuing to bend the path smoothly. This produces a circular arc (part of a circle) while the particle is inside the field region.

diagram-b-iii

Key Takeaways

In a uniform magnetic field, a charge moving perpendicular to $\vec{B}$ follows a circular path.
The curvature direction is set by the initial force direction.

Common Mistakes

Drawing a straight line (would require zero net force).
Drawing a sharp corner instead of a smooth curve (force changes direction continuously).
Curving the path the wrong way (downwards) after having the correct force direction.

Things to Be Careful About

“Possible path” means any reasonable circular arc consistent with the force direction; it does not have to hit a specific point unless stated.
Keep the curve smooth and starting tangent to the initial velocity direction at Y.

Techniques used

recognise that a perpendicular magnetic force provides centripetal accelerationsketch the resulting circular arc trajectory

(c)

5M

(i)

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

3M Similar

DifficultyMedium

Worked solution

Answer

Apply a uniform electric field across the region, perpendicular to the motion, so that the electric force on $+Q$ opposes the magnetic force.

Here the magnetic force is upwards, so choose $\vec{E}$ downward so that $\vec{F}_E = QE$ is downward.

Adjust $E$ (or select the correct $v$ ) so that

QE = BQv

so resultant force is zero and the particle travels straight to Z.

Final answer

Use a uniform E-field to provide an electric force opposite to the magnetic force; set QE = BQv so net force is zero and the particle goes straight to Z.

Detailed explanation

Background Concept

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

Electric force:

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

Magnetic force:

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

If the forces are equal in magnitude and opposite in direction, the resultant force is zero, so the particle continues in a straight line at constant velocity. This is the principle of a velocity selector.

Understanding the Question

In part (b), the particle entering the magnetic field is deflected (curves) and so would not reach the point Z straight ahead. You are asked how to use an electric field together with the magnetic field so that the particle now passes through Z (i.e. goes straight through without deflection).

Approach

Determine the direction of magnetic force (from part b: upwards for a positive charge).
Choose an electric field so the electric force is opposite (downwards).
State the condition for straight-line motion: forces balance, so net force is zero.

Step-by-Step Reasoning

From (b)(ii), at entry the magnetic force on $+Q$ is upwards. So to cancel this, we need an electric force downwards.

For a positive charge, electric force is in the direction of $\vec{E}$ . Therefore set up an electric field downward (e.g. with parallel plates: top plate positive, bottom plate negative, giving $\vec{E}$ downward).

Then the magnitudes can be balanced:

F_E = QE, \qquad F_B = BQv

When $QE = BQv$ , the forces are equal and opposite, resultant force is zero, and the particle does not accelerate sideways. It continues straight through to point Z.

Key Takeaways

Electric and magnetic forces can oppose each other in crossed fields.
Straight-line motion occurs when the net transverse force is zero.

Common Mistakes

Choosing $\vec{E}$ in the wrong direction (which would increase the deflection).
Saying “make $E$ equal to $B$ ” (wrong quantities; you must compare forces).
Forgetting that the charge is positive (direction of $\vec{F}_E$ depends on sign of charge).

Things to Be Careful About

Be explicit that the electric field must be perpendicular to the particle’s motion (so it produces a sideways force).
The balancing condition is about forces ( $QE$ and $BQv$ ), not about field strengths alone.

Techniques used

identify the magnetic deflection direction and required compensationapply electric force QE to oppose magnetic force BQvstate the condition for zero net force (straight-line path)

(ii)

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

$v$ = ______

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For no deflection,

QE = BQv

Cancel $Q$ :

E = Bv

So

v = \frac{E}{B}

Answer

v = \frac{E}{B}

Final answer

v = E/B

Detailed explanation

Background Concept

In a velocity selector, perpendicular electric and magnetic fields are adjusted so that a charged particle passes straight through. The condition is that the transverse forces cancel:

F_E = F_B

with $F_E = QE$ and (for $\vec{v} \perp \vec{B}$ ) $F_B = BQv$ .

Understanding the Question

You have introduced an electric field (part c(i)) so that the particle is not deflected and passes through point Z. You are now asked to derive the speed $v$ in terms of $B$ and the electric field strength $E$ .

Approach

Write expressions for electric and magnetic force magnitudes, set them equal for zero net force, and rearrange for $v$ .

Step-by-Step Reasoning

For a positive charge $Q$ :

Electric force magnitude:

F_E = QE

Magnetic force magnitude (since $\theta = 90^\circ$ ):

F_B = BQv

No deflection requires $F_E = F_B$ :

QE = BQv

Cancel $Q$ (provided $Q \neq 0$ ):

E = Bv

Rearrange:

v = \frac{E}{B}

Key Takeaways

Balance forces, not fields: $QE$ balances $BQv$ .
The selected speed is $v = E/B$ .

Common Mistakes

Writing $v = B/E$ (inverting the ratio).
Keeping $Q$ in the final expression (it cancels).
Using $F_B = BQv\sin\theta$ but forgetting $\sin\theta = 1$ here.

Things to Be Careful About

This result assumes $\vec{v}$ is perpendicular to $\vec{B}$ and $\vec{E}$ is set perpendicular to $\vec{v}$ .
$E$ must be in $\text{V m}^{-1}$ and $B$ in $\text{T}$ for $v$ to come out in $\text{m s}^{-1}$ .

Techniques used

equate electric and magnetic forces for zero deflectioncancel common factors to isolate the speedrearrange to obtain v in terms of E and B

Q8MediumQuantum PhysicsNuclear Physics

(a)

State what is meant by the de Broglie wavelength.

1M Similar

(b)

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

wavelength = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

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

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

Answer

$1.49 \times 10^{-11}\ \text{m}$ (≈ $1.5 \times 10^{-11}\ \text{m}$ )

Final answer

1.49 × 10^−11 m

Detailed explanation

Background Concept

The de Broglie wavelength is

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

For speeds that are not extremely close to $c$ , the momentum can be taken as the classical value

p = mv.

Understanding the Question

You are given an electron speed $v = 4.9 \times 10^{7}\ \text{m s}^{-1}$ and asked to calculate its de Broglie wavelength. You need $h$ and the electron mass $m$ .

Approach

Find the electron momentum using $p = mv$ .
Substitute into $\lambda = h/p$ .
Give the answer in metres with sensible significant figures.

Step-by-Step Reasoning

Momentum:

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

De Broglie wavelength:

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

This is about $1.5 \times 10^{-11}\ \text{m}$ to 2 s.f.

Key Takeaways

Higher momentum means smaller de Broglie wavelength.
Use SI units throughout so the result comes out in metres.

Common Mistakes

Using $p = mv^2$ instead of $p=mv$ .
Forgetting that $h$ is in $\text{J s}$ and momentum is in $\text{kg m s}^{-1}$ .
Dropping powers of ten when dividing.

Things to Be Careful About

This speed is about $0.16c$ ; most A-level mark schemes accept the non-relativistic momentum here.
Quote the final answer in $\text{m}$ as requested, and to a reasonable number of significant figures.

Techniques used

apply $\lambda = h/p$use non-relativistic momentum $p = mv$substitute values with correct SI unitsround to appropriate significant figures

(c)

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

similarity: ______

difference: ______

2M Similar

(d)

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

8M

(i)

State the name of this type of reaction.

1M Similar

(ii)

State what happens to the electron and to the positron.

2M Similar

(iii)

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

1M Similar

DifficultyMedium

Worked solution

Answer

The initial total momentum is zero (equal and opposite momenta). A single photon cannot have zero momentum ( $p = E/c$ ), so two photons must be produced with equal and opposite momenta to conserve momentum.

Final answer

To conserve momentum: one photon cannot have zero momentum; two opposite photons can.

Detailed explanation

Background Concept

Momentum is conserved in all interactions. Photons carry momentum as well as energy. For a photon,

p = \frac{E}{c}.

So if a photon has energy $E$ , its momentum cannot be zero.

Understanding the Question

The electron and positron have the same speed in opposite directions, so the total momentum before collision is zero. After the collision, photons are produced. The question asks why there must be two photons rather than one.

Approach

Use conservation of momentum:

Work out the total initial momentum.
Consider whether one photon could carry that momentum.
Conclude that two photons in opposite directions can give zero net momentum.

Step-by-Step Reasoning

diagram-diii-1

Before collision: electron momentum and positron momentum are equal in magnitude and opposite in direction, so

\vec{p}_\text{total} = \vec{p}_e + \vec{p}_{\text{pos}} = 0.

Suppose only one photon were produced. Any photon with energy $E$ has momentum $p = E/c$ , so its momentum is non-zero.
That would mean the total momentum after the collision would be non-zero, contradicting conservation of momentum.
With two photons, they can be emitted in opposite directions with equal momentum magnitudes, so the vector sum of their momenta is zero, matching the initial zero momentum.

Key Takeaways

Conservation of momentum is the key reason for two photons.
Photons always carry momentum if they carry energy.

Common Mistakes

Saying “two photons are produced to conserve energy” (energy conservation alone does not force two; momentum conservation does).
Claiming a photon can have zero momentum.

Things to Be Careful About

It is the vector nature of momentum that matters: equal and opposite momenta can sum to zero.
In this scenario the initial momenta cancel because the speeds are equal and directions opposite.

Techniques used

apply conservation of momentumuse photon momentum $p = E/c$deduce need for two photons with opposite momenta

(iv)

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

1M Similar

DifficultyMedium-Easy

Worked solution

Working

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

E_k = \frac{1}{2}(9.11 \times 10^{-31})(4.9 \times 10^{7})^2 = 1.09 \times 10^{-15}\ \text{J} \approx 1.1 \times 10^{-15}\ \text{J}

Answer

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

Final answer

1.1 × 10^−15 J

Detailed explanation

Background Concept

For a particle of mass $m$ moving with speed $v$ (non-relativistic), the kinetic energy is

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

Understanding the Question

You are asked to verify (“show that”) the kinetic energy of an electron with speed $4.9 \times 10^7\ \text{m s}^{-1}$ is $1.1 \times 10^{-15}\ \text{J}$ . This is a direct substitution check.

Approach

Substitute the electron mass and the given speed into $E_k = \tfrac{1}{2}mv^2$ , then round to match the stated value.

Step-by-Step Reasoning

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

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

Square the speed:

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

Multiply through:

E_k = \frac{1}{2}(9.11 \times 10^{-31})(2.401 \times 10^{15}) = \frac{1}{2}(2.19 \times 10^{-15}) = 1.09 \times 10^{-15}\ \text{J} \approx 1.1 \times 10^{-15}\ \text{J}.

Key Takeaways

“Show that” means your working should end at the given value (within rounding).

Common Mistakes

Forgetting the factor $\tfrac{1}{2}$ .
Squaring the power of ten incorrectly: $(10^{7})^2 = 10^{14}$ .

Things to Be Careful About

Keep track of significant figures: the given speed is 2 s.f., so $1.1 \times 10^{-15}\ \text{J}$ is appropriate.
At higher speeds you might need relativistic kinetic energy, but this question is clearly intended to use the classical expression.

Techniques used

use kinetic energy formula $E_k = \tfrac{1}{2}mv^2$substitute values with correct unitsround to the stated value

(v)

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

wavelength = ______ $\text{m}$

3M Similar

DifficultyMedium

Worked solution

Working

Energy per photon (momenta equal and opposite so energies equal):

E_\gamma = mc^2 + E_k

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

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

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

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

Answer

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

Final answer

2.39 × 10^−12 m

Detailed explanation

Background Concept

In electron–positron annihilation, the total energy available becomes photon energy. The important energy terms are:

Rest energy of each particle: $E_0 = mc^2$ .
Kinetic energy of each particle: $E_k$ .

For a photon,

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

Understanding the Question

An electron and a positron move with equal speeds in opposite directions and annihilate to produce two gamma photons. You are given (from part (iv)) the electron kinetic energy before collision: $E_k = 1.1 \times 10^{-15}\ \text{J}$ . You must find the wavelength of the emitted gamma radiation (to 3 s.f.).

Because the initial total momentum is zero, the two photons must have equal and opposite momenta; that implies they have equal energies (and therefore equal wavelengths).

Approach

Find the energy of one photon using energy conservation and the fact that there are two identical photons.
Include rest mass energy and kinetic energy.
Convert photon energy to wavelength with $\lambda = hc/E$ .

Step-by-Step Reasoning

Total initial energy:

Electron: $mc^2 + E_k$
Positron: $mc^2 + E_k$
So

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

Two photons are produced. With zero net momentum, they are emitted in opposite directions with equal energies, so

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

Calculate rest energy of electron:

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

Add the kinetic energy (from (iv)):

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

Convert energy to wavelength:

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

Key Takeaways

In annihilation, rest mass energy dominates over the kinetic energy at these speeds.
With equal and opposite initial momenta, the photons share the energy equally.
Use $E = hc/\lambda$ to get wavelength from photon energy.

Common Mistakes

Forgetting the rest mass energy $mc^2$ and using only kinetic energy (gives a much longer wavelength).
Dividing by 2 twice (once for two particles and again for two photons), leading to $E_\gamma = \tfrac{1}{2}(mc^2 + E_k)$ .
Using $E = h/\lambda$ (missing the factor $c$ ).

Things to Be Careful About

The energy per photon here is $mc^2 + E_k$ (not $2mc^2 + 2E_k$ ), because there are two photons.
Quote to three significant figures as requested.
Keep $c = 3.00 \times 10^8\ \text{m s}^{-1}$ consistent with the precision expected.

Techniques used

apply energy conservation in annihilationuse mass-energy equivalence $E = mc^2$use photon energy relation $E = hc/\lambda$account for two-photon sharing of energy

Q9MediumNuclear Physics

(a)

Define activity of a radioactive sample.

1M Similar

(b)

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

3M Similar

DifficultyMedium-Easy

Worked solution

Answer

Each nucleus has a constant probability per unit time of decaying, so the rate of decay is proportional to the number $N$ of undecayed nuclei:

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

Solving gives

N = N_0 e^{-\lambda t}

and since $A = \lambda N$ ,

A = A_0 e^{-\lambda t}

so activity varies exponentially with time.

Final answer

Because dN/dt = -λN (constant decay probability), giving N = N0 e^{-λt} and hence A = A0 e^{-λt}.

Detailed explanation

Background Concept

Radioactive decay is random for a single nucleus, but for a large sample we can describe it statistically.

Key idea: in a given short time $\Delta t$ , each nucleus has the same constant probability of decaying. This probability per unit time is the decay constant $\lambda$ (units $\text{s}^{-1}$ or $\text{min}^{-1}$ ).

If there are $N$ undecayed nuclei present, then more nuclei means more potential decays per second. So the overall decay rate must be proportional to $N$ .

Understanding the Question

The question asks why the activity-time graph is exponential (not linear). You need to connect:

random decay with constant probability,
rate proportional to how many are left,
resulting exponential solution.

Approach

Use “constant probability per unit time” to justify that the number decaying per second is proportional to $N$ .
Write the differential equation $\mathrm{d}N/\mathrm{d}t = -\lambda N$ .
State that solving it gives an exponential, and then relate activity to $N$ via $A = \lambda N$ .

Step-by-Step Reasoning

Constant chance of decay: each nucleus is equally likely to decay in any short time interval.
More nuclei (\Rightarrow) more decays per second: if you double $N$ , you expect double the decays per second. So the decay rate is proportional to $N$ .
Write the rate equation:

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

The minus sign shows $N$ decreases.
4. This differential equation has the standard solution:

N = N_0 e^{-\lambda t}

Activity is defined by

A = \lambda N

so substituting $N(t)$ gives

A = \lambda N_0 e^{-\lambda t} = A_0 e^{-\lambda t}

Therefore the activity decreases exponentially.

Key Takeaways

Exponential decay happens because the rate is proportional to the amount remaining.
The physical reason is a constant probability per unit time for each nucleus.
$N(t)$ and $A(t)$ both follow $e^{-\lambda t}$ .

Common Mistakes

Claiming it is exponential “because the half-life is constant” without explaining why the half-life is constant.
Writing $A = A_0 e^{+\lambda t}$ (wrong sign).
Mixing up $A$ and $N$ without using $A = \lambda N$ .

Things to Be Careful About

Be explicit that it is the probability per unit time that is constant, not the number of decays per second.
Ensure you connect the argument to activity: either through $A \propto N$ or $A = \lambda N$ .
Use correct units for $\lambda$ (inverse time).

Techniques used

use the assumption of constant decay probability per unit timeset up the proportionality between decay rate and number of undecayed nucleisolve the differential equation dN/dt = -lambda N to obtain an exponential

(c)

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

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

6M

(i)

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

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

For activity,

A = A_0 e^{-\lambda t}

So

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

\ln\left(\frac{A}{A_0}\right) = -\lambda t

\lambda = \frac{1}{t}\ln\left(\frac{A_0}{A}\right)

\lambda = \frac{1}{8.4}\ln\left(\frac{180}{120}\right)=0.0483\ \text{min}^{-1}

Answer

$\lambda = 0.0483\ \text{min}^{-1}$

Final answer

0.0483 min^-1

Detailed explanation

Background Concept

For a single isotope undergoing radioactive decay,

N = N_0 e^{-\lambda t}

where:

$N$ is the number of undecayed nuclei at time $t$ ,
$N_0$ is the initial number,
$\lambda$ is the decay constant (probability per unit time for one nucleus).

Activity is proportional to $N$ :

A = \lambda N

So activity also decays exponentially:

A = A_0 e^{-\lambda t}

Understanding the Question

You are told:

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

You must find $\lambda$ in units of $\text{min}^{-1}$ , so you should keep $t$ in minutes.

Approach

Use the exponential law for activity, rearrange to make $\lambda$ the subject using natural logs, then substitute the two activity values and the time.

Step-by-Step Reasoning

Start with

A = A_0 e^{-\lambda t}

Divide by $A_0$ :

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

Take natural logs of both sides (this is the standard way to bring down the power):

\ln\left(\frac{A}{A_0}\right) = -\lambda t

Rearrange:

\lambda = -\frac{1}{t}\ln\left(\frac{A}{A_0}\right)=\frac{1}{t}\ln\left(\frac{A_0}{A}\right)

Substitute $A_0=180$ , $A=120$ , $t=8.4\ \text{min}$ :

\lambda = \frac{1}{8.4}\ln\left(\frac{180}{120}\right)=\frac{1}{8.4}\ln(1.5)

Since $\ln(1.5) \approx 0.405$ ,

\lambda \approx \frac{0.405}{8.4} = 0.0483\ \text{min}^{-1}

The units are $\text{min}^{-1}$ because the logarithm is dimensionless and we divide by time in minutes.

Key Takeaways

Use $A = A_0 e^{-\lambda t}$ for activity decay.
Rearranging exponentials typically requires taking $\ln$ .
Keep time units consistent with the requested unit for $\lambda$ .

Common Mistakes

Using $\log_{10}$ instead of $\ln$ (unless handled consistently, it gives the wrong numerical value).
Forgetting the negative sign and getting a negative decay constant.
Converting $8.4\ \text{min}$ into seconds but still stating $\text{min}^{-1}$ .

Things to Be Careful About

The ratio must be $A_0/A$ (or handle the minus sign correctly).
$\ln(180/120)$ is $\ln(1.5)$ , not $\ln(0.666\ldots)$ unless you keep the minus sign.
Quote $\lambda$ to a sensible number of significant figures (typically 2 or 3).

Techniques used

use the exponential decay model for activityrearrange using natural logarithmssubstitute numerical values with consistent time units

(ii)

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

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

1M Similar

DifficultyEasy

Worked solution

Working

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

t_{1/2} = \frac{0.693}{0.0483} = 14.4\ \text{min}

Answer

$14.4\ \text{min}$

Final answer

14.4 min

Detailed explanation

Background Concept

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

Starting from

A = A_0 e^{-\lambda t}

At $t=t_{1/2}$ , $A = \frac{A_0}{2}$ :

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

Taking $\ln$ :

\ln\left(\frac{1}{2}\right) = -\lambda t_{1/2}

Since $\ln(1/2) = -\ln 2$ ,

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

Understanding the Question

You have already found the decay constant in (c)(i): $\lambda \approx 0.0483\ \text{min}^{-1}$ . Now you must calculate the half-life in minutes.

Approach

Use the standard relationship

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

and substitute the value of $\lambda$ .

Step-by-Step Reasoning

Substitute:

t_{1/2} = \frac{0.693}{0.0483}

Calculate:

t_{1/2} \approx 14.4\ \text{min}

Units: $\lambda$ is $\text{min}^{-1}$ , so dividing by it gives minutes.

Key Takeaways

The half-life is related to $\lambda$ by $t_{1/2} = \ln 2 / \lambda$ .
You only need one substitution once $\lambda$ is known.

Common Mistakes

Using $t_{1/2} = 2\lambda$ or other incorrect rearrangements.
Using $\ln(1/2)$ but forgetting it is negative.
Rounding too aggressively (e.g. giving $14\ \text{min}$ without justification).

Things to Be Careful About

Ensure you use the same time unit as in $\lambda$ .
Keep enough significant figures from $\lambda$ so the final half-life is accurate (typically 3 s.f.).

Techniques used

use the half-life relation t_1/2 = ln 2 / lambdasubstitute a calculated decay constantquote the result with appropriate significant figures

(iii)

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

Fig. 9.1

3M Similar

DifficultyMedium

Worked solution

Answer

Sketch an exponential decay curve from $A=180\ \text{Bq}$ at $t=0$ , passing through $(8.4\ \text{min},\ 120\ \text{Bq})$ , decreasing with decreasing gradient and approaching $A=0$ .

(Using $\lambda=0.0483\ \text{min}^{-1}$ , $A(24\ \text{min}) \approx 57\ \text{Bq}$ .)

Final answer

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

Detailed explanation

Background Concept

An exponentially decaying activity follows

A = A_0 e^{-\lambda t}

The key features of an exponential decay graph are:

it starts at $A_0$ when $t=0$ ,
it decreases rapidly at first,
the slope (rate of decrease) becomes less steep with time,
it approaches zero but never becomes negative.

A useful checkpoint is the half-life: every $t_{1/2}$ , the activity halves.

Understanding the Question

You must sketch $A$ against $t$ from $0$ to $24\ \text{min}$ on the provided axes.

Given points from the stem:

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

You can use your calculated $\lambda$ (or half-life) to place additional points to guide the sketch up to $24\ \text{min}$ .

Approach

Mark the two given data points on the axes.
(Optionally) calculate a value near $t=24\ \text{min}$ to anchor the tail of the curve.
Draw a smooth exponential curve: steep at the start, flattening out, never crossing below zero.

Step-by-Step Reasoning

Plot $(0, 180)$ .
Plot $(8.4, 120)$ .
Use $\lambda \approx 0.0483\ \text{min}^{-1}$ to estimate the activity at $t=24\ \text{min}$ :

A(24) = 180 e^{-0.0483\times 24}

Compute the exponent:

0.0483\times 24 \approx 1.16

So

A(24) \approx 180 e^{-1.16} \approx 180\times 0.314 \approx 56.6\ \text{Bq}

This tells you the curve should be around $\approx 57\ \text{Bq}$ at $24\ \text{min}$ .

Draw a smooth curve through the plotted points that flattens as $t$ increases and tends towards $A=0$ .

(Alternative check using half-life: from (c)(ii), $t_{1/2} \approx 14.4\ \text{min}$ , so at $t\approx 14.4\ \text{min}$ the activity should be about $90\ \text{Bq}$ ; at $t\approx 28.8\ \text{min}$ it would be about $45\ \text{Bq}$ , consistent with $A(24)\approx 57\ \text{Bq}$ .)

Key Takeaways

An exponential decay sketch must show a curve that decreases and then levels off.
Use known points (including ones you calculate) to keep the sketch quantitatively sensible.
The curve approaches zero asymptotically.

Common Mistakes

Drawing a straight line decrease (linear decay).
Drawing a curve that reaches zero at a finite time and then stays at zero.
Not passing through the given point $(8.4\ \text{min}, 120\ \text{Bq})$ .
Making the curve concave up (it should be concave up in the sense that the magnitude of the negative gradient decreases; on standard axes it appears to flatten).

Things to Be Careful About

Ensure the curve starts exactly at $180\ \text{Bq}$ when $t=0$ .
The activity must stay positive; do not let the sketch cross below the time axis.
The tail value near $t=24\ \text{min}$ should be in the right region (around $\sim 50$ to $60\ \text{Bq}$ using your $\lambda$ ), not still near $120\ \text{Bq}$ or near zero.

Techniques used

use exponential decay shape to sketch a curveplot and pass through given data pointsextrapolate using the decay constant or half-life

Q10Medium-EasyAstronomy and Cosmology

(a)

State Hubble’s law.

2M Similar

DifficultyEasy

Worked solution

Answer

Hubble’s law: the recessional speed $v$ of a galaxy is proportional to its distance $d$ from the observer,

v = H_0 d

where $H_0$ is the Hubble constant.

Final answer

Recessional speed is proportional to distance: v = H0 d.

Detailed explanation

Background Concept

Hubble’s law is an observational relationship in cosmology. For distant galaxies, the spectral lines are shifted to longer wavelengths (redshift), and the inferred recessional speed $v$ is found to increase with distance $d$ .

The law is written as

v = H_0 d

where:

$v$ is the recessional speed (speed the galaxy is moving away along the line of sight),
$d$ is the distance to the galaxy,
$H_0$ is the Hubble constant (units of $\text{s}^{-1}$ ).

Understanding the Question

The question simply asks for the statement of Hubble’s law (2 marks). That means giving the proportionality and (usually) the equation.

Approach

State: “speed is proportional to distance” and write $v = H_0 d$ . Define symbols briefly.

Step-by-Step Reasoning

Identify what Hubble’s law connects: recessional speed and distance.
State the proportionality: $v \propto d$ .
Convert proportionality into an equation with constant of proportionality $H_0$ :

v = H_0 d

Key Takeaways

Hubble’s law links cosmic recession speed to distance.
The constant of proportionality is the Hubble constant $H_0$ .

Common Mistakes

Writing an inverse relationship (e.g. $v \propto 1/d$ ).
Forgetting to mention that the galaxy is receding (moving away).
Giving only words with no equation when the mark scheme expects the mathematical form too.

Things to Be Careful About

Use the correct symbols: $v$ for speed, $d$ for distance, $H_0$ for the constant.
$H_0$ has units $\text{s}^{-1}$ (equivalently $\text{m s}^{-1}\ \text{m}^{-1}$ ).

Techniques used

state the proportionality between recessional speed and distanceexpress a physical law as an equation with defined quantities

(b)

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

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

6M

(i)

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

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

The detected wavelength is larger than the emitted wavelength, so the light has been redshifted.

This is due to the star/galaxy moving away from the Earth (expansion of the Universe), giving a Doppler shift to longer wavelengths.

Final answer

Because the galaxy is receding, the light is redshifted (Doppler/expansion), so the observed wavelength is longer.

Detailed explanation

Background Concept

When a source of waves moves relative to an observer, the observed wavelength and frequency change (Doppler effect). For light from distant galaxies, the dominant reason is the expansion of the Universe: space stretches while the light travels, increasing its wavelength. This is observed as a redshift (shift to longer wavelength).

Qualitatively:

source receding (\rightarrow) observed wavelength increases (redshift), observed frequency decreases.

Understanding the Question

You are told the wavelength at maximum intensity is emitted at $4.62 \times 10^{-7}\ \text{m}$ , but is observed at $4.91 \times 10^{-7}\ \text{m}$ . The question asks why these are different.

The key clue is that the observed wavelength is larger than the emitted one.

Approach

Compare observed and emitted wavelengths:

if $\lambda_{\text{obs}} > \lambda_{\text{emit}}$ , this indicates redshift.
Then state the physical cause: recession of the galaxy/star due to cosmic expansion (equivalently Doppler effect for light).

Step-by-Step Reasoning

Note that

\lambda_{\text{obs}} = 4.91\times 10^{-7}\ \text{m} > \lambda_{\text{emit}} = 4.62\times 10^{-7}\ \text{m}.

Larger observed wavelength means the light has been shifted towards the red end of the spectrum (redshift).
Redshift occurs when the source is moving away from the observer; for distant galaxies this is because the Universe is expanding, so the galaxy is receding from Earth.

Key Takeaways

$\lambda$ increases (\rightarrow) redshift.
Redshift implies recession (moving away) and is evidence for universal expansion.

Common Mistakes

Saying it is blueshift when $\lambda_{\text{obs}} > \lambda_{\text{emit}}$ .
Claiming the wavelength changes because of absorption in space (not the main effect here).
Mixing up cause and effect: the motion causes the shift, not the other way round.

Things to Be Careful About

Be explicit about direction: “moving away” is essential.
You do not need to mention “maximum intensity” here; the same redshift idea applies to any spectral feature.

Techniques used

identify Doppler/expansion redshift from a wavelength increaselink a redshift to recession of the source relative to the observer

(ii)

Calculate the speed of the star relative to the Earth.

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

2M Similar

DifficultyMedium-Easy

Worked solution

Working

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

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

v = c\frac{\Delta \lambda}{\lambda} = 3.00\times 10^{8}\times \frac{2.9\times 10^{-8}}{4.62\times 10^{-7}}

v = 1.88\times 10^{7}\ \text{m s}^{-1}

Answer

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

Final answer

1.9 × 10^7 m s^-1

Detailed explanation

Background Concept

For electromagnetic waves, a shift in wavelength can be related to recessional speed.

Define the redshift $z$ :

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

For relatively small redshifts (non-relativistic approximation),

z \approx \frac{v}{c}

where $v$ is the line-of-sight recessional speed and $c$ is the speed of light.

Understanding the Question

You are given:

emitted wavelength: $\lambda_{\text{emit}} = 4.62 \times 10^{-7}\ \text{m}$
observed wavelength: $\lambda_{\text{obs}} = 4.91 \times 10^{-7}\ \text{m}$

You must find the speed of the star relative to Earth. Since the observed wavelength is longer, the star is receding.

Approach

Find the wavelength change $\Delta \lambda = \lambda_{\text{obs}} - \lambda_{\text{emit}}$ .
Compute the fractional change $\Delta \lambda / \lambda_{\text{emit}}$ .
Use $v = c(\Delta \lambda/\lambda_{\text{emit}})$ .

Step-by-Step Reasoning

Compute the change in wavelength:

\Delta \lambda = (4.91-4.62)\times 10^{-7}\ \text{m} = 0.29\times 10^{-7}\ \text{m} = 2.9\times 10^{-8}\ \text{m}.

Find the redshift:

z = \frac{\Delta \lambda}{\lambda_{\text{emit}}} = \frac{2.9\times 10^{-8}}{4.62\times 10^{-7}} \approx 0.0628.

Convert to speed using $v \approx zc$ :

v = (0.0628)(3.00\times 10^8) = 1.88\times 10^7\ \text{m s}^{-1}.

To 2 significant figures (matching the data),

v \approx 1.9\times 10^7\ \text{m s}^{-1}.

Key Takeaways

Redshift is a fractional wavelength increase: $z = \Delta\lambda/\lambda_{\text{emit}}$ .
For small $z$ , use $v/c \approx z$ .

Common Mistakes

Using $\lambda_{\text{obs}}$ in the denominator instead of $\lambda_{\text{emit}}$ (usually the definition uses emitted/rest wavelength).
Forgetting to subtract wavelengths (using $4.91/4.62$ directly without minus 1).
Dropping powers of ten when subtracting numbers in standard form.
Giving the answer without unit $\text{m s}^{-1}$ .

Things to Be Careful About

This is a non-relativistic approximation; it is acceptable here because the question structure suggests using $\Delta\lambda/\lambda = v/c$ .
Quote an appropriate number of significant figures consistent with the input data.

Techniques used

calculate redshift using fractional wavelength changeuse the low-redshift approximation v/c = Δλ/λsubstitute values and express the result with correct units

(iii)

The wavelength of maximum intensity of emission is used to determine a value for the surface temperature of the star.

Explain how the temperature determined using the observed wavelength compares with the true value of temperature determined using the emitted wavelength.

2M Similar

DifficultyMedium-Easy

Worked solution

Answer

Using Wien’s law $\lambda_{\max}T = b$ , a larger (observed) $\lambda_{\max}$ gives a smaller calculated $T$ .

So the temperature found using the observed wavelength is less than the true surface temperature (it is underestimated).

Final answer

It would be lower than the true temperature (underestimated).

Detailed explanation

Background Concept

Wien’s displacement law relates the wavelength at which a black-body spectrum peaks to the absolute temperature:

\lambda_{\max}T = b

where $b \approx 2.90\times 10^{-3}\ \text{m K}$ . The key point is the inverse relationship:

T = \frac{b}{\lambda_{\max}}.

So if the peak wavelength appears larger, the inferred temperature becomes smaller.

Understanding the Question

The star’s radiation has a peak wavelength that is emitted at $4.62\times 10^{-7}\ \text{m}$ but observed at $4.91\times 10^{-7}\ \text{m}$ because of redshift.

The question asks: if we use the observed (redshifted) peak wavelength to calculate temperature, how does that temperature compare with the true temperature based on the emitted wavelength?

Approach

Use the proportionality from Wien’s law:

T \propto \frac{1}{\lambda_{\max}}.

Then compare the two wavelengths: observed is larger, so inferred temperature must be smaller.

Step-by-Step Reasoning

Start with Wien’s law:

\lambda_{\max}T = b.

Rearrange:

T = \frac{b}{\lambda_{\max}}.

Since $\lambda_{\text{obs}} > \lambda_{\text{emit}}$ , it follows that

\frac{1}{\lambda_{\text{obs}}} < \frac{1}{\lambda_{\text{emit}}}

so

T_{\text{from obs}} < T_{\text{true}}.

Therefore the temperature calculated using the observed wavelength is an underestimate.

Key Takeaways

Wien’s law gives an inverse relationship between peak wavelength and temperature.
Redshift increases observed wavelength, so it makes the star appear cooler if uncorrected.

Common Mistakes

Saying the temperature would be higher because “the wavelength increased”. (It’s the opposite due to the inverse law.)
Confusing intensity with temperature here; the question is about the wavelength at maximum intensity, not the maximum intensity value itself.

Things to Be Careful About

Wien’s law uses absolute temperature in kelvin.
The comparison is qualitative; no calculation is necessary unless asked.

Techniques used

apply Wien’s displacement law qualitativelycompare proportional changes using an inverse relationship

(c)

A value for the Hubble constant is $2.3 \times 10^{-18}\ \text{s}^{-1}$ .

Use your answer in (b)(ii) to determine the distance of the star in (b) from the Earth.

distance = ______ $\text{m}$

2M Similar

DifficultyMedium-Easy

Worked solution

Working

Hubble’s law:

v = H_0 d \Rightarrow d = \frac{v}{H_0}

d = \frac{1.88\times 10^{7}}{2.3\times 10^{-18}} = 8.17\times 10^{24}\ \text{m}

Answer

$8.2\times 10^{24}\ \text{m}$

Final answer

8.2 × 10^24 m

Detailed explanation

Background Concept

Hubble’s law connects recessional speed and distance:

v = H_0 d.

If you know $v$ and $H_0$ , you can find distance by rearranging:

d = \frac{v}{H_0}.

The units check is helpful:

$v$ is in $\text{m s}^{-1}$ ,
$H_0$ is in $\text{s}^{-1}$ ,
so $d$ comes out in $\text{m}$ .

Understanding the Question

You are given $H_0 = 2.3\times 10^{-18}\ \text{s}^{-1}$ and instructed to use your result from (b)(ii) (the star’s recessional speed relative to Earth). You must calculate the distance from Earth.

Approach

Start with $v = H_0 d$ .
Rearrange to $d = v/H_0$ .
Substitute $v$ from part (b)(ii) and the given $H_0$ .

Step-by-Step Reasoning

Rearrange Hubble’s law:

d = \frac{v}{H_0}.

Substitute $v = 1.88\times 10^{7}\ \text{m s}^{-1}$ and $H_0 = 2.3\times 10^{-18}\ \text{s}^{-1}$ :

d = \frac{1.88\times 10^{7}}{2.3\times 10^{-18}}.

Handle powers of ten:

\frac{10^{7}}{10^{-18}} = 10^{25}.

Divide the leading numbers:

\frac{1.88}{2.3} \approx 0.817.

So

d \approx 0.817\times 10^{25} = 8.17\times 10^{24}\ \text{m} \approx 8.2\times 10^{24}\ \text{m}.

Key Takeaways

Use $d = v/H_0$ to convert recessional speed into distance.
Check units: dividing by $\text{s}^{-1}$ is equivalent to multiplying by seconds, giving metres.

Common Mistakes

Multiplying instead of dividing ( $d = vH_0$ ).
Using the wrong $v$ (e.g. using $\Delta\lambda$ rather than the calculated speed).
Incorrect power-of-ten handling when dividing by $10^{-18}$ .

Things to Be Careful About

Keep your $v$ value consistent with (b)(ii); if your (b)(ii) differs slightly, follow through (error carried forward is usually allowed).
Give the distance in standard form and include $\text{m}$ as the unit.

Techniques used

apply Hubble’s law to relate speed and distancerearrange an equation to make distance the subjectsubstitute numerical values and use standard form

Physics 9702/41 — May/June 2025

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