The capacitance of the capacitor is 22 mF. Calculate the energy stored in the capacitor when it is fully charged.

The resistance of the wire is 8.0 Ω. Determine the time taken for the capacitor to discharge through the resistance wire. Assume that the capacitor is completely discharged when the potential difference across it has fallen to 0.24 V.

The mass of the resistance wire is 0.61 g and its observed temperature rise is 28 K. Estimate the specific heat capacity of the wire. Include an appropriate unit for your answer.

Suggest one other energy loss in the experiment and the effect it will have on the value for the specific heat capacity of the wire.

«$\frac{1}{2}C{V}^2=\frac{1}{2}\times 0.22\times {24}^2$» = «J»

$\frac{1}{100}=e^{-\frac{t}{8.0\times 0.022}}$

$\ln 0.01=-\frac{t}{8.0\times 0.022}$

0.81 «s»

c = $\frac{Q}{m\times \mathrm{\Delta }T}$

OR

$\frac{6.3}{0.00061\times 28}$

370 J kg^–1 K^–1

Allow ECF from 3(a) for energy transferred.

Correct answer only to include correct unit that matches answer power of ten.

Allow use of g and kJ in unit but must match numerical answer, eg: 0.37 J kg^–1 K^–1 receives [1]

ALTERNATIVE 1

some thermal energy will be transferred to surroundings/along connecting wires/to
thermometer

estimate «of specific heat capacity by student» will be larger «than accepted value»

ALTERNATIVE 2

not all energy transferred as capacitor did not fully discharge

so estimate «of specific heat capacity by student» will be larger «than accepted value»

0.46 mole of an ideal monatomic gas is trapped in a cylinder. The gas has a volume of 21 m³ and a pressure of 1.4 Pa.

(i) State how the internal energy of an ideal gas differs from that of a real gas.

(ii) Determine, in kelvin, the temperature of the gas in the cylinder.

(iii) The kinetic theory of ideal gases is one example of a scientific model. Identify two reasons why scientists find such models useful.

i
«intermolecular» potential energy/PE of an ideal gas is zero/negligible

ii
THIS IS FOR USE WITH AN ENGLISH SCRIPT ONLY
use of $T=\frac{PV}{nR}$ or $T=\frac{1.4\times 21}{0.46\times 8.31}$
Award mark for correct re-arrangement as shown here not for quotation of Data Booklet version.
Award [2] for a bald correct answer in K.
Award [2 max] if correct 7.7 K seen followed by –265°C and mark BOD. However, if only –265°C seen, award [1 max].

7.7K
Do not penalise use of “°K”

ii
THIS IS FOR USE WITH A SPANISH SCRIPT ONLY
$T=\frac{PV}{nR}$
Award mark for correct re-arrangement as shown here not for quotation of Data Booklet version.

$T=\frac{1.4\times 2.1\times {10}^{-6}}{0.46\times 8.31}$
Uses correct unit conversion for volume

T = 7.7×10^-6K
Award [2] for a bald correct answer in K. Finds solution. Allow an ECF from MP2 if unit not converted, ie candidate uses 21m3 and obtains 7.7 K
Do not penalise use of “°K”

iii
«models used to»
predict/hypothesize / lead to further theories
Response needs to identify two different reasons. (N.B. only one in SL).

explain / help with understanding / help to visualize
Do not allow any response that is gas specific. The question is couched in general, nature of science terms and must be answered as such.

simulate
simplify/approximate

Write down the nuclear equation for this decay.

Deduce that the activity of the radium-226 is almost constant during the experiment.

Show that about 3 x 10¹⁵ alpha particles are emitted by the radium-226 in 6 days.

The wall of cylinder A is made from glass. Outline why this glass wall had to be very thin.

The experiment was carried out at a temperature of 18 °C. The volume of cylinder B was 1.3 x 10^–5 m³ and the volume of cylinder A was negligible. Calculate the pressure of the helium gas that was collected in cylinder B over the 6 day period. Helium is a monatomic gas.

${}_2^4\alpha$

OR

${}_2^4\text{He}$

${}_{86}^{222}\text{Rn}$

These must be seen on the right-hand side of the equation.

ALTERNATIVE 1

6 days is 5.18 x 10⁵ s

activity after 6 days is $A_0{e}^{-1.4\times {10}^{-11}\times 5.8\times {10}^5}\approx A_0$

OR

A = 0.9999927 A₀ or 0.9999927 $\lambda$N₀

OR

states that index of e is so small that $\frac{A}{A_0}$ is ≈ 1

OR

A – A₀ ≈ 10^–15 «s^–1»

ALTERNATIVE 2
shows half-life of the order of 10¹¹ s or 5.0 x 10¹⁰ s

converts this to year «1600 y» or days and states half-life much longer than experiment compared to experiment

Award [1 max] if calculations/substitutions have numerical slips but would lead to correct deduction.

eg: failure to convert 6 days to seconds but correct substitution into equation will give MP2.

Allow working in days, but for MP1 must see conversion of $\lambda$ or half-life to day^–1.

ALTERNATIVE 1 

use of A = $\lambda$N₀

conversion to number of molecules = nN_A = 3.7 x 10²⁰

OR

initial activity = 5.2 x 10⁹ «s^–1»

number emitted = (6 x 24 x 3600) x 1.4 x 10^–11 x 3.7 x 10²⁰ or 2.7 x 10¹⁵ alpha particles

ALTERNATIVE 2
use of N = N₀$e^{-\lambda t}$

N₀ = n x N_A = 3.7 x 10²⁰

alpha particles emitted «= number of atoms disintegrated = N – N₀ =» N₀$\left(1-e^{-\lambda \times 6\times 24\times 3600}\right)$ or 2.7 x 10¹⁵ alpha particles 

Must see correct substitution or answer to 2+ sf for MP3

alpha particles highly ionizing
OR
alpha particles have a low penetration power
OR
thin glass increases probability of alpha crossing glass
OR
decreases probability of alpha striking atom/nucleus/molecule

Do not allow reference to tunnelling.

conversion of temperature to 291 K

p = 4.5 x 10^–9 x 8.31 x «$\frac{291}{1.3\times {10}^{-5}}$»

OR

p = 2.7 x 10¹⁵ x 1.3 x 10^–23 x «$\frac{291}{1.3\times {10}^{-5}}$»

0.83 or 0.84 «Pa»

Allow ECF for 2.7 x 10¹⁵ from (b)(ii).

Determine, in kJ, the total kinetic energy of the particles of the gas.

Explain, with reference to the kinetic model of an ideal gas, how an increase in temperature of the gas leads to an increase in pressure.

ALTERNATIVE 1

average kinetic energy = $\frac{3}{2}$1.38 × 10^–23 × 313 = 6.5 × 10^–21 «J»

number of particles = 3.0 × 6.02 × 10²³ = 1.8 × 10²⁴

total kinetic energy = 1.8 × 10²⁴ × 6.5 × 10^–21 = 12 «kJ»

ALTERNATIVE 2

ideal gas so U = KE

KE = $\frac{3}{2}$8.31 × 131 × 3

total kinetic energy = 12 «kJ»

[3 marks]

larger temperature implies larger (average) speed/larger (average) KE of molecules/particles/atoms

increased force/momentum transferred to walls (per collision) / more frequent collisions with walls

increased force leads to increased pressure because P = F/A (as area remains constant)

Ignore any mention of PV = nRT.

[3 marks]

Distinguish between the internal energy of the oxygen at the boiling point when it is in its liquid phase and when it is in its gas phase.

Calculate, in kW, the heater power required.

Calculate the volume of the oxygen produced in one second when it is allowed to expand to a pressure of 0.11 MPa and to reach a temperature of –13 °C.

State one assumption of the kinetic model of an ideal gas that does not apply to oxygen.

Internal energy is the sum of all the PEs and KEs of the molecules (of the oxygen) ✔

PE of molecules in gaseous state is zero ✔

(At boiling point) average KE of molecules in gas and liquid is the same ✔

gases have a higher internal energy ✔

Molecules/particles/atoms must be included once, if not, award [1 max]

ALTERNATIVE 1:

flow rate of oxygen = 8 «g s⁻¹» ✔

«2.1 × 10⁵ × 8 × 10⁻³» = 1.7 «kW» ✔

ALTERNATIVE 2:

Q = «0.25 × 32 × 10⁻³ × 2.1 × 10⁵ =» 1680 «J» ✔

power = «1680 W =» 1.7 «kW» ✔

T = 260 «K» ✔

V = «$\frac{nRT}{p}=$» 4.9 × 10⁻³ «m³» ✔

ideal gas has point objects ✔

no intermolecular forces ✔

non liquefaction ✔

ideal gas assumes monatomic particles ✔

the collisions between particles are elastic ✔

Allow the opposite statements if they are clearly made about oxygen eg oxygen/this can be liquified

Show that the speed of the loop is 20 cm s⁻¹.

Sketch, on the axes, a graph to show the variation with time of the magnetic flux linkage $\Phi$ in the loop.

Sketch, on the axes, a graph to show the variation with time of the magnitude of the emf induced in the loop.

There are 85 turns of wire in the loop. Calculate the maximum induced emf in the loop.

The resistance of the loop is 2.4 Ω. Calculate the magnitude of the magnetic force on the loop as it enters the region of magnetic field.

Show that the energy dissipated in the loop from t = 0 to t = 3.5 s is 0.13 J.

The mass of the wire is 18 g. The specific heat capacity of copper is 385 J kg⁻¹ K⁻¹. Estimate the increase in temperature of the wire.

$\frac{70}{3.5}$ ✓

shape as above ✓

shape as above ✓

Vertical lines not necessary to score.

Allow ECF from (b)(i).

ALTERNATIVE 1

maximum flux at «$5.0\times 5.0\times {10}^{-4}\times 85\times 0.94$» $=0.19975\approx 0.20$«Wb» ✓

emf = «$\frac{0.20}{0.25}=$» $0.80$«V» ✓

ALTERNATIVE 2

emf induced in one turn = BvL = $0.94\times 0.20\times 0.05=0.0094$«V» ✓

emf $=85\times 0.0094=0.80$«V» ✓

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

$I=«\frac{V}{R}=»\frac{0.8}{2.4}$  OR  $0.33$ «A» ✓

$F=«NBIL=85\times 0.94\times 0.33\times 0.05=»=1.3$ «N» ✓

Allow ECF from (c)(i).

Award [2] marks for a bald correct answer.

Energy is being dissipated for 0.50 s ✓

$E=Fvt=1.3\times 0.20\times 0.50=$«$0.13$ J»

OR

$E=Vlt=0.80\times 0.33\times 0.50=$«$0.13$ J» ✓

Allow ECF from (b) and (c).

Watch for candidates who do not justify somehow the use of 0.5 s and just divide by 2 their answer.

$∆T=\frac{0.13}{0.018\times 385}$ ✓

$∆T=1.9\times {10}^{-2}$ «K» ✓

Allow [2] marks for a bald correct answer.

Award [1] for a POT error in MP1.

The mass of a helium atom is 6.6 × 10^-27 kg. Estimate the average speed of the helium atoms in the container.

Show that the number of helium atoms in the container is 4 × 10²⁰.

Calculate the ratio $\frac{\text{volume of helium atoms}}{\text{volume of helium gas}}$.

Discuss, by reference to the kinetic model of an ideal gas and the answer to (c)(i), whether the assumption that helium behaves as an ideal gas is justified.

$\frac{\text{1}}{2}m{v}^2=\frac{3}{2}kT/v=\sqrt{\frac{3kT}{m}}/\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 320}{6.6\times {10}^{-27}}}$   ✔

v = 1.4 × 10³«ms^–1»  ^✔

$N=\frac{pV}{kT}/\frac{5.1\times {10}^5\times 3.2\times {10}^{-6}}{1.38\times {10}^{-23}\times 320}$

OR

$N=\frac{pV{N}_A}{RT}/\frac{5.1\times {10}^5\times 3.2\times {10}^{-6}\times 6.02\times {10}^{23}}{8.31\times 320}$  ✔

$N=3.7\times {10}^{20}$   ✔

«$\frac{4\times {10}^{20}\times 4.9\times {10}^{-31}}{3.2\times {10}^{-6}}=$»$6\times {10}^{-5}$   ✔

«For an ideal gas» the size of the particles is small compared to the distance between them/size of the container/gas

OR

«For an ideal gas» the volume of the particles is negligible/the volume of the particles is small compared to the volume of the container/gas

OR

«For an ideal gas» particles are assumed to be point objects ✔

calculation/ratio/result in (c)(i) shows that volume of helium atoms is negligible compared to/much smaller than volume of helium gas/container «hence assumption is justified» ✔

At HL this was very well answered but at SL many just worked out E=3/2kT and left it as a value for KE.

Again at HL this was very well answered with the most common approach being to calculate the number of moles and then multiply by N_A to calculate the number of atoms. At SL many candidates calculated n but stopped there. Also at SL there was some evidence of candidates working backwards and magically producing a value for ‘n’ that gave a result very close to that required after multiplying by N_A.

This was well answered with the most common mistake being to use the volume of a single atom rather than the total volume of the atoms.

At HL candidates seemed more able to focus on the key part feature of the question, which was the nature of the volumes involved. Examiners were looking for an assumption of the kinetic theory related to the volume of the atoms/gas and then a link to the ratio calculated in ci). The command terms were slightly different at SL and HL, giving slightly more guidance at SL.

State what is meant by an ideal gas.

Calculate the number of atoms in the gas.

Calculate, in J, the internal energy of the gas.

Calculate, in Pa, the new pressure of the gas.

Explain, in terms of molecular motion, this change in pressure.

a gas in which there are no intermolecular forces

OR

a gas that obeys the ideal gas law/all gas laws at all pressures, volumes and temperatures

OR

molecules have zero PE/only KE

Accept atoms/particles.

[1 mark]

N = «$\frac{pV}{kT}=\frac{5.3\times {10}^5\times 2.1\times {10}^{-4}}{1.38\times {10}^{-23}\times 310}$» 2.6 × 10²²

[1 mark]

«For one atom U = $\frac{3}{2}$kT» $\frac{3}{2}$ × 1.38 × 10^–23 × 310 / 6.4 × 10^–21 «J»

U = «2.6 × 10²² × $\frac{3}{2}$ × 1.38 × 10^–23 × 310» 170 «J»

 Allow ECF from (a)(ii)

Award [2] for a bald correct answer

Allow use of U = $\frac{3}{2}$pV

[2 marks]

p₂ = «5.3 × 10⁵ × $\frac{2.1\times {10}^{-4}}{6.8\times {10}^{-4}}$» 1.6 × 10⁵ «Pa»

[1 mark]

«volume has increased and» average velocity/KE remains unchanged

«so» molecules collide with the walls less frequently/longer time between collisions with the walls

«hence» rate of change of momentum at wall has decreased

«and so pressure has decreased»

The idea of average must be included

Decrease in number of collisions is not sufficient for MP2. Time must be included.

Accept atoms/particles.

[2 marks]

Show that the intensity of the solar radiation at the location of Titan is 16 W m⁻².

Titan has an atmosphere of nitrogen. The albedo of the atmosphere is 0.22. The surface of Titan may be assumed to be a black body. Explain why the average intensity of solar radiation absorbed by the whole surface of Titan is 3.1 W m⁻².

Show that the equilibrium surface temperature of Titan is about 90 K.

The mass of Titan is 0.025 times the mass of the Earth and its radius is 0.404 times the radius of the Earth. The escape speed from Earth is 11.2 km s⁻¹. Show that the escape speed from Titan is 2.8 km s⁻¹.

The orbital radius of Titan around Saturn is $R$ and the period of revolution is $T$.

Show that $T^2=\frac{4{\mathrm{\pi }}^2{R}^{ 3}}{GM}$ where $M$ is the mass of Saturn.

The orbital radius of Titan around Saturn is 1.2 × 10⁹m and the orbital period is 15.9 days. Estimate the mass of Saturn.

Show that the mass of a nitrogen molecule is 4.7 × 10⁻²⁶ kg.

Estimate the root mean square speed of nitrogen molecules in the Titan atmosphere. Assume an atmosphere temperature of 90 K.

Discuss, by reference to the answer in (b), whether it is likely that Titan will lose its atmosphere of nitrogen.

incident intensity $\frac{1360}{9.3^2}$ OR $15.7\approx 16$ «W m⁻²» ✓

Allow the use of 1400 for the solar constant.

exposed surface is ¼ of the total surface ✓

absorbed intensity = (1−0.22) × incident intensity ✓

0.78 × 0.25 × 15.7  OR  3.07 «W m⁻²» ✓

Allow 3.06 from rounding and 3.12 if they use 16 W m⁻².

σT ⁴ = 3.07

OR

T = 86 «K» ✓

$v=«\sqrt{\frac{2GM}{R}}=»\sqrt{\frac{0.025}{0.404}}\times 11.2$

OR

$2.79$ «km s⁻¹» ✓

correct equating of gravitational force / acceleration to centripetal force / acceleration ✓

correct rearrangement to reach the expression given ✓

Allow use of $\sqrt{\frac{GM}{R}}=\frac{2\mathrm{\pi }R}{T}$ for MP1.

$T=15.9\times 24\times 3600$ «s» ✓

$M=\frac{4{\mathrm{\pi }}^2{\left(1.2\times {10}^9\right)}^3}{6.67\times {10}^{-11}\times {\left(15.9\times 24\times 3600\right)}^2}=5.4\times {10}^{26}$«kg» ✓

Award [2] marks for a bald correct answer.

Allow ECF from MP1.

$m=\frac{28\times {10}^{-3}}{6.02\times {10}^{23}}$

OR

$4.65\times {10}^{-26}$ «kg» ✓

$«\frac{1}{2}m{v}^2=\frac{3}{2}kT\Rightarrow »v=\sqrt{\frac{3kT}{m}}$ ✓

$v=«\sqrt{\frac{3\times 1.38\times {10}^{-23}\times 90}{4.651\times {10}^{-26}}}=»283\approx 300$ «ms⁻¹» ✓

Award [2] marks for a bald correct answer.

Allow 282 from a rounded mass.

no, molecular speeds much less than escape speed ✓

Allow ECF from incorrect (d)(ii).

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

Draw, on the axes, a graph to show the variation with nucleon number $A$ of the binding energy per nucleon, $\frac{\text{BE}}{A}$. Numbers are not required on the vertical axis.

Identify, with a cross, on the graph in (a)(ii), the region of greatest stability.

Some unstable nuclei have many more neutrons than protons. Suggest the likely decay for these nuclei.

Show that the energy released in this decay is about 6 MeV.

The plutonium nucleus is at rest when it decays.

Calculate the ratio $\frac{\text{kinetic energy of alpha particle}}{\text{kinetic energy of uranium}}$.

Estimate the power, in kW, that is available from the plutonium at launch.

The spacecraft will take 7.2 years (2.3 × 10⁸ s) to reach a planet in the solar system. Estimate the power available to the spacecraft when it gets to the planet.

 State and explain what happens to the kinetic energy of an emitted photoelectron.

 State and explain what happens to the rate at which charge leaves the metallic surface.

the energy needed to «completely» separate the nucleons of a nucleus

OR

the energy released when a nucleus is assembled from its constituent nucleons ✓

Accept reference to protons and neutrons.

curve rising to a maximum between 50 and 100 ✓

curve continued and decreasing ✓

Ignore starting point.

Ignore maximum at alpha particle.

At a point on the peak of their graph ✓

beta minus «decay» ✓

correct mass numbers for uranium (234) and alpha (4) ✓

$234\times 7.600+4\times 7.074-238\times 7.568$ «MeV» ✓

energy released 5.51 «MeV» ✓

Ignore any negative sign.

«$\frac{K{E}_{\alpha }}{K{E}_U}=$»$\frac{\frac{p^2}{2m_{\alpha }}}{\frac{p^2}{2m_U}}$  OR  $\frac{m_U}{m_{\alpha }}$ ✓

«$\frac{234}{4}=$» $58.5$ ✓

Award [2] marks for a bald correct answer.

Accept $\frac{117}{2}$ for MP2.

number of nuclei present $=\frac{33\times {10}^3}{238}\times 6.02\times {10}^{23}«=8.347\times {10}^{25}»$ ✓

initial activity is $\lambda N_0=2.5\times {10}^{-10}\times 8.347\times {10}^{25}«=2.08\times {10}^{16} \text{Bq}»$ ✓

power is $2.08\times {10}^{16}\times 5.51\times {10}^6\times 1.6\times {10}^{-19}\approx 18$ «kW» ✓

Allow a final answer of 20 kW if 6 MeV used.

Allow ECF from MP1 and MP2.

available power after time t is $P_0{e}^{-\lambda t}$ ✓

$18e^{-2.50\times {10}^{-10}\times 2.3\times {10}^8}=17.0$ «kW» ✓

MP1 may be implicit.

Allow ECF from (c)(i).

Allow 17.4 kW from unrounded power from (c)(i).

Allow 18.8 kW from 6 MeV.

stays the same ✓

as energy depends on the frequency of light ✓

Allow reference to wavelength for MP2.

Award MP2 only to answers stating that KE decreases due to Doppler effect.

decreases ✓

as number of photons incident decreases ✓

The switch S is initially open. Calculate the total power dissipated in the circuit.

The switch is now closed. State, without calculation, why the current in the cell will increase.

The switch is now closed. $\text{Deduce the ratio }\frac{\text{power dissipated in Y with S open}}{\text{power dissipated in Y with S closed}}$.

The cell is used to charge a parallel-plate capacitor in a vacuum. The fully charged capacitor is then connected to an ideal voltmeter.

The capacitance of the capacitor is 6.0 μF and the reading of the voltmeter is 12 V.

Calculate the energy stored in the capacitor.

Calculate the change in the energy stored in the capacitor.

Suggest, in terms of conservation of energy, the cause for the above change.

total resistance of circuit is 8.0 «Ω» ✔

$P=\frac{{12}^2}{8.0}=18$«W» ✔

«a resistor is now connected in parallel» reducing the total resistance

OR

current through YZ unchanged and additional current flows through X ✔

evidence in calculation or statement that pd across Y/current in Y is the same as before ✔

so ratio is 1 ✔

$E=«\frac{1}{2}C{V}^2=\frac{1}{2}\times 6\times {10}^{-6}\times {12}^2=»4.3\times {10}^{-4}$«$\text{J}$» ✔

ALTERNATIVE 1

capacitance doubles and voltage halves ✔

since $E=\frac{1}{2}C{V}^2$ energy halves   ✔

so change is «–»2.2×10^–4 «J»  ✔

ALTERNATIVE 2

$E=\frac{1}{2}C{V}^2\text{ and }Q=CV\text{ so }E=\frac{Q^2}{2C}$  ✔ 

capacitance doubles and charge unchanged so energy halves ✔

so change is «−»2.2 × 10⁻⁴ «J» ✔

it is the work done when inserting the dielectric into the capacitor ✔

Most candidates scored both marks. ECF was awarded for those who didn’t calculate the new resistance correctly. Candidates showing clearly that they were attempting to calculate the new total resistance helped examiners to award ECF marks.

Most recognised that this decreased the total resistance of the circuit. Answers scoring via the second alternative were rare as the statements were often far too vague.

Very few gained any credit for this at both levels. Most performed complicated calculations involving the total circuit and using 12V – they had not realised that the question refers to Y only.

Most answered this correctly.

By far the most common answer involved doubling the capacitance without considering the change in p.d. Almost all candidates who did this calculated a change in energy that scored 1 mark.

Very few scored on this question.

Write down the equation for this decay.

Show that the initial quantity of potassium-40 in the rock sample was about 450 µmol.

The half-life of potassium-40 is 1.3 × 10⁹ years. Estimate the age of the rock sample.

Outline how the decay constant of potassium-40 was determined in the laboratory for a pure sample of the nuclide.

${}_{20}{}^{40}\text{Ca}$ ✓

${}_{-1}{}^0\text{e}+{\overline{\nu }}_{\text{e}}$  OR  ${}_{-1}{}^0\beta +{\overline{\nu }}_{\text{e}}$  ✓

Full equation ${}_{19}{}^{40}\text{K}\to {}_{20}{}^{40}\text{Ca}+{}_{-1}{}^0\text{e}+{\overline{\nu }}_{\text{e}}$

total K-40 decayed = $\frac{\text{12\hspace{0.05em}μmol}}{0.11}=109$ «μmol» ✓

so total K-40 originally was 109 + 340 = 449 «μmol»✓ 

ALTERNATIVE 1

$\lambda =\frac{\text{ln}\left(2\right)}{t_{{\displaystyle \frac{1}{2}}}}$ used to give 𝜆 = 5.3 x 10^-10 per year ✓

$340=\left(449\right)\left(e^{-5.3\times {10}^{-10}\times t}\right)$

OR

$\ln \left(\frac{340}{449}\right)=-5.3\times {10}^{-10}\times t$  ✓

t = 5.2 x 10⁸ «years» ✓

ALTERNATIVE 2

$p=\frac{340}{449}=0.76$ «remaining» ✓

$n=\frac{\ln \left(p\right)}{0.693}=\frac{\ln \left(0.76\right)}{0.693}=0.40$ ✓

t = 0.40 x 1.3 x 10⁹ = 5.2 x 10⁸ «years» ✓

ALTERNATIVE 3

$p=\frac{340}{449}=0.76$ «remaining» ✓

$0.76={\left(\frac{1}{2}\right)}^{\frac{t}{1.3\times {10}^9}}$ ✓

t = 0.40 x 1.3 x 10⁹= 5.2 x 10⁸ «years» ✓

Allow 5.3 x 10⁸ years for final answer.

Allow ECF for MP3 for an incorrect number of half-lives.

«use the mass of the sample to» determine number of potassium-40 atoms / nuclei in sample ✓

«use a counter to» determine (radio)activity / A of sample ✓

use A = λN «to determine the decay constant / λ» ✓

This question was very well done by candidates. The majority were able to identify the correct nuclide of Calcium and many correctly included an electron/beta particle and a properly written antineutrino.

This was a "show that" question that was generally well done by candidates.

This was a more challenging question for candidates. Many were able to calculate the decay constant and recognized that the ratio of initial and final quantities of the potassium-40 was important. A very common error was mixing the two common half-life equations up and using the wrong values in the exponent (using half life instead of the decay constant, or using the decay constant instead of the half life). Examiners were generous with ECF for candidates who clearly showed an incorrect number of half-lives multiplied by the time for one half-life.

Describing methods of determining half-life continues to be a struggle for candidates with very few earning all three marks. Many candidates described a method more appropriate to measuring a short half- life, but even those descriptions fell far short of being acceptable.

Outline how this diagram shows that the gravitational field strength of planet X decreases with distance from the surface.

The diagram shows part of the surface of planet X. The gravitational potential at the surface of planet X is –3V and the gravitational potential at point Y is –V.

Sketch on the grid the equipotential surface corresponding to a gravitational potential of –2V.

A meteorite, very far from planet X begins to fall to the surface with a negligibly small initial speed. The mass of planet X is 3.1 × 10²¹ kg and its radius is 1.2 × 10⁶ m. The planet has no atmosphere. Calculate the speed at which the meteorite will hit the surface.

At the instant of impact the meteorite which is made of ice has a temperature of 0 °C. Assume that all the kinetic energy at impact gets transferred into internal energy in the meteorite. Calculate the percentage of the meteorite’s mass that melts. The specific latent heat of fusion of ice is 3.3 × 10⁵ J kg^–1.

the field lines/arrows are further apart at greater distances from the surface

circle centred on Planet X
three units from Planet X centre

loss in gravitational potential = $\frac{6.67\times {10}^{-11}\times 3.1\times {10}^{21}}{1.2\times {10}^6}$

«= 1.72 × 10⁵ JKg⁻¹»

equate to $\frac{1}{2}$v²

v = 590 «m s⁻¹»

Allow ECF from MP1.

available energy to melt one kg 1.72 × 10⁵ «J»

fraction that melts is $\frac{1.72\times {10}^5}{3.3\times {10}^5}$ = 0.52 OR 52%

Allow ECF from MP1.

Allow 53% from use of 590 ms^-1.