Explain how each observation provides support for the particle theory but not the wave theory of light.

Determine a value for Planck’s constant.

State what is meant by the work function of a metal.

Calculate the work function of barium in eV.

The experiment is repeated with a metal surface of cadmium, which has a greater work function. Draw a second line on the graph to represent the results of this experiment.

Observation 1:
particle – photon energy is below the work function
OR
E = hf and energy is too small «to emit electrons»
wave – the energy of an em wave is independent of frequency

Observation 2:
particle – a single electron absorbs the energy of a single photon «in an almost instantaneous interaction»
wave – it would take time for the energy to build up to eject the electron

attempt to calculate gradient of graph = «$\frac{4.2\times {10}^{-19}}{6.2\times {10}^{14}}$»

$=6.8-6.9\times {10}^{-34}$ «Js»

Do not allow a bald answer of 6.63 x 10^-34 Js or 6.6 x 10^-34 Js.

ALTERNATIVE 1
minimum energy required to remove an electron «from the metal surface»

ALTERNATIVE 2
energy required to remove the least tightly bound electron «from the metal surface»

ALTERNATIVE 1
reading of y intercept from graph in range 3.8 − 4.2 × 10^–19 «J»
conversion to eV = 2.4 – 2.6 «eV»

ALTERNATIVE 2
reading of x intercept from graph «5.8 − 6.0 × 10¹⁴ Hz» and using hf₀ to get 3.8 − 4.2 × 10^–19 «J»
conversion to eV = 2.4 – 2.6 «eV»

line parallel to existing line
to the right of the existing line

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).

Show that the wavelength of an electron in the beam is about $3\times {10}^{-15} \mathrm{m}$.

Discuss how the results of the experiment provide evidence for matter waves.

The accepted value of the diameter of the carbon-12 nucleus is $4.94\times {10}^{-15} \mathrm{m}$. Estimate the angle ${\theta }_0$ at which the minimum of the intensity is formed.

Outline why electrons with energy of approximately ${10}^7 \mathrm{eV}$ would be unsuitable for the investigation of nuclear radii.

Experiments with many nuclides suggest that the radius of a nucleus is proportional to $A^{\frac{1}{3}}$, where $A$ is the number of nucleons in the nucleus. Show that the density of a nucleus remains approximately the same for all nuclei.

$\lambda =\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{1.60\times {10}^{-19}\times 4.2\times {10}^8}$ OR $=2.96\times {10}^{-15}«\mathrm{m}»$ ✓

Answer to at least $\mathit{2}$ s.f. (i.e. 3.0)

«the shape of the graph suggests that» electrons undergo diffraction «with carbon nuclei» ✓

only waves diffract ✓

$\sin {\theta }_0=\frac{2.96\times {10}^{-15}}{4.94\times {10}^{-15}}«=0.599»$ ✓

$37 «\mathrm{degrees}»$ OR $0.64/0.65 «\mathrm{rad}»$ ✓

the de Broglie wavelength of electrons is «much» longer than the size of a nucleus ✓

hence electrons would not undergo diffraction
OR
no diffraction pattern would be observed ✓

volume of a nucleus proportional to ${\left(A^{\frac{1}{3}}\right)}^3=A$ AND mass proportional to $A$ ✓

the ratio $\frac{\mathrm{mass}}{\mathrm{volume}}$ independent of $A$ «hence density the same for all nuclei» ✓

Both needed for MP1

An easy calculation with only one energy conversion to consider and a 'show' answer to help. 

This question was challenging for candidates many of whom seemed to have little idea of the experiment. Many answers discussed deflection, with the idea that forces between the electron and the nucleus causing it to deflect at a particular angle. This was often combined with the word interference to suggest evidence of matter waves. A number of answers described a demonstration the candidates remembered seeing so answers talked about fuzzy green rings.

This was answered reasonably well with only the odd omission of the sine in the equation.

Candidates generally scored poorly on this question. There was confusion between this experiment and another diffraction one, so often the new wavelength was compared to the spacing between atoms. Also, in line with answers to b(i) there were suggestions that the electrons did not have sufficient energy to reach the nucleus or would be deflected by too great an angle to be seen.

This question proved challenging and it wasn't common to find answers that scored both marks. Of those that had the right approach some missed out on both marks by describing A as the mass of the nucleus rather than proportional to the mass of the nucleus.

Calculate the wavelength of the light.

Electrons emitted from the surface of the photocell have almost no kinetic energy. Explain why this does not contradict the law of conservation of energy.

Radiation of photon energy 5.2 x 10^–19 J is now incident on the photocell. Calculate the maximum velocity of the emitted electrons.

Describe the change in the number of photons per second incident on the surface of the photocell.

State and explain the effect on the maximum photoelectric current as a result of increasing the photon energy in this way.

wavelength = «$\frac{hc}{E}=\frac{1.99\times {10}^{-25}}{\text{3}\text{.5}\times \text{1}{\text{0}}^{-19}}=$» 5.7 x 10^–7 «m»

If no unit assume m.

«potential» energy is required to leave surface

Do not allow reference to “binding energy”.
Ignore statements of conservation of energy.

all/most energy given to potential «so none left for kinetic energy»

energy surplus = 1.7 x 10^–19 J

v_max = $\sqrt{\frac{2\times 1.7\times {10}^{-19}}{9.1\times {10}^{-31}}}=6.1\times {10}^5$ «m s^–1»

Award [1 max] if surplus of 5.2 x 10^–19J used (answer: 1.1 x 10⁶ m s^–1)

«same intensity of radiation so same total energy delivered per square metre per second» 

light has higher photon energy so fewer photons incident per second

Reason is required

1:1 correspondence between photon and electron

so fewer electrons per second

current smaller

Allow ECF from (c)(i)
Allow ECF from MP2 to MP3.

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 ✓

State how the density of a nucleus varies with the number of nucleons in the nucleus.

Show that the nuclear radius of phosphorus-31 (${}_{15}^{31}\text{P}$) is about 4 fm.

State the maximum distance between the centres of the nuclei for which the production of ${}_{15}^{32}\text{P}$ is likely to occur.

Determine, in J, the minimum initial kinetic energy that the deuterium nucleus must have in order to produce ${}_{15}^{32}\text{P}$. Assume that the phosphorus nucleus is stationary throughout the interaction and that only electrostatic forces act.

${}_{15}^{32}\text{P}$ undergoes beta-minus (β^–) decay. Explain why the energy gained by the emitted beta particles in this decay is not the same for every beta particle.

State what is meant by decay constant.

In a fresh pure sample of ${}_{15}^{32}\text{P}$ the activity of the sample is 24 Bq. After one week the activity has become 17 Bq. Calculate, in s^–1, the decay constant of ${}_{15}^{32}\text{P}$.

it is constant ✔

R = $\text{1}\text{.20}\times {10}^{-15}\times {31}^{\frac{1}{3}}=3.8\times {10}^{-15}$ «m» ✔

Must see working and answer to at least 2SF

separation for interaction = 5.3 or 5.5 «fm» ✔

energy required = $\frac{15e^2}{4\pi {\epsilon }_0\times 5.3\times {10}^{-15}}$ ✔

= 6.5 / 6.6 ×10⁻¹³ OR 6.3 ×10⁻¹³«J» ✔

Allow ecf from (b)(i)

«electron» antineutrino also emitted ✔

energy split between electron and «anti»neutrino ✔

probability of decay of a nucleus ✔

OR

the fraction of the number of nuclei that decay

in one/the next second

OR

per unit time ✔

1 week = 6.05 × 10⁵ «s»

17 = $24{\text{e}}^{-\lambda \times 6.1\times {10}^5}$ ✔

5.7 × 10⁻⁷ «s^–1» ✔

Award [2 max] if answer is not in seconds

If answer not in seconds and no unit quoted award [1 max] for correct substitution into equation (MP2)

State and explain the nature of the particle labelled X.

Outline how these experiments are carried out.

Outline why the particles must be accelerated to high energies in scattering experiments.

State and explain one example of a scientific analogy.

Plot the position of magnesium-24 on the graph.

Draw a line on the graph, to show the variation of nuclear radius with nucleon number.

«electron» neutrino

it has a lepton number of 1 «as lepton number is conserved»

it has a charge of zero/is neutral «as charge is conserved»
OR
it has a baryon number of 0 «as baryon number is conserved»

Do not allow antineutrino

Do not credit answers referring to energy

«high energy particles incident on» thin sample

detect angle/position of deflected particles

reference to interference/diffraction/minimum/maximum/numbers of particles

Allow “foil” instead of thin

λ $\propto \frac{1}{\sqrt{E}}$ OR λ $\propto \frac{1}{E}$

so high energy gives small λ

to match the small nuclear size

Alternative 2

E = hf/energy is proportional to frequency

frequency is inversely proportional to wavelength/c = fλ

to match the small nuclear size

Alternative 3

higher energy means closer approach to nucleus

to overcome the repulsive force from the nucleus

so greater precision in measurement of the size of the nucleus

Accept inversely proportional

Only allow marks awarded from one alternative

two analogous situations stated

one element of the analogy equated to an element of physics

eg: moving away from Earth is like climbing a hill where the contours correspond to the equipotentials

Atoms in an ideal gas behave like pool balls

The forces between them only act during collisions

correctly plotted

Allow ECF from (d)(i)

single smooth curve passing through both points with decreasing gradient

through origin

Describe the photoelectric effect.

Show that the maximum velocity of the photoelectrons is $700 {\text{km\hspace{0.05em}\hspace{0.05em}s}}^{-1}$.

The photoelectrons are emitted from a sodium surface. Sodium has a work function of 2.3 eV.

Calculate the wavelength of the radiation incident on the sodium. State an appropriate unit for your answer.

electrons are ejected from the surface of a metal ✓

after gaining energy from photons/electromagnetic radiation ✓

there is a minimum «threshold» energy/frequency
OR
maximum «threshold» wavelength ✓

«$eV = \frac{1}{2} m{v}^2$» and manipulation to get $v$ ✓

$v=\sqrt{\frac{2\times 1.6\times {10}^{-19}\times 1.4}{9.1\times {10}^{-31}}}$  OR  $702«{\text{km\hspace{0.05em}s}}^{-1}»$✓

Must see either complete substitution or calculation to at least 3 s.f. for MP2

$E=2.3+1.4$ ✓

$\lambda =\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{3.7\times 1.6\times {10}^{-19}}$ ✓

$=3.4\times {10}^{-7} \text{m }\mathit{\text{OR}}\text{ 340 nm}$ ✓

Must see an appropriate unit to award MP3.

Outline the cause of the electron emission for radiation A.

Outline why electrons are never emitted for radiation C.

Outline why radiation B gives different results.

Explain why there is no effect on the table of results when the intensity of source B is doubled.

Photons with energy 1.1 × 10⁻¹⁸J are incident on a third metal surface. The maximum energy of electrons emitted from the surface of the metal is 5.1 × 10⁻¹⁹J.

Calculate, in eV, the work function of the metal.

photon transfers «all» energy to electron ✓

photon energy is less than both work functions
OR
photon energy is insufficient «to remove an electron» ✓

Answer must be in terms of photon energy.

Identifies P work function lower than Q work function ✓

changing/doubling intensity «changes/doubles number of photons arriving but» does not change energy of photon ✓

$5.1\times {10}^{-19}=1.1\times {10}^{-18}-\mathrm{\varphi }$ ✓

work function $\text{= «}\frac{(11.0-5.1)\times {10}^{-19}}{1.6\times {10}^{-19}}\text{= » 3.7 «eV»}$  ✓

Award [2] marks for a bald correct answer.

A current is observed on the ammeter when violet light illuminates C. With V held constant the current becomes zero when the violet light is replaced by red light of the same intensity. Explain this observation.

The graph shows the variation of photoelectric current I with potential difference V between C and A when violet light of a particular intensity is used.

The intensity of the light source is increased without changing its wavelength.

(i) Draw, on the axes, a graph to show the variation of I with V for the increased intensity.

(ii) The wavelength of the violet light is 400 nm. Determine, in eV, the work function of caesium.

(iii) V is adjusted to +2.50V. Calculate the maximum kinetic energy of the photoelectrons just before they reach A.

reference to photon
OR
energy = hf or =$\frac{hc}{\lambda }$

violet photons have greater energy than red photons

when hf > Φ or photon energy> work function then electrons are ejected

frequency of red light < threshold frequency «so no emission»
OR
energy of red light/photon < work function «so no emission»

i
line with same negative intercept «–1.15V»

otherwise above existing line everywhere and of similar shape with clear plateau

Award this marking point even if intercept is wrong.

ii
$\frac{hc}{\lambda e}=$ «$\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{40\times {10}^{-9}\times 1.6\times {10}^{-19}}=$» 3.11 «eV»

Intermediate answer is 4.97×10⁻¹⁹ J.

Accept approach using f rather than c/λ

«3.10 − 1.15 =» 1.96 «eV»
Award [2] for a bald correct answer in eV.
Award [1 max] if correct answer is given in J (3.12×10⁻¹⁹ J).

iii

«KE = qVs =» 1.15 «eV»

OR

1.84 x 10⁻¹⁹ «J»

Allow ECF from MP1 to MP2.

adds 2.50 eV = 3.65 eV

OR

5.84 x 10⁻¹⁹ J

Must see units in this question to identify energy unit used.
Award [2] for a bald correct answer that includes units.
Award [1 max] for correct answer without units.

Write down the equation to represent this decay.

The unstable lead nuclide has a half-life of 15 × 10⁶ years. A sample initially contains 2.0 μmol of the lead nuclide. Calculate the number of thallium nuclei being formed each second 30 × 10⁶ years later.

The neutron number N and the proton number Z are not equal for the nuclide ${}_{81}{}^{205}\text{Tl}$. Explain, with reference to the forces acting within the nucleus, the reason for this.

Thallium-205 (${}_{81}{}^{205}\text{Tl}$) can also form from successive alpha (α) and beta-minus (β⁻) decays of an unstable nuclide. The decays follow the sequence α β⁻ β⁻ α. The diagram shows the position of ${}_{81}{}^{205}\text{Tl}$ on a chart of neutron number against proton number.

Draw four arrows to show the sequence of changes to N and Z that occur as the ${}_{81}{}^{205}\text{Tl}$ forms from the unstable nuclide.

${}_{82}{}^{205}\text{Pb}$ ✓

${}_{-1}{}^0\text{e }\mathit{\text{AND }} {}_0{}^0\nu _{\text{e}}$ ✓

calculates $\lambda =\frac{\ln 2}{15\times {10}^6} «= 4.62\times {10}^{-8}{\text{ year}}^{-1}»$ ✓

calculates nuclei remaining $\text{= }0.50\times {10}^{-6}\times 6.0\times {10}^{23} «=3.0\times {10}^{17}»$ ✓

activity $=«\lambda N=1.4\times {10}^{10} \text{nuclei per year}»=440 «\text{nuclei per second}»$ ✓

Accept conversion to seconds at any stage.

Award [3] marks for a bald correct answer.

Allow ECF from MP1 and MP2

Allow use of decay equation.

Reference to proton repulsion OR nucleon attraction ✓

strong force is short range OR electrostatic/electromagnetic force is long range ✓

more neutrons «than protons» needed «to hold nucleus together»  ✓

any α change correct ✓

any β change correct ✓

diagram fully correct ✓

Award [2] max for a correct diagram without arrows drawn.

For MP1 accept a (−2, −2 ) line with direction indicated, drawn at any position in the graph.

For MP2 accept a (1, −1) line with direction indicated, drawn at any position in the graph.

Award [1] max for a correct diagram with all arrows in the opposite direction.

Uranium-238 decays into a nuclide of thorium-234 (Th).

Write down the complete equation for this radioactive decay.

Thallium-206 $\left({}_{81}{}^{206}\text{Tl}\right)$ decays into lead-206 $\left({}_{82}{}^{206}\text{Pb}\right)$.

Identify the quark changes for this decay.

The half-life of uranium-238 is about 4.5 × 10⁹ years. The half-life of thallium-206 is about 4.2 minutes.

Compare and contrast the methods to measure these half-lives.

Outline why high temperatures are required for fusion to occur.

Outline, with reference to the graph, why energy is released both in fusion and in fission.

Uranium-235 $\left({}_{92}{}^{235}\text{U}\right)$ is used as a nuclear fuel. The fission of uranium-235 can produce krypton-89 and barium-144.

Determine, in MeV and using the graph, the energy released by this fission.

${}_{92}{}^{238}\text{U→«}{}_{90}{}^{234}»\text{Th+}«{}_2{}^4»\alpha$ ✓ 

Allow He for alpha.

udd→uud
OR
down quark changes to up quark ✓ 

measure «radio»activity/«radioactive» decay/A for either
OR
take measurements with a Geiger counter. ✓

for Uranium measure number/N of radioactive atoms/OWTTE ✓

for Thalium measure «rate of» change in activity over time. ✓

correct connection for either Uranium or Thalium to determine half life ✓ 

links temperature to kinetic energy/speed of particles ✓ 

energy required to overcome «Coulomb» electrostatic repulsion ✓ 

«energy is released when» binding energy per nucleon increases

any use of (value from graph) x (number of nucleons) ✓

«235 × 7.6 – (89 × 8.6 + 144 × 8.2) =» 160 «MeV» ✓ 

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.

Identify a property of electrons demonstrated by this experiment.

Show that the energy E of each electron in the beam is about 7 × 10⁻¹¹J.

The de Broglie wavelength for an electron is given by $\frac{hc}{E}$. Show that the diameter of an oxygen-16 nucleus is about 4 fm.

Estimate, using the result in (a)(iii), the volume of a tin-118 $\left({}_{50}{}^{118}\text{Sn}\right)$ nucleus. State your answer to an appropriate number of significant figures.

wave properties ✓

Accept reference to diffraction or interference.

440 x 10⁶ x 1.6 x 10^-19  OR  7.0 × 10^-11 «J» ✓

$\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{7\times {10}^{-11}}$  OR  $\frac{1.24\times {10}^{-6}}{440\times {10}^6}$  OR  2.8 × 10^-15«m» seen ✓

read off graph as 46° ✓

«Use of $D=\frac{\lambda }{\sin \theta }$=» 3.9 × 10^-15 m ✓

Accept an angle between 45 and 47 degrees.

Allow ECF from MP2

ALTERNATIVE 1

use of $R\propto A^{\frac{1}{3}}$   OR  $V\propto A$ ✓

volume of $\text{Sn}=\frac{4}{3}\pi \left(\frac{A_{Sn}}{A_O}\right)r_O^{ 3}$ or equivalent working ✓

2.3 to 2.5 × 10^-43«m³»✓

answer to 1 or 2sf ✓

ALTERNATIVE 2

use of $R=R_{\text{o}}\times A^{\frac{1}{3}}$ ✓

volume of $\text{Sn}=\frac{4}{3}\pi R^3$  OR  5.9 x 10^-15 seen ✓

8.5 × 10^-43 «m³»✓

answer to 1 or 2sf ✓

Although the question expects candidates to work from the oxygen radius found, allow ALT 2 working from the Fermi radius.

MP4 is for any answer stated to 1 or 2 significant figures.

ai) Well answered.

aii) Well answered.

aiii) This was generally well done but quite a few attempted the small angle approximation. Probably worth a mention in the report.

b) Most gained credit from the first alternative solution, trying to use the data as the question intended. There were the inevitable slips and calculator mistakes. Most got the fourth mark.

Bohr modified the Rutherford model by introducing the condition mvr = n$\frac{h}{2\pi }$. Outline the reason for this modification.

Show that the speed v of an electron in the hydrogen atom is related to the radius r of the orbit by the expression

$$v=\sqrt{\frac{k{e}^2}{m_{\text{e}}r}}$$

where k is the Coulomb constant.

Using the answer in (b) and (c)(i), deduce that the radius r of the electron’s orbit in the ground state of hydrogen is given by the following expression.

$$r=\frac{h^2}{4{\pi }^{2}k{m}_{\text{e}}{e}^2}$$

Calculate the electron’s orbital radius in (c)(ii).

Explain what may be deduced about the energy of the electron in the β^– decay.

Suggest why the β^– decay is followed by the emission of a gamma ray photon.

Calculate the wavelength of the gamma ray photon in (d)(ii).

the electrons accelerate and so radiate energy

they would therefore spiral into the nucleus/atoms would be unstable

electrons have discrete/only certain energy levels

the only orbits where electrons do not radiate are those that satisfy the Bohr condition «mvr = n$\frac{h}{2\pi }$»

[3 marks]

$\frac{m_{\text{e}}{v}^2}{r}=\frac{k{e}^2}{r^2}$

OR

KE = $\frac{1}{2}$PE hence $\frac{1}{2}$m_ev² = $\frac{1}{2}\frac{k{e}^2}{r}$

«solving for v to get answer»

Answer given – look for correct working

[1 mark]

combining v = $\sqrt{\frac{k{e}^2}{m_{\text{e}}r}}$ with m_evr = $\frac{h}{2\pi }$ using correct substitution

«eg ${m_e}^2\frac{k{e}^2}{m_{\text{e}}r}{r}^2=\frac{h^2}{4{\pi }^2}$»

correct algebraic manipulation to gain the answer

Answer given – look for correct working

Do not allow a bald statement of the answer for MP2. Some further working eg cancellation of m or r must be shown

[2 marks]

« r = $\frac{{(6.63\times {10}^{-34})}^2}{4{\pi }^2\times 8.99\times {10}^9\times 9.11\times {10}^{-31}\times {(1.6\times {10}^{-19})}^2}$»

r = 5.3 × 10^–11 «m»

[1 mark]

the energy released is 3.54 – 0.48 = 3.06 «MeV»

this is shared by the electron and the antineutrino

so the electron’s energy varies from 0 to 3.06 «MeV»

[3 marks]

the palladium nucleus emits the photon when it decays into the ground state «from the excited state»

[1 mark]

Photon energy

E = 0.48 × 10⁶ × 1.6 × 10^–19 = «7.68 × 10^–14 J»

λ = «$\frac{hc}{E}=\frac{6.63\times {10}^{-34}\times 3\times {10}^8}{7.68\times {10}^{-14}}$ =» 2.6 × 10^–12 «m»

Award [2] for a bald correct answer

Allow ECF from incorrect energy

[2 marks]

Identify the missing information for this decay.

On the graph, sketch how the number of boron nuclei in the sample varies with time.

After 4.3 × 10⁶ years,

$$\frac{\text{number of produced boron nuclei}}{\text{number of remaining beryllium nuclei}}=7.$$

Show that the half-life of beryllium-10 is 1.4 × 10⁶ years.

Beryllium-10 is used to investigate ice samples from Antarctica. A sample of ice initially contains 7.6 × 10¹¹ atoms of beryllium-10. The present activity of the sample is 8.0 × 10⁻³ Bq.

Determine, in years, the age of the sample.

State what is meant by thermal radiation.

Discuss how the frequency of the radiation emitted by a black body can be used to estimate the temperature of the body.

Calculate the peak wavelength in the intensity of the radiation emitted by the ice sample.

The temperature in the laboratory is higher than the temperature of the ice sample. Describe one other energy transfer that occurs between the ice sample and the laboratory.

${}_4^{10}\text{Be}{\to }_5^{10}\text{B}{+}_{-1}^0\text{e}+{\overline{\text{V}}}_{\text{e}}$

antineutrino AND charge AND mass number of electron ${}_{-1}^0\text{e}$, $\overline{\text{V}}$

conservation of mass number AND charge ${}_5^{10}\text{B}$, ${}_4^{10}\text{Be}$

Do not accept V.

Accept $\overline{V}$ without subscript e.

[2 marks]

correct shape ie increasing from 0 to about 0.80 N₀

crosses given line at 0.50 N₀

[2 marks]

ALTERNATIVE 1

fraction of Be = $\frac{1}{8}$, 12.5%, or 0.125

therefore 3 half lives have elapsed

$t_{\frac{1}{2}}=\frac{4.3\times {10}^6}{3}=1.43\times {10}^6$ «≈ 1.4 × 10⁶» «y»

ALTERNATIVE 2

fraction of Be = $\frac{1}{8}$, 12.5%, or 0.125

$\frac{1}{8}={\text{e}}^{-\lambda }(4.3\times {10}^6)$ leading to λ = 4.836 × 10^–7 «y»^–1

$\frac{\ln 2}{\lambda }$ = 1.43 × 10⁶ «y»

Must see at least one extra sig fig in final answer.

[3 marks]

λ «= $\frac{\ln 2}{1.4\times {10}^6}$» = 4.95 × 10^–7 «y^–1»

rearranging of A = λN₀e^–λt to give –λt = ln $\frac{8.0\times {10}^{-3}\times 365\times 24\times 60\times 60}{4.95\times {10}^{-7}\times 7.6\times {10}^{11}}$ «= –0.400»

t = $\frac{-0.400}{-4.95\times {10}^{-7}}=8.1\times {10}^5$ «y»

Allow ECF from MP1

[3 marks]

emission of (infrared) electromagnetic/infrared energy/waves/radiation.

[1 mark]

the (peak) wavelength of emitted em waves depends on temperature of emitter/reference to Wein’s Law

so frequency/color depends on temperature

[2 marks]

$\lambda =\frac{2.90\times {10}^{-3}}{253}$

= 1.1 × 10^–5 «m»

Allow ECF from MP1 (incorrect temperature).

[2 marks]

from the laboratory to the sample

conduction – contact between ice and lab surface.

OR

convection – movement of air currents

Must clearly see direction of energy transfer for MP1.

Must see more than just words “conduction” or “convection” for MP2.

[2 marks]

Show that the energy of photons from the UV lamp is about 10 eV.

Calculate, in J, the maximum kinetic energy of the emitted electrons.

Suggest, with reference to conservation of energy, how the variable voltage source can be used to stop all emitted electrons from reaching the collecting plate.

The variable voltage can be adjusted so that no electrons reach the collecting plate. Write down the minimum value of the voltage for which no electrons reach the collecting plate.

On the diagram, draw and label the equipotential lines at –0.4 V and –0.8 V.

An electron is emitted from the photoelectric surface with kinetic energy 2.1 eV. Calculate the speed of the electron at the collecting plate.

E₁ = –13.6 «eV» E₂ = – $\frac{13.6}{4}$ = –3.4 «eV»

energy of photon is difference E₂ – E₁ = 10.2 «≈ 10 eV»

Must see at least 10.2 eV.

[2 marks]

10 – 5.1 = 4.9 «eV»

4.9 × 1.6 × 10^–19 = 7.8 × 10^–19 «J»

Allow 5.1 if 10.2 is used to give 8.2×10⁻¹⁹ «J».

EPE produced by battery

exceeds maximum KE of electrons / electrons don’t have enough KE

For first mark, accept explanation in terms of electric potential energy difference of electrons between surface and plate.

[2 marks]

4.9 «V»

Allow 5.1 if 10.2 is used in (b)(i).

Ignore sign on answer.

[1 mark]

two equally spaced vertical lines (judge by eye) at approximately 1/3 and 2/3

labelled correctly

[2 marks]

kinetic energy at collecting plate = 0.9 «eV»

speed = «$\sqrt{\frac{2\times 0.9\times 1.6\times {10}^{-19}}{9.11\times {10}^{-31}}}$» = 5.6 × 10⁵ «ms^–1»

Allow ECF from MP1

[2 marks]

does not support the wave nature of light.

does support the photon nature of light.

Calculate, in eV, the work function of the metal surface.

The intensity of the light incident on the surface is reduced by half without changing the wavelength. Draw, on the graph, the variation of the current $I$ with potential $V$ after this change.

«low intensity light would» transfer energy to the electron at a low rate/slowly ✔
time would be required for the electron «to absorb the required energy» to escape/be emitted ✔

NOTE: OWTTE

«in the photon theory of light» the electron interacts with a single photon ✔
and absorbs all the energy OR and can leave the metal immediately ✔

NOTE: Reference to photon-electron collision scores MP1

$\varphi =\frac{hc}{\lambda }-E_{\mathrm{K}}$ ✔

$E_{\mathrm{K}}=1.5«\mathrm{eV}»$ ✔

$\varphi =«\frac{1.24\times {10}^{-6}}{480\times {10}^{-9}}-1.5=» 1.1 «\mathrm{eV}»$✔

NOTE: Allow reading from the graph of $E_k=\mathit{1}\mathit{.4}$leading to an answer of 1.2 «eV».

similar curve lower than original ✔

with same horizontal intercept ✔

Use the graph to show that the nuclear radius of silicon-30 is about 4 fm.

Estimate, using the result from (a)(i), the nuclear radius of thorium-232 $\left({}_{90}^{232}\text{Th}\right)$.

Suggest one reason why a beam of electrons is better for investigating the size of a nucleus than a beam of alpha particles of the same energy.

Outline why deviations from Rutherford scattering are observed when high-energy alpha particles are incident on nuclei.

read off between 17 and 19 «deg» ✔

correct use of d = $\frac{\lambda }{\sin \theta }$ = 7.8 × 10⁻¹⁵ «m» ✔

so radius = $\frac{7.8}{2}$ «fm» = 3.9 «fm» ✔

Award ecf for wrong angle in MP1.

Answer for MP3 must show at least 2 sf.

R_Th = R_si  ${\left(\frac{A_{\text{Th}}}{A_{\text{Si}}}\right)}^{\frac{1}{3}}$ or substitution ✔

7.4 «fm» ✔

electron wavelength shorter than alpha particles (thus increased resolution)
OR
electron is not subject to strong nuclear force ✔

nuclear forces act ✔

nuclear recoil occurs ✔

significant penetration into nucleus / probing internal structure of individual nucleons ✔

incident particles are relativistic ✔

This question was left blank by many candidates and many of those who attempted it chose an angle that when used with the correct equation gave an answer close to the given answer of 4 fm. Very few selected the correct angle, calculated the correct diameter, and divided by two to get the correct radius.

This question was also left blank by many candidates. Many who did answer simply used the ratio of the of the mass numbers of the two elements and failed to take the cube root of the ratio. It should be noted that the question specifically stated that candidates were expected to use the result from 2ai, and not just simply guess at the new radius.

This question was very poorly answered with the vast majority of candidates simply listing differences between alpha particles and electrons (electrons have less mass, electrons have less charge, etc) rather than considering why high speed electrons would be better for studying the nucleus.

Candidates struggled with this question. The vast majority of responses were descriptions of Rutherford scattering with no connection made to the deviations when high-energy alpha particles are used. Many of the candidates who did appreciate that this was a different situation from the traditional experiment made vague comments about the alpha particles “hitting” the nucleus.

A particular K meson has a quark structure $\overline{\mathrm{u}}$s. State the charge, strangeness and baryon number for this meson.

The Feynman diagram shows the changes that occur during beta minus (β^–) decay.

Label the diagram by inserting the four missing particle symbols and the direction of the arrows for the decay particles.

C-14 decay is used to estimate the age of an old dead tree. The activity of C-14 in the dead tree is determined to have fallen to 21% of its original value. C-14 has a half-life of 5700 years.

(i) Explain why the activity of C-14 in the dead tree decreases with time.

(ii) Calculate, in years, the age of the dead tree. Give your answer to an appropriate number of significant figures.

charge: –1«e» or negative or K⁻

strangeness: –1  

baryon number: 0

Negative signs required.
Award [2] for three correct answers, [1 max] for two correct answer and [0] for one correct answer.

correct symbols for both missing quarks

exchange particle and electron labelled W or W^– and e or e^–

Do not allow W⁺ or e⁺ or β⁺. Allow β or β^–.

arrows for both electron and anti-neutrino correct

Allow ECF from previous marking point.

i
number of C-14 atoms/nuclei are decreasing
OR
decreasing activity proportional to number of C-14 atoms/nuclei
OR
A = A₀e^–λt so A decreases as t increases
Do not allow “particles”
Must see reference to atoms or nuclei or an equation, just “C-14 is decreasing” is not enough.

ii
0.21 = (0.5)ⁿ
OR
$0.21=e^{-\left(\frac{\ln 2\times t}{5700}\right)}$

n = 2.252 half-lives or t =1 2834 «y»
Early rounding to 2.25 gives 12825 y

13000 y rounded correctly to two significant figures:
Both needed; answer must be in year for MP3.
Allow ECF from MP2.
Award [3] for a bald correct answer.

Show that the speed $v$ of the electron with mass $m$, is given by $v=\sqrt{\frac{2k{e}^2}{mr}}$.

Hence, deduce that the total energy of the electron is given by $E_{\mathrm{TOT}}=-\frac{k{e}^2}{r}$.

In this model the electron loses energy by emitting electromagnetic waves. Describe the predicted effect of this emission on the orbital radius of the electron.

Show that the de Broglie wavelength $\lambda$ of the electron in the $n=3$ state is  $\lambda =5.1\times {10}^{-10}$ m.

The formula for the de Broglie wavelength of a particle is $\lambda =\frac{h}{mv}$.

Estimate for $n=3$, the ratio $\frac{\mathrm{circumference} \mathrm{of} \mathrm{orbit}}{\mathrm{de} \mathrm{Broglie} \mathrm{wavelength} \mathrm{of} \mathrm{electron}}$.

State your answer to one significant figure.

The description of the electron is different in the Schrodinger theory than in the Bohr model. Compare and contrast the description of the electron according to the Bohr model and to the Schrodinger theory.

equating centripetal to electrical force $\frac{2k{e}^2}{r^2}=\frac{m{v}^2}{r}$ to get result ✔

uses (a)(i) to state $E_{\mathrm{k}}=\frac{k{e}^2}{r}$ OR states $E_{\mathrm{p}}=-\frac{2k{e}^2}{r}$ ✔

adds « $E_{\mathrm{TOT}}=E_{\mathrm{k}}+E_{\mathrm{p}}=\frac{k{e}^2}{r}-\frac{2k{e}^2}{r}$» to get the result ✔

the total energy decreases
OR
by reference to $E_{\mathrm{TOT}}=-\frac{k{e}^2}{r}$ ✔

the radius must also decrease ✔

NOTE: Award [0] for an answer concluding that radius increases

with $n=3, v=«\sqrt{\frac{2\times 8.99\times {10}^9\times {\left(1.6\times {10}^{-19}\right)}^2}{9.11\times {10}^{-31}\times 9\times 2.7\times {10}^{-11}}}=» 1.44\times {10}^6 «\mathrm{m} {\mathrm{s}}^{-1}»$✔

$\lambda =\frac{6.63\times {10}^{-34}}{9.11\times {10}^{-31}\times 1.44\times {10}^6}$  OR  $\lambda =5.05\times {10}^{-10} «\mathrm{m}»$✔

$\frac{{\displaystyle 2\pi r}}{{\displaystyle \lambda }}=«\frac{{\displaystyle 2\pi \times 9\times 2.7\times {10}^{-11}}}{{\displaystyle 5.1\times {10}^{-10}}}=2.99»\cong 3$ ✔

NOTE: Allow ECF from (b)(i)

reference to fixed orbits/specific radii OR quantized angular momentum in Bohr model ✔

electron described by a wavefunction/as a wave in Schrödinger model OR as particle in Bohr model ✔

reference to «same» energy levels in both models ✔

reference to «relationship between wavefunction and» probability «of finding an electron in a point» in Schrödinger model ✔