A positive pion decays into a positive muon and a neutrino.

\[{\pi ^ + } \to {\mu ^ + } + {v_\mu }\]

The momentum of the muon is measured to be 29.8 MeV c^{–1} in a laboratory reference frame in which the pion is at rest. The rest mass of the muon is 105.7 MeV c^{–2} and the mass of the neutrino can be assumed to be zero.

For the laboratory reference frame

a.i.

write down the momentum of the neutrino.

show that the energy of the pion is about 140 MeV.

State the rest mass of the pion with an appropriate unit.

**Answer/Explanation**

## Markscheme

a.i.

**«**–**»**29.8 **«**MeVc^{–1}**»**

**[1 mark]**

*E _{π}* = \(\sqrt {p_\mu ^2{c^2} + m_\mu ^2{c^4}} \) +

*p*

_{v}c

*OR**E*= 109.8

_{μ}**«**MeV

**»**

*E _{π}* =

**«**\(\sqrt {{{29.8}^2} + {{105.7}^2}} \) + 29.8 =

**»**139.6

**«**MeV

**»**

*Final value to at least 3 sig figs required for mark.*

*[2 marks]*

139.6 MeVc^{–2}

*Units required.*

*Accept 140 MeVc ^{–}*

^{2}

*.*

*[1 mark]*

A student pours a canned carbonated drink into a cylindrical container after shaking the can violently before opening. A large volume of foam is produced that fills the container. The graph shows the variation of foam height with time.

a.

Determine the time taken for the foam to drop to

(i) half its initial height.

(ii) a quarter of its initial height.

The change in foam height can be modelled using ideas from other areas of physics. Identify **one** other situation in physics that is modelled in a similar way.

**Answer/Explanation**

## Markscheme

a.

i

18 «s»

*Allow answer in the range of 17 «s» to 19 «s».Ignore wrong unit.*

ii

36 «s»

*Allow answer in the range of 35 «s» to 37 «s».*

radioactive/nuclear decay* OR*capacitor discharge

*cooling*

**OR**

*Accept any relevant situation, eg: critically damping, approaching terminal velocity*

This question is about the hydrogen atom.

The diagram shows the three lowest energy levels of a hydrogen atom.

a.

An electron is excited to the *n*=3 energy level. On the diagram, draw arrows to show the possible electron transitions that can lead to the emission of a photon.

Show that a photon of wavelength 656 nm can be emitted from a hydrogen atom.

**Answer/Explanation**

## Markscheme

a.

energy / 10–18 J

two correct arrows

a third correct arrow;*Award [1] for three upward arrows.*

energy of photon is \(E = \left( {\frac{{6.63 \times {{10}^{ – 34}} \times 3 \times {{10}^8}}}{{656 \times {{10}^{ – 9}}}} = } \right)3.03 \times {10^{ – 19}}\left( {\rm{J}} \right)\);

((0.545-0.242)×10–18=3.03×10–19 J) is the difference in energy between *n=*3 and *n*=2;

## Examiners report

a.

In part (a) almost everyone drew just two transitions out of the three possible.

In part (b) many candidates were able to identify the transition.

This question is about radioactive decay.

Meteorites contain a small proportion of radioactive aluminium-26 \(\left( {_{{\text{13}}}^{{\text{26}}}{\text{Al}}} \right)\) in the rock.

The amount of \(_{{\text{13}}}^{{\text{26}}}{\text{Al}}\) is constant while the meteorite is in space due to bombardment with cosmic rays.

After reaching Earth, the number of radioactive decays per unit time in a meteorite sample begins to diminish with time. The half-life of aluminium-26 is \(7.2 \times {10^5}\) years.

a.

Aluminium-26 decays into an isotope of magnesium (Mg) by \({\beta ^ + }\) decay.

\[_{{\text{13}}}^{{\text{26}}}{\text{Al}} \to _{\text{Y}}^{\text{X}}{\text{Mg}} + {\beta ^ + } + {\text{Z}}\]

Identify X, Y and Z in this nuclear decay process.

X:

Y:

Z:

Explain why the beta particles emitted from the aluminium-26 have a continuous range of energies.

State what is meant by half-life.

A meteorite which has just fallen to Earth has an activity of 36.8 Bq. A second meteorite of the same mass, which arrived some time ago, has an activity of 11.2 Bq. Determine, in years, the time since the second meteorite arrived on Earth.

**Answer/Explanation**

## Markscheme

a.

*X*: 26 and *Y*: 12; *(both needed for **[1]**)*

*Z*: *v*/neutrino;

*Do not allow the antineutrino.*

total energy released is fixed;

neutrino carries some of this energy;

(leaving the beta particle with a range of energies)

the time taken for half the radioactive nuclides to decay / the time taken for the activity to decrease to half its initial value;

*Do not allow reference to change in weight.*

\(\lambda = \left( {\frac{{\ln 2}}{{7.2 \times {{10}^5}}} = } \right){\text{ }}9.63 \times {10^{ – 7}}\);

\(11.2 = 36.8{e^{ – (9.63 \times {{10}^{ – 7}})t}}\);

\(t = 1.24 \times {10^6}{\text{ (yr)}}\);

## Examiners report

a.

This was generally well answered, although a significant minority insisted that nuclear half-life is defined by a loss of mass.

This was generally well answered, although a significant minority insisted that nuclear half-life is defined by a loss of mass.

This was generally well answered, although a significant minority insisted that nuclear half-life is defined by a loss of mass.

This question is about atomic spectra and energy states.

a.

Outline how atomic absorption spectra provide evidence for the quantization of energy states in atoms.

The diagram shows some atomic energy levels of hydrogen.

A photon of energy 2.86 eV is emitted from a hydrogen atom. Using the diagram, draw an arrow to indicate the electron transitions that results in the emission of this photon.

**Answer/Explanation**

## Markscheme

a.

an atom will only absorb a photon if the photon energy corresponds to an energy difference between two of its energy states;

the absorption of energy takes places in discrete quantities (quanta);

arrow drawn downwards from \( – 0.54\) level to \( – 3.40\) level;

## Examiners report

a.

In (a), candidates rarely described absorption spectra, rather explaining emission spectra. There were very poor explanations of quantized energy states.

(b) was generally well answered, although a common mistake was to have the change in the wrong direction.

This question is about radioactive decay.

In a particular nuclear medical imaging technique, carbon-11 \((_{\;6}^{11}{\text{C}})\) is used. It is radioactive and decays through \({\beta ^ + }\) decay to boron (B).

The half-life of carbon-11 is 20.3 minutes.

a.i.

Identify the numbers and the particle to complete the decay equation.

State the nature of the \({\beta ^ + }\) particle.

Outline a method for measuring the half-life of an isotope, such as the half-life of carbon-11.

State the law of radioactive decay.

Derive the relationship between the half-life \({T_{\frac{1}{2}}}\) and the decay constant \(\lambda \) , using the law of radioactive decay.

Calculate the number of nuclei of carbon-11 that will produce an activity of \(4.2 \times {10^{20}}{\text{ Bq}}\).

**Answer/Explanation**

## Markscheme

a.i.

\(\left( {_{\;6}^{11}{\text{C}} \to _{\;5}^{11}{\text{B}} + _{ + 1}^{\;\;0}{\beta ^ + } + v{\text{ (or neutrino)}}} \right)\)

\(_{\;6}^{11}{\text{C}} \to _{\;5}^{11}{\text{B}} + _{ + 1}^{\;\;0}{\beta ^ + }\);

v (or neutrino);

*Award **[1] **for all the correct numbers and **[1] **for the neutrino.*

positron / antielectron / lepton;

measure activity as a function of time;

create a graph of activity with time, and estimate half-life from the graph;

make at least three estimates of half-life from the graph and take mean;

*or*

measure activity as a function of time;

create a graph of ln(A) with time, find the decay constant \(\lambda \) from the gradient;

estimate the half-life using \({T_{\frac{1}{2}}} = \frac{{\ln 2}}{\lambda }\);

the rate of decay is proportional to the amount of (radioactive) material remaining;

the number of undecayed nuclei at time \(t\) is given by \(N = {N_0}{e^{ – \lambda t}}\), where \({N_0}\) is the number of undecayed nuclei at time \(t = 0\) and \(\lambda \) is the decay constant;

\(\frac{{{N_0}}}{2} = {N_0}{e^{ – {T_{\frac{1}{2}}}}}\);

\(\ln \left( {\frac{1}{2}} \right) = – \lambda {T_{\frac{1}{2}}}\) so \({T_{\frac{1}{2}}} = \frac{{\ln 2}}{\lambda }\);

\(\lambda = \frac{{\ln 2}}{{60 \times 20.3}}{\text{ }}( = 5.69 \times {10^{ – 4}}{s^{ – 1}})\);

\(\frac{A}{\lambda } = \frac{{4.2 \times {{10}^{20}}}}{{5.69 \times {{10}^{ – 4}}}} = 7.4 \times {10^{23}}\);

## Examiners report

a.i.

(a)(i) was well answered.

(a)(ii) was well answered.

(b)(i) was poorly answered, with many referring to measurement of the loss of mass of the sample.

(b)(ii) was very poorly answered.

Most did not use the law of radioactive decay, as required in (b)(iii).

(b)(iv) was either very well answered or very poorly answered.

This question is about atomic spectra.

The diagram shows some of the energy levels of a hydrogen atom.

a.

Explain how atomic line spectra provide evidence for the existence of discrete electron energy levels in atoms.

(i) Calculate the wavelength of the photon that will be emitted when an electron moves from the –3.40 eV energy level to the –13.6 eV energy level.

(ii) State and explain if it is possible for a hydrogen atom in the ground state to absorb a photon with an energy of 12.5 eV.

**Answer/Explanation**

## Markscheme

a.

each line represents a single frequency/wavelength;

which corresponds to a specific photon energy / \(E = hf\);

energy of photon determined by energy change of electrons;

electrons transition between energy levels (so discrete energy levels);

*Award **[3 max] **for reverse argument that discrete energy levels produce line spectra.*

(i) \(\Delta E = [13.6 – 3.40] \times 1.60 \times {10^{ – 19}}{\text{ (}} = 1.63 \times {10^{ – 18}}{\text{ J)}}\);

\(E = \frac{{hc}}{\lambda }\); *(accept implicit use of this equation)*

\(\lambda = \frac{{6.63 \times {{10}^{ – 34}} \times 3.00 \times {{10}^8}}}{{1.63 \times {{10}^{ – 18}}}} = 1.22 \times {10^{ – 7}}{\text{ (m)}}\);

*Award **[3] **for a correct bald answer.*

(ii) photon absorbed when its energy is equal to the difference between two energy levels;

so absorption not possible;

## Examiners report

a.

In (a) the logical sequence is: line spectra ➡ discrete photon energy ➡ discrete electron transitions ➡ discrete electron energy levels. However, very few were able to sequence their answers in this way. Despite this there were many reasonable answers.

In (i) many correct answers were seen, but there were some answers where the de Broglie formula was mistakenly used. (ii) was poorly answered as few could explain that a 12.5eV photon did not match any of the possible transition energies.

This question is about atomic energy levels.

a.

Explain how atomic spectra provide evidence for the quantization of energy in atoms.

Outline how the de Broglie hypothesis explains the existence of a **discrete** set of wavefunctions for electrons confined in a box of length* L*.

The diagram below shows the shape of two allowed wavefunctions *ѱ _{A}* and

*ѱ*for an electron confined in a one-dimensional box of length

_{B}*L*.

(i) With reference to the de Broglie hypothesis, suggest which wavefunction corresponds to the larger electron energy.

(ii) Predict and explain which wavefunction indicates a larger probability of finding the electron near the position \(\frac{L}{2}\) in the box.

(iii) On the graph in (c) on page 7, sketch a possible wavefunction for the **lowest** energy state of the electron.

**Answer/Explanation**

## Markscheme

a.

atomic spectra have discrete line structures / only discrete frequencies/wavelengths;

photon energy is related to frequency/wavelength;

photons have discrete energies;

photons arise from electron transitions between energy levels;

which must have discrete values of energy;

de Broglie suggests that electrons/all particles have an associated wavelength; this wave will be a stationary wave which meets the boundary conditions of the box; the stationary wave has wavelength \(\frac{{2L}}{n}\) (where *L* is the length of the box and where *n* is an integer);

(i) wavelength of *ψ _{A}* larger than

*ψ*;

_{B}therefore momentum of

*ψ*larger than

_{B}*ψ*(from de Broglie hypothesis); therefore

_{A}*ψ*has larger energy;

_{B}*Award*

**[1 max]**for a bald correct answer.**or**

*ψ _{B}* has

*n*=3,

*ψ*has

_{A}*n*=2;

E

_{K}∝

*n*

^{2};

so

*ψ*corresponds to the larger energy;

_{B}(ii) *ψ _{A}*=0,

*ψ*≠0 in the middle of the box/at \(\frac{L}{2}\);

_{B}so

*ψ*

_{B}corresponds to the larger probability since probability ∝Ι

*ψ*Ι

^{2};

*Accept ∝ ψ*

^{2}.**or**

the probability (of finding the electron) is related to the amplitude;

amplitude of *ψ _{B}* is greater than amplitude of

*ψ*so

_{A}*ψ*is more likely to be found;

_{B}*Award [1 max] for a bald correct answer.*

(iii)

correct sketch; (accept –*ψ*)*Accept wavefunction with any amplitude.*

## Examiners report

a.

(a) Candidates struggled with this question. Although they demonstrated some familiarity with the idea, they could not clearly describe the connection between atomic structure and the emission spectra, usually discussing electrons without photons. The arguments leading from atomic spectra to energy levels were not logically organised.

There were very few correct answers to (b).

(i) was reasonably well done by many, although many did not refer to the de Broglie hypothesis explicitly and thus relate wavelength to momentum and so to energy.

(ii) was poorly answered. Not many candidates understood the relation between amplitude and probability of locating the particle.

(iii) was well done by most.

This question is about radioactive decay.

Sodium-22 undergoes *β*+ decay.

a.

Identify the missing entries in the following nuclear reaction.

\[{}_{11}^{22}{\rm{Na}} \to {}_ \ldots ^{22}{\rm{Ne}} + {}_ \ldots ^0e + {}_0^0 \ldots \]

Define *half-life*.

Sodium-22 has a decay constant of 0.27 yr^{–1}.

(i) Calculate, in years, the half-life of sodium-22.

(ii) A sample of sodium-22 has initially 5.0 × 10^{23} atoms. Calculate the number of sodium-22 atoms remaining in the sample after 5.0 years.

**Answer/Explanation**

## Markscheme

a.

\({}_{11}^{22}{\rm{Na}} \to {}_{10}^{22}{\rm{Ne}} + {}_{ + 1}^0e + {}_0^0v\)

\({}_{10}^{22}{\rm{Ne}}\);

\({}_{ + 1}^0e\) (*accep*t \({}_ + ^0e\))

\({}_0^0v\); (*award [0] for* \({}_0^0\bar v\))

time taken for half/50% of the nuclei to decay / activity to drop by half/50%;

(i) \({T_{\frac{1}{2}}} = \frac{{\ln 2}}{\lambda }\);

\(\frac{{0.693}}{{0.27{\rm{y}}{{\rm{r}}^{ – 1}}}}\)=2.6 (years);*Award [2] for a bald correct answer.*

(ii) *N*=5.0 x 10^{23 }x e^{-0.27×5.0};*N*=1.3 x 10^{23;}*Award [2] for a bald correct answer.*

## Examiners report

a.

This question was well done in general.

Many candidates referred to mass halving rather than activity.

This question is about radioactive decay.

a.

A nuclide of the isotope potassium-40 \(\left( {{}_{19}^{40}{\rm{K}}} \right)\) decays into a stable nuclide of the isotope

argon-40 \(\left( {{}_{18}^{40}{\rm{Ar}}} \right)\). Identify the particles X and Y in the nuclear equation below.

\[{}_{19}^{40}{\rm{K}} \to {}_{18}^{40}{\rm{Ar + X + Y}}\]

The half-life of potassium-40 is 1.3×10^{9}yr. In a particular rock sample it is found that 85 % of the original potassium-40 nuclei have decayed. Determine the age of the rock.

State the quantities that need to be measured in order to determine the half-life of a long-lived isotope such as potassium-40.

**Answer/Explanation**

## Markscheme

a.

neutrino/*ν*;

positron / e^{+} / \({}_{ + 1}^0{\rm{e}}\) / β^{+};*Award [1 max] for wrongly stating electron and antineutrino. Both needed for the ECF.*

*Order of answers is not important.*

\(\lambda = \left( {\frac{{\ln 2}}{{1.3 \times {{10}^9}}} = } \right)5.31 \times {10^{ – 10}}{\rm{y}}{{\rm{r}}^{ – 1}}\);

\(0.15 = {{\rm{e}}^{\left[ { – 5.31 \times {{10}^{ – 10}} \times t} \right]}}\);*t*=3.6×10^{9}yr;*Award [3] for a bald correct answer.*

**or**

(0.5)^{n}=0.15;

\(n = \frac{{\log \left( {0.15} \right)}}{{\log \left( {0.5} \right)}} = 2.74{\rm{half – lives}}\);

2.74×1.3×10^{9}=3.6×10^{9}yr;*Award [3] for a bald correct answer.*

the count rate/activity of a sample;

the mass/number of atoms in the sample;

This question is about atomic energy levels.

a.

Outline a laboratory procedure for producing and observing the atomic absorption spectrum of a gas.

(i) Describe the appearance of an atomic absorption spectrum.

(ii) Explain why the spectrum in (a) provides evidence for quantization of energy in atoms.

The principal energy levels of the hydrogen atom in electronvolt (eV) are given by

\[{E_n} = \frac{{13.6}}{{{n^2}}}\]

where *n* is a positive integer.

Determine the wavelength of the absorption line that corresponds to an electron transition from the energy level given by *n*=1 to the level given by *n*=3.

**Answer/Explanation**

## Markscheme

a.

shine white light through;

a tube of the gas;

then observe with spectroscope/grating/prism;

(i) continuous spectrum crossed by dark lines;

(ii) dark lines formed by the absorption of photons;

the absorbed photons have specific/discreet wavelengths;

indicating discreet differences in energy;

which can only be explained by existence of energy levels;

\(E = 13.6\left[ {\frac{1}{{{1^2}}} – \frac{1}{{{3^2}}}} \right] = 12.1{\rm{eV}}\);

12.1eV=12.1×1.6×10^{–19}J =\(\frac{{hc}}{\lambda }\);*λ*=102nm * or* 103nm;

*Award*

**[3]**for a bald correct answer.This question is about atomic spectra.

Diagram 1 shows some of the energy levels of the hydrogen atom. Diagram 2 is a representation of part of the emission spectrum of atomic hydrogen. The lines shown represent transitions involving the – 3.40 eV level.

a.

Deduce that the energy of a photon of wavelength 658 nm is 1.89 eV.

(i) On **diagram 1**, draw an arrow to show the electron transition between energy levels that gives rise to the emission of a photon of wavelength 658 nm. Label this arrow with the letter A.

(ii) On** diagram 1**, draw arrows to show the electron transitions between energy levels that give rise to the emission of photons of wavelengths 488 nm, 435 nm and 411 nm. Label these arrows with the letters B, C and D.

Explain why the lines in the emission spectrum of atomic hydrogen, shown in **diagram 2**, become closer together as the wavelength of the emitted photons decreases.

**Answer/Explanation**

## Markscheme

a.

\(E = \frac{{hc}}{\lambda }\);

\( = \frac{{6.63 \times {{10}^{ – 34}} \times 3.00 \times {{10}^8}}}{{658 \times {{10}^{ – 9}}}} = 3.02 \times {10^{ – 19}}\);

=\(\frac{{3.02 \times {{10}^{ – 19}}}}{{1.60 \times {{10}^{ – 19}}}}\);

=1.89eV

**or**

the photon of wavelength 658nm is the longest (in the emission graph);

therefore it has the shortest frequency and lowest energy (from *E*=*hf* );

therefore it arises from the transition between the –1.51eV and the –3.40eV energy levels which have a difference of 1.89eV;

(i) see diagram below;

(ii) see diagram below;*All three must be correct for the mark.*

at higher energy levels, energy levels become closer together;

the energy differences between higher energy levels and the lower level (*n*=2) become more equal;

hence the difference in wavelength of emitted photons decreases / *OWTTE*;

This question is about radioactive decay.

Iodine-124 (I-124) is an unstable radioisotope with proton number 53. It undergoes beta plus decay to form an isotope of tellurium (Te).

a.

State the reaction for the decay of the I-124 nuclide.

The graph below shows how the activity of a sample of iodine-124 changes with time.

(i) State the half-life of iodine-124.

(ii) Calculate the activity of the sample at 21 days.

(iii) A sample of an unknown radioisotope has a half-life twice that of iodine-124 and the same initial activity as the sample of iodine-124. On the axes opposite, draw a graph to show how the activity of the sample would change with time. Label this graph X.

(iv) A second sample of iodine-124 has half the initial activity as the original sample of iodine-124. On the axes opposite, draw a graph to show how the activity of this sample would change with time. Label this graph Y.

**Answer/Explanation**

## Markscheme

a.

\({}_{53}^{124}{\rm{I}} \to {}_{52}^{124}{\rm{Te + }}{}_1^0\beta ^+ \);

\({}_0^0v/v\);*Do not allow an antineutrino.**Award [1 max] for *\({}_{53}^{124}{\rm{I}} \to {}_{54}^{124}{\rm{Te + }}{}_1^0\beta^- + \bar v\).

(i) 4 days;

(ii) \(\lambda = \frac{{\ln 2}}{{{T_{\frac{1}{2}}}}} = \frac{{\ln 2}}{4} = \left( {0.173{\rm{da}}{{\rm{y}}^{ – 1}}} \right)\);

\(A = {A_0}{e^{ – \lambda t}} = 16 \times {10^7} \times {e^{ – 0.173 \times 21}}\left( {{\rm{Bq}}} \right)\);*A*=4.2×10^{6}Bq; *Award [2 max] for bald answer in range *4.2−4.5×10

^{6}Bq

*, or linear*

*interpolation between half lives giving*4.4×10

^{6}Bq.

(iii) graph passing through or near (0,16), (8,8) and (16,4) – see below;

(iv) graph passing through or near (0,8), (4,4) and (8,2) – see below; *Do not penalize if graph does not pass through (12,1) and (16,0.5).*

This question is about the spectrum of atomic hydrogen.

Calculate the difference in energy in eV between the energy levels in the hydrogen atom that give rise to the red line in the spectrum.

**Answer/Explanation**

## Markscheme

\(E = \left( {\frac{{hc}}{\lambda } = } \right){\text{ }}3.03 \times {10^{ – 19}}{\text{ J}}\);

\( = 1.90{\text{ eV}}\);

*Award **[2] **for bald correct answer.*

## Examiners report

The calculation in (b) was often correctly done.

This question is about radioactive decay.

The half-life of Au-189 is 8.84 minutes. A freshly prepared sample of the isotope has an activity of 124Bq.

a.

A nucleus of a radioactive isotope of gold (Au-189) emits a neutrino in the decay to a nucleus of an isotope of platinum (Pt).

In the nuclear reaction equation below, state the name of the particle X and identify the nucleon number \(A\) and proton number \(Z\) of the nucleus of the isotope of platinum.

\[_{\;79}^{189}Au \to _Z^APt + X + v\]

X:

*A*:

*Z*:

(i) Calculate the decay constant of Au-189.

(ii) Determine the activity of the sample after 12.0 min.

**Answer/Explanation**

## Markscheme

a.

*X*: positron ** or** \({\beta ^ + }\);

*A*: 189 and *Z*: 78; *(both responses needed)*

(i) \(0.0784{\text{ mi}}{{\text{n}}^{ – 1}}\);

(ii) recognize to use \(A = {A_0}{e^{ – \lambda t}}\);

\(A = 48.4{\text{ Bq}}\);

*Award **[2] **for bald correct answer.*

## Examiners report

a.

A surprisingly large number of candidates were unable to correctly identify the products of beta plus decay.

Unit errors were often made in (i) and in (ii) there was often some very strange arithmetic to be seen.