This question is about radioactive decay.

A nucleus of magnesium-23 decays forming a nucleus of sodium-23 with the emission of an electron neutrino and a β^{+} particle.

a.

Outline why the existence of neutrinos was hypothesized to account for the energy spectrum of beta decay.

The decay constant for magnesium-23 is 0.061 s^{−1}. Calculate the time taken for the number of magnesium-23 nuclei to fall to 12.5% of its initial value.

**Answer/Explanation**

## Markscheme

a.

spectrum of beta decay is continuous;

with a maximum value of energy;

the resulting energy difference between energy of any β(+) and maximum β(+) energy is accounted for by the energy of the neutrino / reference to energy difference between parent energy level and excited energy level of daughter;

\({{\text{T}}_{\frac{1}{2}}} = \frac{{{\text{In}}2}}{{0.061}} = 11.4\left( {\text{s}} \right)\);

\(\left( {N = \frac{1}{8}{N_0}{\text{ so}}} \right)t = \left( {3{T_{\frac{1}{2}}} = } \right)34\left( {\text{s}} \right)\)

**or**

\(t = – \frac{{{\text{In}}0.125}}{{0.061}}\);

t=34(s);

This question is about radioactive decay.

Nuclide X has a half-life that is estimated to be in the thousands of years.

a.

Outline how the half-life of X can be determined experimentally.

A pure sample of X has a mass of 1.8 kg. The half-life of X is 9000 years. Determine the mass of X remaining after 25000 years.

**Answer/Explanation**

## Markscheme

a.

measurement of mass of sample / determination of molar mass;

determination of number of nuclei *N*;

measurement of activity *A*;

determination of decay constant from \(\lambda = \frac{A}{N}\);

half-life from \({T_{\frac{1}{2}}} = \frac{{1{\rm{n}}2}}{\lambda }\);

\(\lambda = \left( {\frac{{l{\rm{n2}}}}{{{T_{\frac{1}{2}}}}} = \frac{{l{\rm{n}}2}}{{9000}} = } \right)7.70 \times {10^{ – 5}}{\rm{y}}{{\rm{r}}^{ – 1}}\);

\(m = \left( {{m_0}{e^{ – \lambda t}} = } \right)1.8 \times {e^{ – 7.70 \times {{10}^{ – 5}} \times 25000}}\);

m=0.26 (kg);

**or**

\(\frac{{25000}}{{9000}} = 2.77\) half-lives;

fractional mass left = \({\left( {\frac{1}{2}} \right)^{2.77}} = 0.15\);

mass left=1.8×0.15=0.26 (kg);*Award [3] *

*for a bald correct answer.*

## Examiners report

a.

Most candidates were very uncertain about determining a very long half-life. Part marks were often obtained for stating how half-life was obtained from the decay constant, but determination of activity and number of sample atoms was not usually mentioned. Most candidates described how the half-life of a nuclide with a short half-life can be found.

In (b) surprisingly few candidates know the easy way to calculate fraction remaining. Find the number of half-lives passed (n). Fraction remaining = 0.5n. This works even when n is non-integer. Most obtained at least 1 mark for finding the decay constant or the number of half-lives. Quite a few candidates assumed a proportional relationship for the non-integer part of n.

This question is about nuclear physics and radioactive decay.

a.

Define *decay constant.*

A sample of 1.6 mol of the radioactive nuclide radon-210 \(\left( {{}_{86}^{210}{\rm{Rn}}} \right)\) decays into polonium-206 \(\left( {{}_{84}^{206}{\rm{Po}}} \right)\) with the production of one other particle.

\[{}_{86}^{210}{\rm{Rn}} \to {}_{84}^{206}{\rm{Po + X}}\]

(i) Identify particle X.

(ii) The radioactive decay constant of radon-210 is 8.0×10^{–5}s^{–1}. Determine the time required to produce 1.1 mol of polonium-206.

Particle X has an initial kinetic energy of 6.2MeV after the decay in (b). In a scattering experiment, particle X is aimed head-on at a stationary gold-197 \(\left( {{}_{76}^{197}{\rm{Au}}} \right)\) nucleus.

Determine the distance of closest approach of particle X to the Au nucleus.

**Answer/Explanation**

## Markscheme

a.

the probability of decay of a nucleus per unit time;

(i) alpha particle / helium nucleus;

(ii) number of Po nuclei produced=number of Rn nuclei decayed *(seen or implied)*;

\(0.5 = 1.6{e^{ – \lambda t}}\);

\(t = \left( { – \frac{{\ln \frac{{0.5}}{{1.6}}}}{\lambda } = } \right)\frac{{1.163}}{{8.0 \times {{10}^{ – 5}}}}\);

1.5×10^{4}(s);

initial kinetic energy=electric potential energy at closest distance;

kinetic energy \(E = \left( {6.2 \times {{10}^6} \times 1.6 \times {{10}^{ – 19}} = } \right)9.9 \times {10^{ – 13}}\left( {\rm{J}} \right)\);

\(d = k\frac{{{q_1}{q_2}}}{E} = 8.99 \times {10^9}\frac{{2 \times 79 \times {{\left[ {1.6 \times {{10}^{ – 19}}} \right]}^2}}}{{9.9 \times {{10}^{ – 13}}}}\left( {\rm{m}} \right)\) * or* \( = 3.7 \times {10^{ – 14}}\left( {\rm{m}} \right)\);

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 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 radioactive decay.

a.

Define the *decay constant *of a radioactive isotope.

Show that the decay constant \(\lambda \) is related to the half-life \({T_{\frac{1}{2}}}\) by the expression

\[\lambda {T_{\frac{1}{2}}} = \ln 2.\]

Strontium-90 is a radioactive isotope with a half-life of 28 years. Calculate the time taken for 65% of the strontium-90 nuclei in a sample of the isotope to decay.

**Answer/Explanation**

## Markscheme

a.

probability of decay (of a nucleus) per unit time;

*Accept *\(\frac{A}{N}\)* with symbols defined.*

\(N = {N_0}{{\text{e}}^{ – \lambda t}}\) and \(N = \frac{{{N_0}}}{2}\) when \(t = {T_{\frac{1}{2}}}\)\(\,\,\,\)** or**\(\,\,\,\)\(\frac{{{N_0}}}{2} = {N_0}{{\text{e}}^{ – \lambda {T_{\frac{1}{2}}}}}\);

\(\frac{1}{2} = {{\text{e}}^{ – \lambda {T_{\frac{1}{2}}}}}\)\(\,\,\,\)** or**\(\,\,\,\)\(2 = {{\text{e}}^{\lambda {T_{\frac{1}{2}}}}}\);

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

*Answer given, award marks for correct working only.*

\(\lambda = \frac{{\ln 2}}{{28}}\)\(\,\,\,\)** or**\(\,\,\,\)\(0.025\left( {{{\text{y}}^{ – 1}}} \right]\);

\(0.35 = {{\text{e}}^{ – 0.025t}}\);

\(t = 42{\text{ (years)}}\);

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

*Award **[2 max] **for an answer of 17 years (using 0.65 instead of 0.35).*

*or*

\(0.35 = {\left[ {\frac{1}{2}} \right]^x}\) where \(x = \frac{t}{{{T_{\frac{1}{2}}}}}\);

\(\frac{t}{{{T_{\frac{1}{2}}}}} = 1.5\);

\(t = 42{\text{ (years)}}\);

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

*Award **[2 max] **for an answer of 17 years (using 0.65 instead of 0.35).*

## Examiners report

a.

(a) was an easy mark.

In (b) about half of the candidates could derive the relationship between half-life and decay constant, but many were completely lost.

The half-life calculation in (c) was generally well done, but a common mistake was to use 0.65 as the fraction remaining.

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 plutonium as a power source.

Plutonium (\({}_{94}^{238}{\rm{Pu}}\)) decays by alpha emission. The energy of the alpha particle emitted is 8.8×10^{–13}J. The decay constant of plutonium-238 is 8.1×10^{–3}yr^{–1}.

a.

Define *decay constant*.

Plutonium-238 is to be used as a power source in a space probe.

(i) Determine the initial activity of plutonium such that the power released by plutonium is 6.0 W.

(ii) The power source becomes useless when the power released decreases to 4.0 W. Determine the time, in years, for which the power source can be used in the space probe.

**Answer/Explanation**

## Markscheme

a.

the probability of decay per unit time / constant of proportionality in the equation relating activity to number of nuclei;

(i) power \(P = {A_0} \times E \Rightarrow {A_0} = \frac{P}{E}\);

\({A_0} = \left( {\frac{{6.0}}{{8.8 \times {{10}^{ – 13}}}} = } \right)6.8 \times {10^{12}}{\rm{Bq}}\);

(ii) realization that power is proportional to activity / \(P = {P_0}{{\rm{e}}^{ – \lambda t}}\);

\(4.0 = 6.0{{\rm{e}}^{ – 8.1 \times {{10}^{ – 3}}t}}\);

taking logs to get \(\ln \frac{4}{6} = – 8.1 \times {10^{ – 3}}t\);

\(t = \left( {\frac{{\ln \frac{4}{6}}}{{ – 8.1 \times {{10}^{ – 3}}}} = } \right)50{\rm{yr}}\);

*First marking point may be implicit in second.**Award [2 max] for using half-life (86 years) and linear fit to give 57 years.*

*Award*

**[4]**for correct answer by other methods.This question is about radioactive decay.

Nitrogen-13 \(\left( {{}_7^{13}{\rm{N}}} \right)\) is an isotope that is used in medical diagnosis. The decay constant of nitrogen-13 is 1.2×10^{–3}s^{–1}.

a.

(i) Define *decay constant*.

(ii) A sample of nitrogen-13 has an initial activity of 800 Bq. The sample cannot be used for diagnostic purposes if its activity becomes less than 150 Bq. Determine the time it takes for the activity of the sample to fall to 150 Bq.

(i) Calculate the half-life of nitrogen-13.

(ii) Outline how the half-life of a sample of nitrogen-13 can be measured in a laboratory.

Nitrogen-13 undergoes β+ decay. Outline the experimental evidence that suggests another particle, the neutrino, is also emitted in the decay.

**Answer/Explanation**

## Markscheme

a.

(i) probability that a nucleus decays in unit time;

(ii) \(150 = 800{e^{ – 1.2 \times {{10}^{ – 3}}t}}\);

1400s;

(i) 580 s;

(ii) activity/count rate measured at regular time intervals/for at least three half-lives;

plot graph activity/count rate versus time;

detail of determination of half-life from graph;

beta energy spectrum is continuous and associated gamma spectrum is discrete;

difference in energies accounted for by existence of another particle;

**or**

if another particle not present;

then momentum not conserved in beta decay;

This question is about radioactive decay.

Nitrogen-13 \(\left( {{}_7^{13}{\rm{N}}} \right)\) is an isotope that is used in medical diagnosis. The decay constant of nitrogen-13 is 1.2×10^{–3}s^{–1}.

a.

(i) Define *decay constant*.

(ii) A sample of nitrogen-13 has an initial activity of 800 Bq. The sample cannot be used for diagnostic purposes if its activity becomes less than 150 Bq. Determine the time it takes for the activity of the sample to fall to 150 Bq.

(i) Calculate the half-life of nitrogen-13.

(ii) Outline how the half-life of a sample of nitrogen-13 can be measured in a laboratory.

Nitrogen-13 undergoes β+ decay. Outline the experimental evidence that suggests another particle, the neutrino, is also emitted in the decay.

**Answer/Explanation**

## Markscheme

a.

(i) probability that a nucleus decays in unit time;

(ii) \(150 = 800{e^{ – 1.2 \times {{10}^{ – 3}}t}}\);

1400s;

(i) 580 s;

(ii) activity/count rate measured at regular time intervals/for at least three half-lives;

plot graph activity/count rate versus time;

detail of determination of half-life from graph;

beta energy spectrum is continuous and associated gamma spectrum is discrete;

difference in energies accounted for by existence of another particle;

**or**

if another particle not present;

then momentum not conserved in beta decay;

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 nuclear energy levels and nuclear decay.

a.

The isotope bismuth-212 undergoes α-decay to an isotope of thallium. In this decay a gamma-ray photon is also produced. The isotope potassium-40 undergoes β^{+} decay to an isotope of argon.

Outline how the

(i) α particle spectrum and the gamma spectrum of the decay of bismuth-212 give evidence for the existence of discrete nuclear energy levels.

(ii) β^{+} spectrum of the decay of potassium-40 led to the existence of the neutrino being postulated.

The isotope potassium-40 occurs naturally in many rock formations. In a particular sample of rock it is found that, out of the total number of argon plus potassium-40 atoms, 23% are potassium-40 atoms.

Determine the age of the rock sample. The decay constant for potassium-40 is 5.3×10^{–10}yr^{–1}.

**Answer/Explanation**

## Markscheme

a.

(i) the α particles produced have discrete energies;

the gamma rays produced have discrete energies;

since the energies of the α particles and of the photons are determined by the difference in nuclear energy levels this implies that nuclear energy levels are also discrete;

(ii) the β^{+} spectrum is continuous;

the neutrino was postulated to account for those β^{+} with less energy than the maximum;

recognize that \(\frac{N}{{{N_0}}}\)=0.23;

\(0.23 = {e^{ – 5.3 \times {{10}^{ – 10}}t}}\);*t*=2.8×10^{9} yr;

This question is about neutrinos.

The spectrum of electron energies emitted in a typical β-decay is continuous. Describe how this observation led physicists to propose the existence of the particles now called neutrinos.

**Answer/Explanation**

## Markscheme

*Look for these points:*

idea that total energy released in the decay is fixed;

beta particle energies are less than this value/continuous;

the neutrino is postulated to account for this “missing” energy;

## Examiners report

This question is about radioactive decay.

a.

Potassium-40 (K-40) is a radioactive isotope that occurs naturally in many different types of rock. A very small percentage of the isotope undergoes β^{+} decay to form an isotope of argon (Ar). Construct and complete the nuclear reaction equation for this decay.

Overall about 10% of a sample of K-40 will decay to argon. In a particular rock sample it is found that there are 1.6×10^{22} atoms of K-40 and 8.4×10^{21} atoms of argon. The half-life of K-40 is 1.2×10^{9} yr. Estimate the time elapsed since the rock sample was formed.

**Answer/Explanation**

## Markscheme

a.

\(\left( {{}_{19}^{40}{\rm{K}} \to {}_{18}^{40}{\rm{Ar}} + {}_1^0{\beta ^ + } + v} \right)\)

\({{}_{18}^{40}{\rm{Ar}}}\);*v*; (do not accept \(\bar v\))

original number of K-40 atoms=(1.6×10^{22}+[8.4×10^{21}×10]=)1.0×10^{23};

decay constant=\(\frac{{\ln 2}}{{1.2 \times {{10}^9}}}\) * or* 5.8×10

^{-10}(yr

^{-1});

\(1.6 \times {10^{22}} = 1.0 \times {10^{23}}{\rm{ }}{{\rm{e}}^{ – {\rm{5.8}} \times {\rm{1}}{{\rm{0}}^{ – {\rm{10}}}}t}}\);

to give

*t*=3.2×10

^{9}(yr);

*Accept any alternative method that leads to the correct answer.*

*Award*

**[3 max]**ECF after incorrect value for N₀ (eg. use of 2.44×10^{22}to give*7.3×10*

^{22}yr).*Award*

**[2 max]**for approximate answers (eg. 3.0×10^{9}yr based on an estimate of*between two and three half-lives.)*

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 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.