IB Physics HL E.3 Radioactive Decay Question Bank

[Maximum number: 10]

The binding energy of the stable nuclide ${ }_{54}^{131} \mathrm{Xe}$ is 1.105 GeV .

(a)

Outline what is meant by binding energy.

[ 1 ]

(b)

Calculate, in $\mathrm{GeV} \mathrm{c}^{-2}$ , the mass of a nucleus of ${ }_{54}^{131} \mathrm{Xe}$ .

[ 2 ]

Use of Binding energy = mass of separate nucleons - mass of nucleus
$\begin{aligned} \text { « } 1105=54 \times 938+77 \times 940-M c^{2} \text { » } M=122 \text { « } \mathrm{GeVc}^{-2} \text { » } \end{aligned}$

Do not allow BCA.

Award ECF from using binding energy with incorrect sign.

An answer that doesn't include binding energy scores [0].

(c)

$\quad{ }_{54}^{133} \mathrm{Xe}$ and ${ }_{54}^{131} \mathrm{Xe}$ are two isotopes of xenon.

[ 3 ]

(i)

Outline what is meant by isotopes.

[ 2 ]

(ii)

$\quad{ }_{54}^{133} \mathrm{Xe}$ is radioactive. Suggest how the binding energy per nucleon for ${ }_{54}^{131} \mathrm{Xe}$ compares with that of ${ }_{54}^{133} \mathrm{Xe}$ .

[ 1 ]

binding energy per nucleon is a measure of stability so ${ }_{54}^{131} \mathrm{Xe}$ has the greater binding
energy per nucleon

Correct mention of stability required to award the mark.

Marking guidance:

Allow opposite comment for Xe 133.

(d)

The graph shows the variation with time of the activity of a pure sample of ${ }_{54}^{133} \mathrm{Xe}$ .
$A / 10^{7} \mathrm{~Bq}$

[ 4 ]

(i)

Estimate the half-life of ${ }_{54}^{133} \mathrm{Xe}$ .

[ 1 ]

(ii)

Calculate the activity of the sample after 25 days as a fraction of the initial activity.

[ 3 ]

ALTERNATIVE 1

decay constant in range 0.128 to 0.133 «days ${ }^{-1}$ »
$\frac{A}{A_{0}}=e^{-\lambda \times 25}$

answer in range 0.036 to 0.040

ALTERNATIVE 2
$n=\frac{25}{5.3}=\ll 4.7 \text { half lives } \gg$ $\frac{A}{A_{0}}=0.5^{4.7}$

answer in range 0.036 to 0.040

Marking guidance:

Allow ECF from (c)(i)

[Maximum number: 2]

The diagram shows a simplified energy-balance model for the Earth surface-atmosphere system.

The following data are given:

\begin{aligned} \text { Average albedo of Earth } & =0.30 \\ \text { Average global temperature of the surface } & =288 \mathrm{~K} \\ \text { Average Earth-Sun distance } & =1.5 \times 10^{11} \mathrm{~m} \end{aligned}

The primary energy source of the Sun is the proton-proton (p-p) chain of fusion reactions. Four protons and two electrons produce a helium nucleus together with neutrinos and gamma photons. The overall reaction is:

4{ }_{1}^{1} \mathrm{p}+2{ }_{-1}^{0} \mathrm{e} \rightarrow{ }_{2}^{4} \mathrm{He}+2{ }_{0}^{0} v_{e}+4 \gamma

The mass of the helium nucleus is 4.001506 u . Calculate, in MeV , the energy released in the reaction.

ALTERNATIVE 1
$\begin{aligned} \Delta m=4 \times 1.007276+2 \times 0.000549-4.001506 «=0.028696 \mathrm{u} » \text { «0.028696×931.5 }=» 26.7 \mathrm{MeV} \text { ALTERNATIVE } 2 \begin{array}{l} \Delta m=4 \times 938+2 \times 0.511-4.001506 \times 931.5 \ll=25.6 \mathrm{MeV} \mathrm{c}^{-2} \gg 25.6 \mathrm{MeV} \end{array} \end{aligned}$

Marks can only be awarded from one Alternative

Unit is not required.

Particle masses in u and $M e V c^{-2}$ lead to differently rounded answers, working must be checked.

Award [1] max if electrons are omitted. This leads to ALT $125.7 \ll \mathrm{MeV} \gg$ , ALT $224.6 \ll \mathrm{MeV} \gg$

Award ECF if incorrect number of particles used e.g. 1 electron rather than 2.

[Maximum number: 10]

A short-range attractive force operates between nucleons.

(a)

State the name of this force.

[ 1 ]

(b)

State and explain one piece of evidence that suggests the presence of this force.

[ 2 ]

ALTERNATIVE 1

Protons repel each other so no nuclei could form

Unless an attractive force exists to overcome this repulsion

ALTERNATIVE 2

Deviation from Rutherford scattering / drop off of scattering at high energies $(\geq 28 \mathrm{MeV})$

This is due to strong nuclear force acting on alpha particle

ALTERNATIVE 3

The binding energy per nucleon is mostly constant above a certain point (A~60)

this is the consequence of an attractive <<short range>> force

ALTERNATIVE 4

Most nuclei have large binding energies

Indicating that a strong force keeps nucleons together

MP1 is required to award MP2

(c)

Radioactive potassium-40 $\left({ }_{19}^{40} \mathrm{~K}\right)$ undergoes $\beta^{-}$ decay to form a nuclide of calcium (Ca).

[ 2 ]

(i)

Write down the complete equation for this radioactive decay.

{ }_{19}^{40} \mathrm{~K} \rightarrow

[ 2 ]

$\begin{aligned} { }_{19}^{40} \mathrm{~K} \rightarrow{ }_{20}^{40} \mathrm{Ca} +{ }_{-1}^{0} \beta+\bar{\nu} \end{aligned}$

(d)

A sample of potassium has a mass of 5.00 g . The sample contains 0.012 % of potassium-40 by mass. The remainder of the sample is a stable nuclide of potassium. The half-life of potassium-40 is $1.28 \times 10^{9}$ years.

[ 5 ]

(i)

Show that the decay constant of potassium-40 is about $1.7 \times 10^{-17} \mathrm{~s}^{-1}$ .

[ 1 ]

$\lambda=\frac{\ln 2}{\left(1.28 \times 10^{9}\right)(365)(24)(3600)}=1.72 \times 10^{-17} \mathrm{~s}^{-1} \quad$

Must see full substitution OR answer to 3 (or more) significant
figures.

(ii)

The mass of one mole of potassium is $39.1 \mathrm{~g} \mathrm{~mol}^{-1}$ .

Determine the initial activity of the sample.

[ 2 ]

$\begin{aligned} N_{0}=\frac{5}{39.1} \times 0.00012 \times 6.02 \times 10^{23}=9.24 \times 10^{18} A_{o}=\lambda N_{o}=\left(1.72 \times 10^{-17}\right)\left(9.24 \times 10^{18}\right)=159 \mathrm{~Bq} \end{aligned}$

(iii)

Estimate the time taken for the potassium-40 to decrease to 98 % of the initial amount.

[ 2 ]

$\begin{aligned} \text { ALTERNATIVE 1 } 0.98=e^{-\lambda t} \text { In }(0.98)=-\left(1.72 \times 10^{-17}\right) \mathrm{t} \text { to give } \mathrm{t}=1.2 \times 10^{15} \mathrm{~s} \text { or } 3.7 \times 10^{7} \text { years } \text { ALTERNATIVE 2 } 0.98=\left(\frac{1}{2}\right)^{n} n=\frac{\ln (0.98)}{\ln (0.50)}=0.029 \text { to give } t=\left(1.28 \times 10^{9}\right)(0.029)=3.7 \times 10^{7} \text { years } \end{aligned}$

[Maximum number: 2]

The binding energy of the stable nuclide ${ }_{54}^{131} \mathrm{Xe}$ is 1.105 GeV .

$\quad{ }_{54}^{133} \mathrm{Xe}$ and ${ }_{54}^{131} \mathrm{Xe}$ are two isotopes of xenon.

Outline what is meant by isotopes.

[Maximum number: 1]

The binding energy of the stable nuclide ${ }_{54}^{131} \mathrm{Xe}$ is 1.105 GeV .

$\quad{ }_{54}^{133} \mathrm{Xe}$ and ${ }_{54}^{131} \mathrm{Xe}$ are two isotopes of xenon.

$\quad{ }_{54}^{133} \mathrm{Xe}$ is radioactive. Suggest how the binding energy per nucleon for ${ }_{54}^{131} \mathrm{Xe}$ compares with that of ${ }_{54}^{133} \mathrm{Xe}$ .

binding energy per nucleon is a measure of stability so ${ }_{54}^{131} \mathrm{Xe}$ has the greater binding
energy per nucleon

Correct mention of stability required to award the mark.

Marking guidance:

Allow opposite comment for Xe 133.

[Maximum number: 1]

A short-range attractive force operates between nucleons.

State the name of this force.

[Maximum number: 2]

Radioactive decay is said to be "random" and "spontaneous". Outline what is meant by each of these terms.

Random:
Spontaneous:

[Maximum number: 1]

Rhodium-106 $\left({ }_{45}^{106} \mathrm{Rh}\right)$ decays into palladium-106 $\left({ }_{46}^{106} \mathrm{Pd}\right)$ by beta minus $\left(\beta^{-}\right)$ decay. The diagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrow represents the $\beta^{-}$ decay.

Suggest why the $\beta^{-}$ decay is followed by the emission of a gamma ray photon.

[Maximum number: 2]

A short-range attractive force operates between nucleons.

Radioactive potassium-40 $\left({ }_{19}^{40} \mathrm{~K}\right)$ undergoes $\beta^{-}$ decay to form a nuclide of calcium (Ca).

Write down the complete equation for this radioactive decay.

{ }_{19}^{40} \mathrm{~K} \rightarrow

$\begin{aligned} { }_{19}^{40} \mathrm{~K} \rightarrow{ }_{20}^{40} \mathrm{Ca} +{ }_{-1}^{0} \beta+\bar{\nu} \end{aligned}$

[Maximum number: 1]

Magnesium-27 nuclei $\left({ }_{12}^{27} \mathrm{Mg}\right)$ decay by beta-minus $\left(\beta^{-}\right)$ decay to form nuclei of aluminium-27 (Al).

A Magnesium-27 nucleus can decay by one of two routes:

Route 1: 70\% of the beta particles are emitted with a maximum kinetic energy of 1.76656 MeV , accompanied by a gamma photon of energy 0.84376 MeV .

Route 2: 30 % of the beta particles have a maximum kinetic energy of 1.59587 MeV with a gamma photon of energy 1.01445 MeV .

The final state of the aluminium-27 nucleus is the same for both routes.

Calculate the difference between the magnitudes of the total energy transfers in parts (a) and (b).
(ii) Explain how the difference in part (b)(ii) arises.

$\text { Difference }=2.6175-(1.76656+0.84376)=2.6175-2.61032=0.007195 \text { «MeV» }$

OR

Difference =2.6175-(1.59587+1.01445)=2.61032=0.007195 «MeV»

9.

b

iii

Another particle/«anti» neutrino is emitted «that accounts for this mass / energy»

[Maximum number: 2]

This question is about the use of radioactive isotopes.

Patients suffering from prostate cancer can experience metastatic bones. As part of the cancer treatment, patients are injected with an isotope of strontium which is readily taken up by the bones.

The table shows the half-life and emission information for strontium-89 (Sr-89) and strontium-90 (Sr-90).

Beta particles have a quality factor of 1 .

State, in terms of range, two reasons why Sr -89 is the more suitable choice of isotope to be used in the treatment of bone cancers.

[Maximum number: 1]

The binding energy of the stable nuclide ${ }_{54}^{131} \mathrm{Xe}$ is 1.105 GeV .

The graph shows the variation with time of the activity of a pure sample of ${ }_{54}^{133} \mathrm{Xe}$ .
$A / 10^{7} \mathrm{~Bq}$

Estimate the half-life of ${ }_{54}^{133} \mathrm{Xe}$ .

[Maximum number: 3]

The radioactive nuclide beryllium-10 (Be-10) undergoes beta minus ( $\beta$ -) decay to form a stable boron (B) nuclide.

The initial number of nuclei in a pure sample of beryllium-10 is $\mathrm{N}_{0}$ . The graph shows how the number of remaining beryllium nuclei in the sample varies with time.

After $4.3 \times 10^{6}$ years,

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

Show that the half-life of beryllium-10 is $1.4 \times 10^{6}$ years.

ALTERNATIVE 1

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

therefore 3 half-lives have elapsed
$t_{\frac{1}{2}}=\frac{4.3\times10^{6}}{3}=1.43\times10^{6}\ \mathrm{y}\approx1.4\times10^{6}\ \mathrm{y}$

ALTERNATIVE 2

fraction of $\mathrm{Be}=\frac{1}{8}, 12.5\%$ , or 0.125
$\frac{1}{8}=e^{-\lambda(4.3\times10^{6})}\text{, leading to }\lambda=4.836\times10^{-7}\ \mathrm{y}^{-1}$ $\frac{\ln 2}{\lambda}=1.43\times10^{6}\ \mathrm{y}$

[Maximum number: 1]

The graph shows the variation with time t of the activity A of a sample of protactinium-234.

Suggest why the activity approaches a non-zero constant value.

[Maximum number: 2]

A short-range attractive force operates between nucleons.

State and explain one piece of evidence that suggests the presence of this force.

ALTERNATIVE 1

Protons repel each other so no nuclei could form

Unless an attractive force exists to overcome this repulsion

ALTERNATIVE 2

Deviation from Rutherford scattering / drop off of scattering at high energies $(\geq 28 \mathrm{MeV})$

This is due to strong nuclear force acting on alpha particle

ALTERNATIVE 3

The binding energy per nucleon is mostly constant above a certain point (A~60)

this is the consequence of an attractive <<short range>> force

ALTERNATIVE 4

Most nuclei have large binding energies

Indicating that a strong force keeps nucleons together

MP1 is required to award MP2

[Maximum number: 3]

This question is about nuclear reactions.

The daughter nuclei of U-235 undergo radioactive decay until eventually a stable isotope of lead is reached.

Explain why the nuclei of U-235 are unstable whereas the nuclei of the lead are stable.

[Maximum number: 2]

Rhodium-106 $\left({ }_{45}^{106} \mathrm{Rh}\right)$ decays into palladium-106 $\left({ }_{46}^{106} \mathrm{Pd}\right)$ by beta minus $\left(\beta^{-}\right)$ decay. The diagram shows some of the nuclear energy levels of rhodium-106 and palladium-106. The arrow represents the $\beta^{-}$ decay.

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

Photon energy

$\begin{aligned} E&=0.48\times10^6\times1.6\times10^{-19}=7.68\times10^{-14}\ \mathrm{J}\\ \lambda&=\frac{hc}{E}=\frac{6.63\times10^{-34}\times3\times10^8}{7.68\times10^{-14}}=2.6\times10^{-12}\ \mathrm{m} \end{aligned}$

[Maximum number: 2]

This question is in two parts.

The alpha particles and gamma rays produced in radioactive decay have discrete energy spectra. This suggests that nuclei also possess discrete energy levels. However, beta particles produced in radioactive decay have continuous energy spectra. Describe how the existence of the antineutrino accounts for the continuous nature of beta spectra.

[Maximum number: 3]

This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.

Part 1 Kinematics and gravitation

A ball is released near the surface of the Moon at time t=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time t of the displacement s of the ball from the point of release.

In a pure sample of lead-212 at one instant, $8.0 \times 10^{-3} \mathrm{~kg}$ of the lead-212 is present. Calculate the mass of lead-212 that remains after a period of 35 hours.

use of $\lambda$ from (h)(i);
correct substitution into $N=N_{0} \mathrm{e}^{-\lambda t}$ ;
8.0 to $8.3 \times 10^{-4} \mathrm{~kg}$ ;

[Maximum number: 2]

State what is meant by decay constant.

[Maximum number: 2]

A short-range attractive force operates between nucleons.

A sample of potassium has a mass of 5.00 g . The sample contains 0.012 % of potassium-40 by mass. The remainder of the sample is a stable nuclide of potassium. The half-life of potassium-40 is $1.28 \times 10^{9}$ years.

The mass of one mole of potassium is $39.1 \mathrm{~g} \mathrm{~mol}^{-1}$ .

Determine the initial activity of the sample.

$\begin{aligned} N_{0}=\frac{5}{39.1} \times 0.00012 \times 6.02 \times 10^{23}=9.24 \times 10^{18} A_{o}=\lambda N_{o}=\left(1.72 \times 10^{-17}\right)\left(9.24 \times 10^{18}\right)=159 \mathrm{~Bq} \end{aligned}$

[Maximum number: 1]

This question is in two parts. Part 1 is about kinematics and gravitation. Part 2 is about radioactivity.

Part 1 Kinematics and gravitation

A ball is released near the surface of the Moon at time t=0. The point of release is on a straight line between the centre of Earth and the centre of the Moon. The graph below shows the variation with time t of the displacement s of the ball from the point of release.

Calculate, in hour ${ }^{-1}$ , the decay constant of lead-212.

$\left.\quad \lambda=\frac{\ln 2}{\mathrm{~T}_{\frac{1}{2}}}=\frac{0.693}{10.6}=\right) 6.5 \times 10^{-2}$ hour $^{-1}$

IB Physics HL E 3 Radioactive Decay Questions

Exam points

Covered syllabus points

Covered syllabus points

Question 1

Question (a)

Question (b)

Question (c)

Question (i)

Question (ii)

Question (d)

Question (i)

Question (ii)

Question 2

Question 3

Question (a)

Question (b)

Question (c)

Question (i)

Question (d)

Question (i)

Question (ii)

Question (iii)

Question 4

Question 5

Question 6

Question 7

Question 8

Question 9

Question 10

Question 11

Question 12

Question 13

Question 14

Question 15

Question 16

Question 17

Question 18

Question 19

Question 20

Question 21

Question 22