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A-Level Physics – Nuclear and Thermal

Radioactivity, binding energy, ideal gases, kinetic theory.

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12 questions  ·  38 marks  ·  answers included  ·  A-Level Physics (Edexcel/AQA/OCR)
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This worksheet covers:

  • Radioactivity
  • Nuclear binding energy
  • Ideal gases
  • Kinetic theory

Use $u=1.661\times10^{-27}\,\text{kg}$, $1\,\text{u}=931.5\,\text{MeV}/c^2$, $k=1.38\times10^{-23}\,\text{J K}^{-1}$ and $R=8.31\,\text{J K}^{-1}\text{mol}^{-1}$. A calculator is required throughout.

Questions

Section A – Radioactivity

1

(a) Define what is meant by the half-life of a radioactive isotope.(b) State two features of radioactive decay that show it is a random process.

Show answer
(a) Half-life is the time taken for the number of undecayed nuclei (or the activity) of a radioactive sample to fall to half of its original value.(b) It is impossible to predict when any individual nucleus will decay; and the decay rate is unaffected by external conditions such as temperature or pressure (any two valid features).
[3]
2

A sample of cobalt-60 initially contains $8.0\times10^{10}$ undecayed nuclei. Cobalt-60 has a half-life of $5.27$ years. Calculate the number of undecayed nuclei remaining after $15.81$ years.

Show answer
$15.81$ years is $3$ half-lives ($15.81/5.27=3$). $N=N_0\left(\frac{1}{2}\right)^3=\dfrac{8.0\times10^{10}}{8}=1.0\times10^{10}$ nuclei.
[3]
3

Calculate the decay constant $\lambda$ of cobalt-60, in $\text{year}^{-1}$, given its half-life of $5.27$ years.

Show answer
$\lambda=\dfrac{\ln2}{T_{1/2}}=\dfrac{0.693}{5.27}=0.132\,\text{year}^{-1}$.
[2]
4

A sample of iodine-131 (half-life $8.02$ days) has an initial activity of $640\,\text{Bq}$. The activity falls to $40\,\text{Bq}$.(a) Calculate the number of half-lives that have elapsed.(b) Calculate the time taken, in days.

Show answer
(a) $\dfrac{640}{40}=16=2^4$, so $4$ half-lives have elapsed.(b) $t=4\times8.02=32.1\,\text{days}$.
[4]

Section B – Nuclear Binding Energy

5

Explain what is meant by the binding energy of a nucleus, and explain why energy is released when a large nucleus such as uranium-235 undergoes fission.

Show answer
Binding energy is the energy required to completely separate a nucleus into its individual, unbound protons and neutrons (equivalently, the energy equivalent of the mass defect, the difference between the mass of the separate nucleons and the mass of the nucleus). Energy is released in fission because the binding energy per nucleon of the two smaller product nuclei is greater than that of the original uranium-235 nucleus, so the total mass decreases and the mass difference is released as energy.
[3]
6

A helium-4 nucleus is formed from 2 protons (mass $1.007276\,\text{u}$ each) and 2 neutrons (mass $1.008665\,\text{u}$ each). The mass of the helium-4 nucleus is $4.001506\,\text{u}$.(a) Calculate the mass defect of the helium-4 nucleus, in u.(b) Calculate the total binding energy of the nucleus, in MeV, using $1\,\text{u}=931.5\,\text{MeV}/c^2$.

Show answer
(a) $\Delta m=2(1.007276)+2(1.008665)-4.001506=0.030376\,\text{u}$.(b) $E=\Delta m\times931.5=0.030376\times931.5=28.3\,\text{MeV}$.
[4]
7

Using your answer to the previous question, calculate the binding energy per nucleon of helium-4, in MeV, and comment on its size compared with iron-56, which has a binding energy per nucleon of about $8.8\,\text{MeV}$.

Show answer
$\dfrac{28.3}{4}=7.07\,\text{MeV per nucleon}$. This is noticeably lower than the value for iron-56, consistent with iron-56 being close to the peak of the binding-energy-per-nucleon curve, where nuclei are most stable.
[3]
8

With reference to the graph of binding energy per nucleon against nucleon number, explain why energy is released in both nuclear fusion (of light nuclei) and nuclear fission (of heavy nuclei).

Show answer
The binding energy per nucleon is low for both very light nuclei and very heavy nuclei, and reaches a maximum around iron-56. In fusion, light nuclei combine to form a nucleus with a higher binding energy per nucleon (closer to the peak). In fission, a heavy nucleus splits into two nuclei with a higher binding energy per nucleon than the original. In both cases the total binding energy increases, meaning the total mass decreases, and this mass difference is released as energy.
[3]

Section C – Thermal Physics and Ideal Gases

9

State three assumptions of the kinetic theory of gases used to derive the ideal gas equation for a model (ideal) gas.

Show answer
Any three of: the gas consists of a very large number of identical molecules in random motion; the volume of the molecules is negligible compared with the volume of the container; intermolecular forces are negligible except during collisions; collisions between molecules (and with the container walls) are perfectly elastic; the duration of a collision is negligible compared with the time between collisions.
[3]
10

A cylinder contains $0.50\,\text{mol}$ of an ideal gas at a temperature of $300\,\text{K}$ in a volume of $0.020\,\text{m}^3$. Calculate the pressure of the gas.

Show answer
Using $pV=nRT$: $p=\dfrac{nRT}{V}=\dfrac{0.50\times8.31\times300}{0.020}=6.23\times10^4\,\text{Pa}$.
[3]
11

For an ideal gas at a temperature of $300\,\text{K}$:(a) Calculate the mean translational kinetic energy of a gas molecule.(b) Calculate the root-mean-square speed of a nitrogen molecule of mass $4.65\times10^{-26}\,\text{kg}$ at this temperature.

Show answer
(a) $\overline{E_K}=\frac{3}{2}kT=\frac{3}{2}(1.38\times10^{-23})(300)=6.21\times10^{-21}\,\text{J}$.(b) $c_{\text{rms}}=\sqrt{\dfrac{3kT}{m}}=\sqrt{\dfrac{3(1.38\times10^{-23})(300)}{4.65\times10^{-26}}}=517\,\text{m s}^{-1}$.
[4]
12

A fixed mass of ideal gas has an initial pressure of $1.0\times10^5\,\text{Pa}$, volume $2.0\,\text{dm}^3$ and temperature $290\,\text{K}$. The gas is compressed to a volume of $1.2\,\text{dm}^3$ and heated to $350\,\text{K}$. Calculate the new pressure of the gas.

Show answer
Using $\dfrac{p_1V_1}{T_1}=\dfrac{p_2V_2}{T_2}$: $p_2=\dfrac{p_1V_1T_2}{T_1V_2}=\dfrac{(1.0\times10^5)(2.0)(350)}{(290)(1.2)}=2.0\times10^5\,\text{Pa}$.
[3]

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