AQA A-Level Physics: Nuclear Physics

The graph of binding energy per nucleon against nucleon number is one of the most important diagrams in nuclear physics. It rises steeply for the lightest nuclei, reaches a broad maximum at iron-56 ($\approx 8.8 \ \text{MeV per nucleon}$≈8.8 MeV per nucleon), and then falls slowly for the heaviest nuclei.

Explain how the shape of this curve accounts for the fact that energy is released in both nuclear fission of a heavy nucleus and nuclear fusion of two light nuclei. Your answer should make clear what is meant by binding energy per nucleon, why iron-56 occupies the peak, and why each process moves nuclei in the direction that releases energy.

(6 marks)

Cobalt-60 is used as a gamma source in industrial radiography. A freshly prepared sealed source contains $1.20 \times 10^{18}$1.20×1018 undecayed cobalt-60 nuclei. The half-life of cobalt-60 is 5.27 years.

Quantity	Value
Initial number of nuclei, $N_0$N0​	$1.20 \times 10^{18}$1.20×1018
Half-life, $t_{1/2}$t1/2​	$5.27 \ \text{years}$5.27 years
1 year	$3.16 \times 10^{7} \ \text{s}$3.16×107 s

(a) Calculate the decay constant of cobalt-60, in $\text{s}^{-1}$s−1. (2 marks)

(b) Calculate the initial activity of the source, in Bq. (2 marks)

(c) Calculate the number of cobalt-60 nuclei remaining after 10.0 years. (2 marks)

A student measures the count rate from a small radioactive source using a Geiger-Müller tube and a counter. Before introducing the source, the student records a steady background count rate of 20 counts per minute. The measured count rate from the source plus background is then recorded over one hour.

Time / min	0	15	30	45	60
Measured count rate / min⁻¹	660	340	180	100	60

(a) Explain why the background count rate must be subtracted from each reading before the half-life is found, and produce a corrected count rate for each time. (2 marks)

(b) Use your corrected values to determine the half-life of the source. (1 mark)

(c) Hence calculate the decay constant of the source, in $\text{s}^{-1}$s−1. (2 marks)

One of the fusion reactions being investigated for future power generation is the fusion of deuterium and tritium:

${}^{2}_{1}\text{H} + {}^{3}_{1}\text{H} \rightarrow {}^{4}_{2}\text{He} + {}^{1}_{0}\text{n}$12​H+13​H→24​He+01​n

The masses of the particles involved are given below.

Particle	Mass / u
$^{2}_{1}\text{H}$12​H (deuterium)	2.01410
$^{3}_{1}\text{H}$13​H (tritium)	3.01605
$^{4}_{2}\text{He}$24​He	4.00260
$^{1}_{0}\text{n}$01​n (neutron)	1.00867

(a) Calculate the energy released in this reaction. Give your answer in MeV. Take $1 \ \text{u} = 931.5 \ \text{MeV}$1 u=931.5 MeV. (4 marks)

(b) A single fission of uranium-235 releases about $200 \ \text{MeV}$200 MeV. By comparing the energy released per nucleon in the two processes, comment on why fusion is often described as a more efficient energy source than fission. (1 mark)

In a scattering experiment, an alpha particle is fired directly at the centre of a stationary gold nucleus. The alpha particle has an initial kinetic energy of $6.0 \ \text{MeV}$6.0 MeV and is brought momentarily to rest by the repulsive electrostatic force before being reflected straight back.

Quantity	Value
Initial kinetic energy of alpha particle	$6.0 \ \text{MeV}$6.0 MeV
Charge on alpha particle	$+2e$+2e
Proton number of gold	$79$79
$\frac{1}{4\pi\varepsilon_0}$4πε0​1​	$8.99 \times 10^{9} \ \text{N m}^2\,\text{C}^{-2}$8.99×109 N m2C−2
$e$e	$1.60 \times 10^{-19} \ \text{C}$1.60×10−19 C

Calculate the distance of closest approach of the alpha particle to the centre of the gold nucleus. (4 marks)

In the alpha-particle scattering experiment, a beam of alpha particles was directed at a very thin gold foil and the angles through which the particles were scattered were observed.

State three observations from this experiment, and for each observation give the conclusion it allowed about the structure of the atom. (3 marks)