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A2 Nuclear Physics 6.4.1(f)6.4.1(g)6.4.1(a)6.4.1(e)6.4.1(d)
~25 min Difficulty: 7/10
Prior knowledge Atomic structurecharge of protons/electronsbasic kinetic energyand the concept of density.
Problem structure
(a) 2, numerical, nuclear radius calculation. (b) 3, numerical, nuclear density calculation. (c) 4, numerical, conservation of energy and electrostatics. (d) 3, qualitative, properties of the strong nuclear force. (e) 3, numerical/qualitative, nuclear equations and isotopes.

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A researcher is investigating the properties of a specific isotope of Gold, $^{197}_{79}\text{Au}$. In a scattering experiment, alpha particles ($^{4}_{2}\text{He}$) are fired at a thin gold foil. (a) Calculate the radius of a $^{197}_{79}\text{Au}$ nucleus. Assume the constant $r_0 = 1.2 \times 10^{-15} \text{ m}$. [2] (b) Calculate the mean density of the gold nucleus. Assume the mass of a nucleon is $1.67 \times 10^{-27} \text{ kg}$. [3] (c) An alpha particle is fired directly towards the centre of a stationary gold nucleus. It has an initial kinetic energy of $8.0 \times 10^{-13} \text{ J}$ when it is very far from the nucleus. Calculate the distance of closest approach to the gold nucleus. Assume the gold nucleus remains stationary and treat both as point charges. Permittivity of free space $\epsilon_0 = 8.85 \times 10^{-12} \text{ F m}^{-1}$; Elementary charge $e = 1.60 \times 10^{-19} \text{ C}$. [4] (d) The strong nuclear force is responsible for holding the nucleons together. Describe how the magnitude and nature of the strong nuclear force change as the separation between two nucleons increases from $0.1 \text{ fm}$ to $5.0 \text{ fm}$. [3] (e) If the gold nucleus were to capture a thermal neutron to become $^{198}_{79}\text{Au}$, and this isotope then underwent $\beta^{-}$ decay, identify the proton number, nucleon number, and the name of the resulting daughter nucleus. [3]
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(a) Using the nuclear radius relation $R = r_0 A^{1/3}$: $A = 197$ $R = (1.2 \times 10^{-15}) \times (197)^{1/3}$ $R = 1.2 \times 10^{-15} \times 5.8186...$ $R = 6.98 \times 10^{-15} \text{ m}$ (or $6.98 \text{ fm}$) Final answer: $7.0 \times 10^{-15} \text{ m}$ (2 SF). (b) Density $\rho = \frac{\text{mass}}{\text{volume}}$. Mass of nucleus $M \approx A \times m_{nucleon} = 197 \times 1.67 \times 10^{-27} = 3.2899 \times 10^{-25} \text{ kg}$. Volume $V = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (r_0 A^{1/3})^3 = \frac{4}{3}\pi r_0^3 A$. $\rho = \frac{A \times m_{nucleon}}{\frac{4}{3}\pi r_0^3 A} = \frac{3 \times m_{nucleon}}{4 \pi r_0^3}$. $\rho = \frac{3 \times 1.67 \times 10^{-27}}{4 \pi (1.2 \times 10^{-15})^3}$ $\rho = \frac{5.01 \times 10^{-27}}{2.171 \times 10^{-44}}$ $\rho = 2.307 \times 10^{17} \text{ kg m}^{-3}$. Final answer: $2.3 \times 10^{17} \text{ kg m}^{-3}$ (2 SF). (c) At the distance of closest approach ($d$), initial $E_k = \text{Electric Potential Energy}$. $E_k = \frac{Q_1 Q_2}{4 \pi \epsilon_0 d}$ $Q_1 (\text{alpha}) = 2e = 2 \times 1.60 \times 10^{-19} = 3.20 \times 10^{-19} \text{ C}$. $Q_2 (\text{gold}) = 79e = 79 \times 1.60 \times 10^{-19} = 1.264 \times 10^{-17} \text{ C}$. $d = \frac{Q_1 Q_2}{4 \pi \epsilon_0 E_k}$ $d = \frac{(3.20 \times 10^{-19}) \times (1.264 \times 10^{-17})}{4 \pi \times 8.85 \times 10^{-12} \times 8.0 \times 10^{-13}}$ $d = \frac{4.0448 \times 10^{-36}}{8.897 \times 10^{-23}}$ $d = 4.546 \times 10^{-14} \text{ m}$. Final answer: $4.5 \times 10^{-14} \text{ m}$ (2 SF). (d) - From $0.1 \text{ fm}$ to approx $0.5 \text{ fm}$: The force is strongly repulsive. - From approx $0.5 \text{ fm}$ to approx $3.0 \text{ fm}$: The force is attractive (reaching a peak and then decreasing). - Beyond $3.0 \text{ fm}$: The force magnitude drops rapidly to zero (short-range). (e) 1. Neutron capture: $^{197}_{79}\text{Au} + ^{1}_{0}\text{n} \rightarrow ^{198}_{79}\text{Au}$. 2. $\beta^{-}$ decay: $^{198}_{79}\text{Au} \rightarrow ^{198}_{80}\text{X} + ^{0}_{-1}\text{e} + \bar{\nu}_e$. Proton number $Z = 80$. Nucleon number $A = 198$. Daughter nucleus: Mercury (Hg).

Marking guidance

(a) - M1: Correct substitution into $R = r_0 A^{1/3}$ [1]. - A1: Correct value $7.0 \times 10^{-15} \text{ m}$ (allow $6.98 \times 10^{-15} \text{ m}$) [1]. (b) - M1: Correct expression for mass ($A \times m_n$) and volume ($\frac{4}{3}\pi R^3$) [1]. - M1: Correct substitution of values (either using $R$ from (a) or the density formula independent of $A$) [1]. - A1: Correct value $2.3 \times 10^{17} \text{ kg m}^{-3}$ [1]. (c) - M1: Equating initial $E_k$ to electric potential energy [1]. - M1: Correct charges for alpha ($+2e$) and gold ($+79e$) [1]. - M1: Correct rearrangement for $d$ [1]. - A1: Correct value $4.5 \times 10^{-14} \text{ m}$ (allow $4.5-4.6 \times 10^{-14} \text{ m}$) [1]. (d) - B1: Repulsive below $0.5 \text{ fm}$ [1]. - B1: Attractive between $0.5 \text{ fm}$ and $3.0 \text{ fm}$ [1]. - B1: Negligible/zero beyond $3.0 \text{ fm}$ [1]. (e) - B1: Correct nucleon number $A = 198$ [1]. - B1: Correct proton number $Z = 80$ [1]. - B1: Correct identification of Mercury (Hg) from $Z=80$ (or stating $Z$ increases by 1) [1].