Nuclear Physics problem-solving question

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]

(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 $0.5 \text{ fm}$:...

(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...

(a) Hint 1: Use the formula relating radius to nucleon number. Hint 2: $A$ is the nucleon number (the top number in the notation). Hint 3: Ensure you use the cube root, not the square root. (b) Hint 1: Density is mass divided by volume. Treat the nucleus as a sphere. Hint 2: The mass of the nucleus is approximately the number of nucleons times the mass of one nucleon. Hint 3: Notice how the $A$ terms might cancel out if you substitute the radius formula into the density formula. (c) Hint 1: Energy is conserved. The kinetic energy is converted entirely into electric potential energy at the point where the particle stops. Hint 2: The charge of an alpha particle is $+2e$ and gold is $+79e$....