Figure 11 shows alpha particles all travelling in the same direction at the same speed - AQA - A-Level Physics - Question 5 - 2020 - Paper 2

Draw an arrow at position X that points radially away from the center of the gold nucleus.

The alpha particle number 2 may be deflected downwards with a scattering angle of 90°. This is because it is closer to the center of the gold nucleus and, therefore, experiences a stronger electromagnetic repulsion, resulting in a greater deflection.

Using conservation of energy, we can equate the potential energy (PE) at a small distance to the kinetic energy (KE) at a large distance:
$KE = -PE$

The potential energy at the distance $5.5 \times 10^{-14}$ m is given by
$PE = \frac{k q_1 q_2}{r}$,
where $k$ is Coulomb's constant, $q_1$ and $q_2$ are the charges of the particles, and $r$ is the distance. Plugging in the values: $KE = \frac{(2.6 \times 10^{-19})(1.6 \times 10^{-19})}{5.5 \times 10^{-14}}$ Calculating gives the speed as ( v = \sqrt{\frac{2 KE}{m}} ).

The nuclear radius can be calculated using the formula: $R = R_0 A^{1/3}$ where $R_0 = 6.98 \times 10^{-15} \text{ m}$ and $A$ is the mass number. Thus, for $^{197}Ag$, the calculation would follow the nuclear radius formula, substituting appropriate values based on $^{197}Ag$'s mass number.

All nucleons are incompressible. This implies that nucleons behave as if they possess similar masses and occupy similar volumes, maintaining a uniform density across different nuclei.