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

An arrow should be drawn at position X pointing radially away from the center of the gold nucleus. This represents the direction in which the momentum change occurs due to the upward deflection of the alpha particle.

Alpha particle number 2 can be suggested for downward deflection at a scattering angle of 90°. This is justified because it is positioned closer to the nucleus, thus experiencing a stronger repulsive force compared to alpha particles further away, resulting in a larger deflection.

Utilizing the conservation of energy, the loss of kinetic energy when the alpha particle comes to rest can be equated to the potential energy at that distance:

ext{Potential Energy (PE)} = rac{(2 imes 6.8 imes 10^{-27} ext{kg})^2}{(5.5 imes 10^{-14} ext{m})}

At a large distance, the kinetic energy (KE) can be expressed as:

KE = rac{1}{2} mv^2

Setting KE equal to PE allows us to solve for the speed, $v$.

The nuclear radius can be calculated using the empirical formula:

$R = R_0 A^{1/3}$

Where $R_0 = 6.98 imes 10^{-15}m$ is the radius constant and $A$ is the mass number. For $^{197}Ag$, with $A = 197$:

$R_{197Ag} = R_0 (197)^{1/3}$

Calculating this gives the radius of the $^{197}Ag$ nucleus.

One conclusion that can be deduced from this fact is that nucleons (protons and neutrons) are incompressible and occupy roughly the same volume, leading to a consistent density regardless of the nucleus size.