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 in a direction radially away from the center of the gold nucleus, indicating the direction of the rate of change in momentum.

Alpha particle number 2 may be deflected downwards with a scattering angle of 90°. This is because it is positioned to experience a stronger repulsive force from the nucleus as it approaches closely, causing a significant change in its trajectory.

To find the speed of alpha particle 4 when it is at a large distance from the nucleus, we use the conservation of energy. The potential energy when it is at rest near the nucleus will equal the kinetic energy when it is at a large distance:

$KE = PE \implies \frac{1}{2} mv^2 = \frac{Z_1 Z_2 e^2}{4\pi \epsilon_0 r}$

Substituting the values:

• Mass $m = 6.8 \times 10^{-27} \text{ kg}$
• Potential energy can be calculated based on the distances and charges involved.

The final calculated speed can be determined from this equation.

The nuclear radius can be calculated using the empirical formula for the radius of a nucleus:

$R = r_0 A^{1/3}$

Where $r_0 = 1.2 \times 10^{-15} m$ and $A$ is the mass number. For $^{197}Ag$, the mass number is 197. Thus, we can find:

$R_{197Ag} = 1.2 \times 10^{-15} (197)^{1/3}$

Calculating this gives the nuclear radius for $^{197}Ag$.

One conclusion that can be deduced is that nucleons are incompressible and have a constant separation between them, as the density of nuclei remains nearly constant across different elements. This suggests that nucleons behave similarly despite differing atomic structures.