Charged plates X and Y have a potential difference 1.5 V between them - AQA - A-Level Physics - Question 22 - 2019 - Paper 1

The change in potential energy (ΔPE) when moving from Y to X can be calculated using the formula:

where q is the charge of the particle and V is the voltage difference. The total energy gained by the particle moving from higher potential (Y) to lower potential (X) will correspond to the kinetic energy gained.

To find which particle gains 3.0 eV of kinetic energy, we need to look at the charges of the options:

1. Proton (q = +1 e):
   • Change in energy: $1.5 ext{ eV} × 1 = 1.5 ext{ eV}$
2. Positron (q = +1 e):
   • Change in energy: $1.5 ext{ eV} × 1 = 1.5 ext{ eV}$
3. Electron (q = -1 e):
   • Change in energy: $1.5 ext{ eV} × (-1) = -1.5 ext{ eV}$ (gains energy when moving towards Y)
4. Alpha Particle (q = +2 e):
   • Change in energy: $1.5 ext{ eV} × 2 = 3.0 ext{ eV}$

Among these, the alpha particle gains 3.0 eV when moving from Y to X.

Thus, the correct answer is: D alpha particle.