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

To determine which particle gains 3.0 eV of kinetic energy moving from plate Y (+1.5 V) to plate X (0 V), we start with the relationship between potential difference and kinetic energy:

The change in potential energy (which is equal to the kinetic energy gained by the particle) can be calculated using the formula:

$KE = q imes V$

where:

• $KE$ is the kinetic energy gained,
• $q$ is the charge of the particle, and
• $V$ is the potential difference.

Since the potential difference is 1.5 V, a particle would gain:

• A proton (charge e): For this particle, the kinetic energy gained would be:

$KE_{proton} = e imes 1.5 ext{ V} = 1.5 eV$

• A positron (charge e): Similar to the proton,

$KE_{positron} = e imes 1.5 ext{ V} = 1.5 eV$

• An electron (charge -e): The kinetic energy gained when moving towards a lower potential is:

$KE_{electron} = -e imes 1.5 ext{ V} = -1.5 eV$

• An alpha particle (charge 2e): Therefore, the kinetic energy for an alpha particle would be:

$KE_{alpha} = 2e imes 1.5 ext{ V} = 3.0 eV$

Thus, the particle that gains 3.0 eV of kinetic energy when moving from Y to X is the alpha particle (D).