Question

The potential difference between two parallel conducting plates in vacuum is 160 V. An alpha particle with mass of 6.50×10-27 kg and charge of 3.20×10-19 C is released from rest near the positive plate. What is the kinetic energy of the alpha particle when it reaches the other plate? The distance between the plates is 24.0 cm.

Answer 1

Here,

Potential difference , Vd = 160 V

mass of particle , me = 6.5 *10^-27 Kg

charge , q = 3.2 *10^-19 C

Now , let the final kinetic energy is KE

as change in kinetic energy = -change in potential energy

final kinetic energy = charge * potential

final kinetic energy = 160 * 3.2 *10^-19 J

final kinetic energy = 5.12 *10^-17 J

the kinetic energy of alpha particle when it reaches the other plate is 5.12 *10^-17 J