Question

An alpha particle of charge +2e and mass 6.64 × 10-27 kg is moving in the +x direction at 1.79×106 m/s. It enters a region where the electric field is 1591 N/C in the -x direction.

a. How far will the alpha particle travel before it comes to a stop? 2.33 * 10^-5 m Calculate the acceleration, then use the appropriate kinematic equation.

b. How much time will elapse from when the alpha particle enters the electric field to when it stops?

Answer 1

a)the force due to electric field on the alpha particle would be:

F = q E ; Also F = ma

ma = q E => a = q E/m

a = 2 x 1.6 x 10^-19 x 1591/(6.64 x 10^-27) = 7.67 x 10^10 m/s^2 (this is actually retardation)

We know from eqn of motion that

v^2 = u^2 + 2 a S

v = 0 ; u = 1.79 x 10^6 m/s ;

0 = (1.79 x 10^6)^2 - 2 x 7.67 x 10^10 x S

S = (1.79 x 10^6)^2/(2 x 7.67 x 10^10) = 20.89 m

Hence, S = 20.89 m

b)v = u + at

v = 0 ; u = 1.79 x 10^6 m/s ;

t = u/a = 1.79 x 10^6/(7.67 x 10^10) = 2.33 x 10^-5 s

Hence, t = 2.33 x 10^-5 s