Question

An alpha particle of charge +2e and mass 6.64 x 1027 kg is moving in the +x direction at 2.15x106 m/s. It enters a region where the electric field is 1465 N/C in the -x direction. How far will the alpha particle travel before it comes to a stop? Submit Answer Tries 0/5 How much time will elapse from when the alpha particle enters the electric field to when it stops? Submit Answer Tries 0/5

Answer 1

q = charge of alpha particle = + 2e = 2 times 1.6 times 10^-19 C = 3.2 times 10^-19 C E = 1465 M/C force an alpha particle = qE/m = 3.2 times 10^-19 times 1465/6.64 times 10^-27 kg = 7.06 times 10^10 m/s^2 in- x direction. a) u = initial velocity = 2.15 times 10^6 m/s a = acceleration = -7.06 times 10^10 m/s^2 v = final velocity = 0 s = distance travelled before stopping from v^2- u^2 = 2as 0- (2.15 times 10^6)^2 = -2 times 7.06 times 10^10 S S = (2.15 times 10^6)^2/2 times 7.06 times 10^10 = 4.6225 times 10^12/14.12 times 10^10 = 32.73 m b) from v = u + at 0 = 2.15 times 10^6- 7.06 times 10^10 t t = 2.15 times 10^6/7.06 times 10^10 = 3.04 times 10^-5 seconds