Question

An alpha particle (q = +2e, m = 4.00 u) travels in a circular path of radius 3.45 cm in a uniform magnetic field with B = 1.67 T. Calculate (a) its speed, (b) its period of revolution, (c) its kinetic energy, and (d) the potential difference through which it would have to be accelerated to achieve this energy.

Answer 1

Step-by-step explanation:

mass of alpha particle, m = 4 u = 4 x 1.67 x 10^-27 kg

charge, q = 2 e = 3.2 x 10^-19 C

radius, r = 3.45 cm

Magnetic field, B = 1.67 T

(a) let v be the speed

`v = (Bqr)/(m)`

`v = (1.67* 3.2* 10^-19* 0.0345)/(4* 1.67* 10^(-27))`

v = 2760000 m/s

(b) Let T be the period of revolution

`T = (2\pi r)/(v)`

`T = (2* 3.14* 0.0345 )/(2760000)`

T = 7.85 x 10^-8 seconds

(c) Kinetic energy, K =0.5 x mv²

K = 0.5 x 4 x 1.67 x 10^-27 x 2760000 x 2760000

K = 6.36 x 10^-15 J

(d) kinetic energy, K = e V

where, V is the potential difference

6.36 x 10^-15 = 1.6 x 10^-19 x V

V = 39754.35 V