Question

A doubly charged helium atom, whose mass is 6.6×10−27kg, is accelerated by a voltage of 3800 V

What will be its radius of curvature in a uniform 0.300-T field?

What is its period of revolution?

Answer 1

given
m = 6.6*10^-27 kg
delta_V = 3800 V
B = 0.3 T
q = 2*e

= 2*1.6*10^-19 C

= 3.2*10^-19 C

let v is the speed of the helium atom after accelerared.

use Work-energy theorem,

Work donw = gain in kinetic energy

W = (1/2)*m*v^2 - 0

q*delta_V = (1/2)*m*v^2

==> v = sqrt(2*q*delta_V/m)

= sqrt(2*3.2*10^-19*3800/(6.6*10^-27))

= 6.07*10^5 m/s

radius of curvature, r = m*v/(B*q)

= 6.6*10^-27*6.07*10^5/(0.3*3.2*10^-19)

= 0.0417 m (or) 4.17 cm <<<<<<<-----------Answer

b) Time period, T = 2*pi*r/v

= 2*pi*0.0417/(6.07*10^5)

= 4.32*10^-7 s <<<<<<<-----------Answer