Singly charged uranium-238 ions are accelerated through a potential difference of 2.90 kV and enter a uniform magnetic field of magnitude 1.29 T directed perpendicular to their velocities. (a) Determine the radius of their circular path. (Give your answer to three significant figures.)

Answer: 0.091 m

Explanation:

r = 1/B * √ (2mV/e), where

r = radius of their circular path

B = magnitude of magnetic field = 1.29 T

m = mass of Uranium - 238 ion = 238 * amu = 238 * 1.6*10^-27 kg

V = potential difference = 2.9 kV

e = charge of the Uranium - 238 ion = 1.6*10^-19 C

r = 1/1.29 * √[ (2 * 238 * 1.6*10^-27 * 2900) / 1.6*10^-19]

r = 1/1.29 * √ (2.21*10^-21 / 1.6*10^-19)

r = 1/1.29 * √0.0138

r = 1/1.29 * 0.117

r = 0.091 m

Therefore, the radius of their circular path is 0.091 m