Question

Singly charged uranium-238 ions are accelerated through a potential difference of 3.10 kV and enter a uniform magnetic field of magnitude 1.38 T directed perpendicular to their velocities. (a) Determine the radius of their circular path. (Give your answer to three significant figures.) (b) Repeat this calculation for uranium-235 ions. (Give your answer to three significant figures.)

Answer 1

A. using energy conservation to find the velocity

dPE = dKE

q*dV = 0.5*m*v^2

v = sqrt(2*q*dV/m)

dV = 3.10 kV = 3100 V

q = +e (since singly charged ion)

v = sqrt(2*1.6*10^-19*3100/(238*1.67*10^-27)) = 49958.47 m/sec

using force balance for circular motion

Fg = Fm

mv^2/r = qvB

r = mv/qB

r = 238*1.67*10^-27*49958.47/(1.6*10^-19*1.38)

r = 0.0899298 m = 8.99*10^-2 m

B. r = mv/qB

v = sqrt(2qV/m)

from both equation

r = sqrt(2Vm/qB^2)

so r is proportional to m

r2/r1 = sqrt(m2/m1)

r235 = r238*sqrt(m235/m238)

r235 = 0.0899298*sqrt(235/238)

r235 = 0.0893612 m = 8.93*10^-2 m

Please Upvote.