Uranium-238 atoms have one electron removed to become singly charged U-238 ions.

Question

Uranium-238 atoms have one electron removed to become singly charged U-238 ions. The ions are accelerated through a potential difference of 2000 V and enter a uniform magnetic field of magnitude 1.20 T directed perpendicular to the ion's velocity.

(a) Determine the radius of the circular path of the ion.

(b) repeat this calculation for uranium-235 ions (singly charged).

(c) How does the ratio of these path radii depend on the accelerating voltage?

(d) How does the ratio of these path radii depend on the magnitude of the magnetic field?

Explanation / Answer

A. using energy conservation to find the velocity

dPE = dKE

qV = 0.5*m*v^2

v = sqrt(2*qV/m)

v = sqrt(2*1.6*10^-19*2000/(238*1.67*10^-27)) = 40127.6 m/sec

using force balance for circular motion

Fg = Fm

mv^2/r = qvB

r = mv/qB

r = 238*1.67*10^-27*40127.6/(1.6*10^-19*1.2)

r = 0.083068 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 = 0.083068*sqrt(235/238)

r235 = 0.082542 m

C. from above formula

r = sqrt(2*V*m/qB^2)

radii of path is proportional to square root of V for the same ion.

D.

radii of path is inverselyproportional to B for the same ion

if ions are different then ratio of these path radii does not depend on V and B.

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