Question

A mass spectrometer is used to examine the isotopes of uranium. Ions in the beam emerge from the velocity selector at a speed of 2.92 ✕ 10⁵ m/s and enter a uniform magnetic field of 0.605 T directed perpendicularly to the velocity of the ions. What is the distance between the impact points formed on the photographic plate by singly charged ions of ²³⁵U and ²³⁸U?
cm

Answer 1

given
v = 2.92*10^5 m/s
B = 0.605 T

let m1 = 235*1.66*10^-27 = 3.901*10^-25 kg

m2 = 238*1.66*10^-27 = 3.9508*10^-25 kg

we know, radius of path followed by a charge in uniform magnetic field,

r = m*v/(B*q)

r1 = m1*v/(B*q)

= 3.901*10^-25*2.92*10^5/(0.605*1.6*10^-19)

= 1.1767 m

r2 = m2*v/(B*q)

= 3.9508*10^-25*2.92*10^5/(0.605*1.6*10^-19)

= 1.1918 m

the distance between the impact points formed on the photographic plate,

d = 2*(r2 - r1)

= 2*(1.1918 - 1.1767)

= 0.0302 m

= 3.02 cm <<<<<<<<<<<---------Answer