Two isotoprs of uranium U235(3.9x10^-25 kg) , U238(3.95x10^-25kg) into spectrometer (1.05x10^5 m/s). Each isotope is singly ionized and B field(0.75T). What is the distance between the two isotopes?

1 Answer
Apr 2, 2016

0.875cm

Explanation:

Given that two isotopes of Uranium "U"235(3.9xx10^-25 kg) and "U"238(3.95xx10^-25kg) into spectrometer.
Velocity of each singly ionized atom is 1.05xx10^5 m//s
Force vecF on a charged particle due to magnetic field =q(vecvxxvecB)

Since the magnetic field is perpendicular to the direction of motion in a spectrometer
|vecF|=q.|vecv|.|vecB|, singly ionized atom has charge 1.60xx 10^-19C

This magnetic force produces circular motion of the ion.
Centripetal force (m.v^2)/r=|vecF|, where r is the radius of circle in which ion moves.
(m.v^2)/r=q.|vecv|.|vecB|, Solving for r
r=(m.|vecv|)/(q.|vecB|)

r=(mxx1.05xx10^5)/(1.60xx 10^-19xx0.75)

r=mxx8.75xx10^23

  1. For "U"235:
    r_235=3.9xx10^-25xx8.75xx10^23
    =0.34125m
  2. For "U"238:
    r_238=3.95xx10^-25xx8.75xx10^23
    =0.345625m

Distance between the two isotopes is equal to difference between the two diameters

buphy.bu.edubuphy.bu.edu
=2xx(0.345625-0.34125)=0.00875m
=0.875cm