Suppose the electric field between the electric plates in the mass spectrometer

Question

Suppose the electric field between the electric plates in the mass spectrometer of the figure below is 2.48 104 V/m and the magnetic fields B = B ' = 0.48 T. The source contains carbon isotopes of mass numbers 12, 13, and 14 from a long-dead piece of a tree. (To estimate atomic masses, multiply by 1.66 10-27 kg.)

http://www.webassign.net/gianpse3/27-32.gif


(a) How far apart are the lines formed by the singly charged ions of each type on the photographic film?


(in mm)

(b) What if the ions were doubly charged?


(in mm)

Explanation / Answer

You are given E = 24800 and B = 0.48

This means the speed of ions that are able to pass through undeflected is

v = E/B = 24800 / 0.48 = 51667

The magnetic force on the ions is qvB = 1.60x10^-19 * 51667 * 0.48 =

= 3.968 x 10^-15 Newtons

This force must equal the centripetal force, or mv^2/r. The distance each travels is 2r (see diagram). So we have to calculate r for each using:

F = mv^2 / r

r = mv^2/F = m * 51667^2/3.968x10^-15 = m * 6.7274 x 10^23

note that in each case, m is the mass of a proton times the atomic number (12, 13, 14) so we can write

r = atomic# * 1.66 x 10^-27 * 6.7274 x 10^23 = atomic# * 0.00111675

this is in meters, so we can convert to millimeters. Also, we'll double it since we want 2r. And...

2r = atomic# * 2.2335 millimeters

So finally,

Part (a)

distance for C-12 = 12*2.2335 = 26.802 mm

distance for C-13 = 13*2.2335 = 29.035 mm

distance for C-14 = 14*2.2335 = 31.269 mm

How far apart are they? 2.2335 millimeters

Part (b) If the ions were doubly charged, the force would be twice as great, so the radius for each ion would be half as much, so 2r would be half as much, and the distance between the lines would be half as much, or...

1.117 millimeters

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