this is the image link: http://img145.imageshack.us/img145/8215/2038.png Suppose

Question

this is the image link: http://img145.imageshack.us/img145/8215/2038.png

Suppose the electric field between the electric plates in the mass spectrometer shown below is 2.04 x 104 V/m and the magnetic fields B = B' = 0.88 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 1.67 x 10-27 kg.)

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

Explanation / Answer

You are given E = 20400 and B = 0.88

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

v = E/B = 20400 / 0.88 = 23181.82

The magnetic force on the ions is qvB = 1.60x10^-19 * 23181.82 * 0.88 =

= 3.264 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 * 23181.82^2)/3.264x10^-15 = m * 1.65x 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 * 1.65 x 10^23 = atomic# * 0.00027495

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

2r = atomic# * 0.5499 millimeters

So finally,

Part (a)

distance for C-12 = 12*0.5499 = 6.5988mm

distance for C-13 = 13*0.5499 =7.1487 mm

distance for C-14 = 14*0.5499 = 7.6986 mm

How far apart are they? 0.499 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...

0.2495 millimeters

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