Question

Suppose the electric field between the electric plates in the mass spectrometer of Figure 27-33 in the textbook is 2.23×104 V/m and the magnetic fields are B=B′=0.38T. 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 1.66×10−27kg.) How far apart are the lines formed by the singly charged ions of each type on the photographic film ( d12−13, d13−14)? What if the ions were doubly charged? d12−13, d13−14

Answer 1

Given :-

E = 2.23 x 10^4 V/m

B = 0.38

v = (2.23 x 10^4) / 0.38

v = 58684.2 m/s

The magnetic force on the ions is

F = qvB

F = 1.6 x 10^-19 x 58684.2 x 0.38

F = 3.568 x 10^-15 N

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

each using:

F = mv^2 / r

r = mv^2/F

r = (1.66 x 10^-27 x 58684.2^2) / (3.568 x 10^-15)

r = 0.0016 m = 1.6 mm

2r = 3.2 mm

Part (a)

distance for C-12 = 12*3.2 = 38.4 mm

distance for C-13 = 13*3.2 = 41.6 mm

distance for C-14 = 14*3.2 = 44.8 mm

they are 3.2 mm apart from each other

(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

it is 1.6 mm