Question

Suppose the electric field between the plates P and P' in the mass spectrometer in following figure is 1.90

Answer 1

First we need the velocity by v = E/B

v = (1.9 X 10⁴)/(.691)

v = 27496 m/s

Then apply r = mv/qB for each...

r = (82)(1.66 X 10^-27)(27496)/(1.6 X 10^-19)(.691)

r = .0339 m and the diameter will be twice that = .0677 m

For the next...

r = (84)(1.66 X 10^-27)(27496)/(1.6 X 10^-19)(.691)

r = .0347 m and the diameter will be twice that = .06936 m

For the last one...

r = (86)(1.66 X 10^-27)(27496)/(1.6 X 10^-19)(.691)

r = .0355 m and the diameter will be twice that = .071 m

The distances between are...

.071 - .06936 = 1.66 X 10^-3 m which is 1.66 mm

.06936 - .0677 = 1.66 X 10^-3 m which is 1.66 mm

So the distance separations are the same.