Question

Consider the mass spectrometer shown schematically in the figure below. The magnitude of the electric field between the plates of the velocity selector is 2.20 103 V/m, and the magnetic field in both the velocity selector and the deflection chamber has a magnitude of 0.030 0 T. Calculate the radius of the path for a singly charged ion having a mass m = 2.34 10-26 kg.

Answer 1

Here, speed of the ion as it leaves the velocity selector.

v = E / B
v = (2200 V/m) / (0.030 T)
v = 73333.33 m/s

Again, when leaving the velocity selector, the magnetic force equals the centripetal force.

Fm = Fc
q v B = m v² / r
r = (m × v) / (q × B)
where
r = radius = ?
m = mass = 2.34 x 10^-26 kg
v = speed = 73333 m/s
q = charge = 1.6x10^-19 C
B = magnetic field = 0.030 T
Putting the given values in the above expression -
r = [ (2.34 x 10^-26 kg) × (73333 m/s) ] / [ (1.6x10^-19 C) × (0.030 T) ] = 0.357 m

Therefore, the requisite radius for a singly charged ion = 0.357 m