Question

In a mass spectrometer a particle of mass m and charge q is accelerated through a potential difference V and allowed to enter a magnetic field B, where it is deflected in a semi-circular path of radius R. The magnetic field is uniform and oriented perpendicular to the velocity of the particle. Derive an expression for the mass of the particle in terms of B, q, V, and R.

Answer 1

m = B²qR² / 2 V

Step-by-step explanation:

If v be the velocity after acceleration under potential difference of V

kinetic energy = loss of electric potential energy

1/2 m v² = Vq ,

v² = 2 Vq / m ----------------------- ( 1 )

In magnetic field , charged particle comes in circular motion in which magnetic force provides centripetal force

magnetic force = centripetal force

Bqv = mv² / R

v = BqR / m

v² = B²q²R² / m² ------------------------- (2)

from (1) and (2)

B²q²R² / m² = 2 Vq / m

m = B²q²R² / 2 Vq

m = B²qR² / 2 V