Question

A mass spectrometer applies a voltage of 2.00 kV to accelerate a singly charged positive ion. A magnetic field of B = 0.400 T then bends the ion into a circular path of radius 0.305 m. What is the mass of the ion?

Answer 1

The mass of the ion is 5.96 X 10⁻²⁵ kg

Step-by-step explanation:

The electrical energy given to the ion Vq will be changed into kinetic energy
`(1)/(2)mv^2`

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force
`(mv^2)/(r)`.

So,

`Vq = (1)/(2)mv^2`

and

`Bqv = (mv^2)/(r)`

Right from these eliminating v, we can derive

`m = (B^2r^2q)/(2V)`

On substituting the value, we get:

`m = ((0.4)^2X (0.305)^2 X1.602X 10^-^1^9)/(2X 2000)\\\\`

m = 5.96 X 10⁻²⁵ kg.