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

Explanation:

The electrical energy given to the ion Vq will be changed into kinetic energy $\frac{1}{2}mv^2$

As the ion moves with velocity v in a magnetic field B then the magnetic Lorentz force Bqv will be balanced by centrifugal force $\frac{mv^2}{r}$.

So,

$Vq = \frac{1}{2}mv^2$

and

$Bqv = \frac{mv^2}{r}$

Right from these eliminating v, we can derive

$m = \frac{B^2r^2q}{2V}$

On substituting the value, we get:

$m = \frac{(0.4)^2X (0.305)^2 X1.602X 10^-^1^9}{2X 2000}$

m = 5.96 X 10⁻²⁵ kg.