Chapter 20, Problem 64P

Solution

To prove: Mass of an ion in a spectrometer is $m=\frac{R^{2}q{B}^2}{2V}$

Given:

Ion accelerates by a voltage V .

R is the radius of the ion and q is the charge of the ion.

Formula used:

The magnetic force is

  $F=qvB$

Centripetal force is

  $F_C=\frac{m{v}^2}{R}$

Calculation:

The magnetic force is equal to the centripetal force for the given situation. So,

  $qvB=\frac{m{v}^2}{R}$

The energy of ion is,

  $E=qV$

Kinetic energy is,

  $K.E.=\frac{1}{2}m{v}^2$

So,

  $qV=\frac{1}{2}m{v}^2$

  $v^2=\frac{2qV}{m}$

(1)

Now,

  $F=qvB$

  $F_c=\frac{m{v}^2}{R}$

  $qvB=\frac{m{v}^2}{R}$

  $R=\frac{mv}{qB}$

Substitute the value of v from equation (1),

  $R=\frac{m\sqrt{2qV}}{qB\sqrt{m}}=\frac{\sqrt{2mqV}}{qB}$

Squaring both sides,

  $R^2=\frac{2mqV}{q^2{B}^2}$

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