An electron gun with its collector at a potential of $100\; V$ fires out electrons in a spherical bulb containing hydrogen gas at low pressure $(\sim 10^{-2} \;mm$ of $Hg$ . A magnetic field of $2.83 \times 10^{-4}\; T$ curves the path of the electrons in a circular orbit of radius $12.0 \;cm .$ (The path can be viewed because the gas tons in the path focus the beam by attracting electrons, and emitting light by electron capture; this method is known as the 'fine beam tube' method.) Determine $e / m$ from the data.

Potential of an anode, $V=100 \,V$

Magnetic field experienced by the electrons, $B =2.83 \times 10^{-4}\, T$

Radius of the circular orbit $r =12.0 \,cm =12.0 \times 10^{-2}\, m$

Mass of each electron $= m ,$

Charge on each electron $= e ,$

Velocity of each electron $= v$

The energy of each electron is equal to its kinetic energy, i.e.,

$\frac{1}{2} m v^{2}=e V$

$v^{2}=\frac{2 e V}{m}\dots(i)$

It is the magnetic field, due to its bending nature, that provides the centripetal force $\left(F=\frac{m v^{2}}{r}\right)$ for the beam. Hence, we can write

Centripetal force $=$ Magnetic force

$\frac{m v^{2}}{r}=e v B$

$e B=\frac{m v}{r}$

$v=\frac{e B r}{m}$

Putting the value of $v$ in equation $(i),$ we get:

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

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

$=\frac{2 \times 100}{\left(2.83 \times 10^{-4}\right)^{2} \times\left(12 \times 10^{-2}\right)^{2}}=1.73 \times 10^{11} \,C kg ^{-1}$

Therefore, the specific charge ratio $(e/m)$ is $1.73 \times 10^{11}\, C k g^{-1}$