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 ions 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.

(A) 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.,
$\frac{1}{2} m v^{2} = e V$
$v^{2} = \frac{2 e V}{m} \dots (i)$
The magnetic field provides the necessary centripetal force for the circular motion:
$\frac{m v^{2}}{r} = e v B$
$v = \frac{e B r}{m}$
Substituting the value of $v$ in equation $(i)$:
$\frac{2 e V}{m} = \left( \frac{e B r}{m} \right)^{2} = \frac{e^{2} B^{2} r^{2}}{m^{2}}$
$\frac{e}{m} = \frac{2 V}{B^{2} r^{2}}$
Substituting the values:
$\frac{e}{m} = \frac{2 \times 100}{(2.83 \times 10^{-4})^{2} \times (12.0 \times 10^{-2})^{2}} = 1.73 \times 10^{11} \; C \; kg^{-1}$
Therefore,the specific charge ratio $(e/m)$ is $1.73 \times 10^{11} \; C \; kg^{-1}$.