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

$\begin{array}{l}\frac{1}{2}{\mathrm{mv}}^2=\mathrm{eV}\\ {\mathrm{v}}^2=\frac{2\mathrm{eV}}{\mathrm{m}} \end{array}$

\( v=\sqrt{\frac{2eV}{m}}\)     .............. (1)
Step 2: Find the force acting on the electron beam and equate to centripetal force.
Force due to magnetic field = evB
Centripetal force = \(\frac{mv^2}{R}\)
\(evB=\frac{mv^2}{R}\\ \Rightarrow v=\frac{RBe}{m}........(2) \)
Step 3: Find the specific charge ratio.
From (1) & (2)
\(\begin{array}{l} R B\left[\frac{e}{m}\right]=\sqrt{2 V\left(\frac{e}{m}\right)} \\ R^{2} B^{2} \frac{e^{2}}{m^{2}}=2 v\left(\frac{e}{m}\right) \\ \left(\frac{e}{m}\right)=\frac{2 V}{R^{2} B^{2}} \end{array}\)
$\begin{array}{l}=\frac{2\times 100}{{(2.83\times {10}^{-4})}^2\times {(12\times {10}^{-2})}^2}=1.73\times {10}^{11}{\mathrm{Ckg}}^{-1}\end{array}$

Therefore, the specific charge ratio (e/m) is 1.73 x 10¹¹ Ckg^-1.