A beam of cathode rays is subjected to crossed Electric ( E ) and...

As the electron beam is not deflected, then or $$\begin{aligned}\mathrm{F}_{\mathrm{m}}& =\mathrm{F}_{\mathrm{e}}\\ \mathrm{Bev}& =\mathrm{Ee}\\ \mathrm{v}& =\frac{\mathrm{E}}{\mathrm{B}} \end{aligned}$$ As the electron moves from cathode to anode, its potential energy at the cathode appears as its kinetic energy at the anode. If $V$ is the potential difference between the anode and cathode, then potential energy of the electron at cathode $=\mathrm{eV}$. Also, kinetic energy of the electron at anode $=\frac{1}{2} \mathrm{mv}^2$. According to law of conservation of energy $$\begin{aligned}\frac{1}{2} \mathrm{mv}^2 & =\mathrm{eV}\\ \mathrm{v}& =\sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}} \end{aligned}$$ From Eqs. (i) and (ii), we have $$\begin{aligned} \sqrt{\frac{2 \mathrm{eV}}{\mathrm{m}}}& =\frac{E}{\mathrm{ B}}\\ \frac{\mathrm{e}}{\mathrm{m}}& =\frac{\mathrm{E}^2}{2 \mathrm{VB}^2} \end{aligned}$$