Physics - Modern Physics Question with Solution | TestHub

As the electron beam is not deflected, then or $$\begin{array}{rl}{\mathrm{F}}_{\mathrm{m}}& ={\mathrm{F}}_{\mathrm{e}}\\ \mathrm{Bev}& =\mathrm{Ee}\\ \mathrm{v}& =\frac{\mathrm{E}}{\mathrm{B}}\end{array}$$ 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 $$\begin{array}{rl}\frac{1}{2}{\mathrm{mv}}^2& =\mathrm{eV}\\ \mathrm{v}& =\sqrt{\frac{2\mathrm{eV}}{\mathrm{m}}}\end{array}$$ From Eqs. (i) and (ii), we have $$\begin{array}{rl}\sqrt{\frac{2\mathrm{eV}}{\mathrm{m}}}& =\frac{E}{\mathrm{B}}\\ \frac{\mathrm{e}}{\mathrm{m}}& =\frac{{\mathrm{E}}^2}{2{\mathrm{VB}}^2}\end{array}$$