A electron (e^ - ) is accelerated by V volts experiences a force F...

When $\mathrm{e}^{-}$is accelerated by V volts, then work done by $\mathrm{e}^{-}$is equals to kinetic energy of $\mathrm{e}^{-}$. or $\quad \mathrm{qV}=\frac{1}{2} \mathrm{mv}^2$ Then, the velocity of $\mathrm{e}^{-}$is $\mathrm{v}=\sqrt{\frac{2 \mathrm{qV}}{\mathrm{m}}}$ or $\mathrm{v}=\mathrm{k} \sqrt{\mathrm{V}}$. When $\mathrm{e}^{-}$accelerated by V volts, then $\mathrm{v}=\mathrm{k} \sqrt{\mathrm{V}}$. Force experiences by $\mathrm{e}^{-}$in a uniform magnetic field, $\mathrm{F}=\mathrm{qvB}$ or $\quad$\mathrm{F}=\mathrm{qk}$\sqrt{\mathrm{V} B}$ ...(i) When e accelerated by 2 V volts, then $\mathrm{v}_1=\mathrm{k} \sqrt{2 \mathrm{~V}}$. $\therefore \quad \mathrm{F}_1=\mathrm{qv}_1 \mathrm{~B}=\mathrm{qk} \sqrt{2 \mathrm{VB}}$ Hence, $\quad \mathrm{F}_1=\sqrt{2} \mathrm{~F} \quad$ [from Eq. (i)]