An electron is accelerated through a potential difference of $45.5 \mathrm{~V}$. The velocity acquired by it (in $\mathrm{ms}^{-1}$ ) is

Given, potential difference, $V=45.5 \mathrm{~V}$ Velocity, $v=$ ? Energy acquired by the electron after getling accelerated through the potential difference $V$ is $\mathrm{eV}$, which is equal to its kinetic energy. Hence, kinetic energy $(\mathrm{KE})=\frac{1}{2} m_{e} v^{2}=e V$ $$ \Rightarrow \quad v=\sqrt{\frac{2 e V}{m_{e}}} \text { [where, } m_{e}=\text { mass of electron] } $$ Substituting the given values, we get $$ \begin{aligned} \Rightarrow \quad v &=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 45.4}{9.11 \times 10^{-31}}} \\ &=\sqrt{16 \times 10^{12}}=4 \times 10^{6} \mathrm{~ms}^{-1} \end{aligned} $$ ###0