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An electron is accelerated through a potential difference of $45.5 \mathrm{~V}$. The velocity acquired by it is $\left(\right.$ in $\mathrm{ms}^{-1}$ )
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Verified Answer
The correct answer is:
$4 \times 10^{6}$
As we know,
Kinetic energy of electron = Energy acquired due to
$$
\begin{aligned}
\Rightarrow \quad \frac{1}{2} m_{e} v^{2}=e V_{0} & \text { potential }\left(V_{0}\right) \\
\Rightarrow \quad v &=\sqrt{\frac{2 e V_{0}}{m_{e}}} \\
&=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 45.5}{9.1 \times 10^{-31}}} \\
&=\sqrt{16 \times 10^{12}}=4 \times 10^{6} \mathrm{~ms}^{-1}
\end{aligned}
$$
Kinetic energy of electron = Energy acquired due to
$$
\begin{aligned}
\Rightarrow \quad \frac{1}{2} m_{e} v^{2}=e V_{0} & \text { potential }\left(V_{0}\right) \\
\Rightarrow \quad v &=\sqrt{\frac{2 e V_{0}}{m_{e}}} \\
&=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 45.5}{9.1 \times 10^{-31}}} \\
&=\sqrt{16 \times 10^{12}}=4 \times 10^{6} \mathrm{~ms}^{-1}
\end{aligned}
$$
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