Q136 / 180 · COMEDK 2013 · Physics

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

  • A
    $10^{6}$
  • B
    zero
  • Correct answer
    C
    $4 \times 10^{6}$
  • D
    $4 \times 10^{4}$

Solution

As we know, Kinetic energy of electron = Energy acquired due to $$ \begin{aligned} \Rightarrow \quad \frac{1}{2} m_{e} v^{2}=e V_{0} & \text { potential }\left(V_{0}\right) \\ \Rightarrow \quad v &=\sqrt{\frac{2 e V_{0}}{m_{e}}} \\ &=\sqrt{\frac{2 \times 1.6 \times 10^{-19} \times 45.5}{9.1 \times 10^{-31}}} \\ &=\sqrt{16 \times 10^{12}}=4 \times 10^{6} \mathrm{~ms}^{-1} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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