MCQ
An electron is accelerated through a potential difference of $200$ volts. If $\mathrm{e} / \mathrm{m}$ for the electron be $1.6 \times 10^{11}$ coulomb $/ \mathrm{kg}$, the velocity acquired by the electron will be
  • $8 \times 10^6 \mathrm{~m} / \mathrm{s}$
  • B
    $8 \times 10^5 \mathrm{~m} / \mathrm{s}$
  • C
    $5.9 \times 10^6 \mathrm{~m} / \mathrm{s}$
  • D
    $5.9 \times 10^5 \mathrm{~m} / \mathrm{s}$

Answer

Correct option: A.
$8 \times 10^6 \mathrm{~m} / \mathrm{s}$
$ \frac{1}{2} m v^2=Q V \Rightarrow v=\sqrt{\frac{2 Q V}{m}}=\sqrt{2\left(\frac{e}{m}\right) V} $
$ \Rightarrow v=\sqrt{2 \times 1.6 \times 10^{11} \times 200}=8 \times 10^6 \mathrm{~m} / \mathrm{s} .$

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