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An electron is accelerated through a potential difference of 200 volts. If e/m for the electron be 1.6 x $10^{11}$ coulomb/kg,
the velocity acquired by the electron will be

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$8 x 10^{6} m / s$

$8 x 10^{5} m / s$

$5.9 x 10^{6} m / s$

$5.9 x 10^{5} m / s$

answer is A.

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Detailed Solution

$\begin{array}{l} \frac{1}{2} {mv}^{2} = QV \Rightarrow v = \sqrt{\frac{2 QV}{m}} = \sqrt{2 (\frac{e}{m}) V} \\ \Rightarrow v = \sqrt{2 \times 1 .6 \times 10^{11} \times 200} = 8 \times 10^{6} m / s \end{array}$

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