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Q.

The mass of an electron is ‘m’. Its charge is e and is accelerated from rest through a potential difference V. The velocity acquired by the electron will be

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a

$\sqrt{\frac{2 eV}{m}}$

b

$\sqrt{\frac{eV}{m}}$

c

$\sqrt{\frac{V}{m}}$

d

$\sqrt{\frac{m}{2 eV}}$

answer is C.

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Detailed Solution

$K E = \frac{1}{2} m v^{2} = e V v = \sqrt{(2 e V / m)}$

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