A potential barrier V volt exists across a P-N junction. The thickness of the depletion region is ‘d’. An electron with velocity ‘V’ approaches P-N junction from N-side. The velocity of the electron crossing the junction is

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$\sqrt{V^{2} - \frac{2 V e}{m}}$

$\sqrt{V^{2} + \frac{2 V e}{m}}$

$\sqrt{\frac{2 V e}{m}}$

answer is B.

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Detailed Solution

$V^{2} = u^{2} - 2 a s$
$\begin{array}{l} \Rightarrow V_{f}^{2} = V_{i}^{2} - 2. \frac{e E}{m} . d \\ P u t i n g v a l u e s o f V_{i} a n d E d w e g e t V_{f} = \sqrt{V^{2} - \frac{2 V e}{m}} \end{array}$

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