First slide

P-N junction diode

Question

A potential barrier V volts 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

Moderate

A

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

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

v
D

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

Solution

There exists an electric field from N side to P side. As electron experience a force opposite to the direction of electric field, it's velocity decreases, as it moves from N to P side; from conservation of energy
$\begin{array}{rcl} \frac{1}{2} m v^{2} - e V & = & \frac{1}{2} m v'^{2} \\ v'^{2} & = & v^{2} - \frac{2 e V}{m} \\ v' & = & \sqrt{v^{2} - \frac{2 e V}{m}} \end{array}$

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