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An electron enters the p-n junction from the n-side with a velocity of $5 \times 10^5 m/s$ . If the depletion region is $5 \times 10^{-7} m/s$ wide and the potential barrier is $0.5 V$ What will be its speed When it enters P side ?

$(a)\;1.4 \times 10^5 m/s \\ (b)\;2.7 \times 10^5 m/s \\ (c)\;0.4 \times 10^5 m/s \\ (d)\;1.8 \times 10^5 m/s $

1 Answer

An electron from the n-side into the p-n junction through the depletion region and When it reaches the p-side it gains an energy equal to exponential barriers.
By conservation of energy,
KE before entering =ev +KE after reaching p side
$ \large\frac{1}{2}$$ \times m \times V_1^2 =e \times 0.5 +\large\frac{1}{2} \times$$ m V_2^2$
Solving we get,
$V_2 =2.7 \times 10^5 m/s$
Hence b is the correct answer.
answered Jan 28, 2014 by meena.p

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