First slide

Diodes

Question

A potential barriers of 0.5 eV exists in p-n junction. If an electron approaches the p-n junction from the n – side with speed of $5 \times 10^{5} m / s$ ,
then electron will enter the P side with velocity (in m/s)

Moderate

A

$25 \times 10^{4}$
B

$26 \times 10^{4}$
C

$27 \times 10^{4}$
D

$28 \times 10^{4}$

Solution

Let electron enters the n region with speed $v_{1}$ and enters the P region with speed $v_{2}$
Applying conservation of energy,
$\frac{1}{2} m {v_{1}}^{2} = e V + \frac{1}{2} m {v_{2}}^{2}$
$\Rightarrow \frac{1}{2} \times 9.1 \times 10^{- 31} \times {(5 \times 10^{5})}^{2} = 1.6 \times 10^{- 19} \times 0.5 + \frac{1}{2} \times 9.1 \times 10^{- 31} {v_{2}}^{2}$
$\Rightarrow 113.75 \times 10^{- 21} = 0.8 \times 10^{- 19} + \frac{1}{2} \times 9.1 \times 10^{- 31} {v_{2}}^{2}$
$\Rightarrow 4.55 \times 10^{- 31} {v_{2}}^{2} = 113.75 \times 10^{- 21} - 80 \times 10^{- 21}$
$\Rightarrow {v_{2}}^{2} = \frac{33.75 \times 10^{- 21}}{4.55 \times 10^{- 31}}$
$\Rightarrow {v_{2}}^{2} = 7.41 \times 10^{10}$
$\Rightarrow v_{2} = 2.7 \times 10^{5}$
$\Rightarrow v_{2} = 27 \times 10^{4} m / s$

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