logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

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 $
Can you answer this question?
 
 

1 Answer

0 votes
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
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...