A potential barrier of 0.4 V exists across a p-n junction. An electron enters the junction from the n-side with a speed of 6.0×105ms−1. The speed with which electron enters the p side will be 3x×105ms−1 the value of x is ______. (Given mass of electron =9×10−31kg, charge on electron =1.6×10−19C.)
Answer: 14
Solution
When an electron moves from the n-side to the p-side across a potential barrier of 0.4 V, it loses kinetic energy equal to the work done against the electric field.
Using the work-energy theorem:
qV=21m(u2−v2)
where u=6.0×105 m/s is the initial speed and v is the final speed.
1.6×10−19×0.4=21×9×10−31×(36×1010−v2)
0.64×10−19=4.5×10−31×(36×1010−v2)
v2=36×1010−91.28×1012=(36−9128)×1010=9196×1010
v=314×105 m/s
Therefore, x=14.
JEE Main · 2022 · 29 Jun 2022 · Shift 2 · Integer answer
