99 Percentile Qs Bank for JEE MainPhysicsSemiconductors
In a p-n junction diode, an electric field of magnitude 2 10^5 ~V / mexists in the depletion region. A particle with charge -3 e can diffuse from n -side to p -side, if it has minimum kinetic energy 0.6 eV . The width of the depletion region of the p-n junction is
Options
- A300 ~nm
- B600 ~nm
- C1000 ~nm
- D1200 ~nm
Correct answer
C. 1000 ~nm
Step-by-step solution
=34 For depletion width d , developed potential is V=E d=2 10^5 d For charged particle, experienced potential is V= Energy Charge = 0 6 eV 3 e =0 2 ~V So, 2 10^5 d=0 2 d=10⁻⁶ ~m or 1000 ~nm