NTA Abhyas JEE Main2020PhysicsSemiconductorsPractice
The conductivity of a semiconductor sample having electron concentration of 5 × 10 18 e m - 3 , hole concentration of 5 × 10 19 holes m - 3 , electron mobility of 2 .0 m 2 V - 1 s - 1 and hole mobility of 0 . 01 m 2 V - 1 s - 1 is (Take charge of an electron as 1 .6 × 10 - 19 C )
Options
- A1 .83 Ω m - 1
- B1 .65 Ω m - 1
- C1 .20   Ω   m - 1
- D0 .59   Ω   m - 1
Correct answer
B. 1 .65 Ω m - 1
Step-by-step solution
The conductivity of a semiconductor is given by σ = e n e μ e + n h μ h = 1 .6 × 10 - 19 5 × 10 18 × 2 + 5 × 10 19 × 0 .01 = 1.6 × 10 - 19 10 19 + 0.05 × 10 19 = 1.6 + 1.05 = 1 .65 ( Ω m ) − 1