JEE Main On 8 April 2017 Question 19
Question: What is the conductivity of a semiconductor sample having electron concentration of $ 5\times 10^{18}{m^{-3}}, $ hole concentration of $ 5\times 10^{19}{m^{-3}}, $ electron mobility of $ 2.0m^{2}{V^{-1}}{s^{-1}} $ and hole mobility of $ 0.01m^{2}{V^{-1}}{s^{-1}} $ (Take charge of electron as $ 1.6\times {10^{-19}}C $ ) [JEE Online 08-04-2017]
Options:
A) $ 1.83{{(\Omega -m)}^{-1}} $
B) $ 1.68{{(\Omega -m)}^{-1}} $
C) $ 1.20{{(\Omega -m)}^{-1}} $
D) $ 0.59{{(W-m)}^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ s=e(n _{e}{\mu_e}+n _{n}{\mu_n}) $ $ =1.6\times {10^{-19}}(5\times 10^{18}\times 2+5\times 10^{10}\times 0.01) $ $ =1.6\times {10^{-19}}(10^{19}+0.05\times 10^{19}) $ $ =1.6\times 1.05 $ $ =1.68 $
