Semiconducting Devices Question 41
Question: In semiconductor the concentrations of electrons and holes are $8 \times 10^{18}/m3 $ and $5 \times 10^{18}/m $respectively. If the mobilities of electrons and hole are 2.3 m2/volt-sec and 0.01 m2/volt-sec respectively, then semiconductor is
Options:
A) N-type and its resistivity is 0.34 ohm-metre
B) P-type and its resistivity is 0.034 ohm-metre
C) N-type and its resistivity is 0.034 ohm-metre
D) P-type and its resistivity is 3.40 ohm-metre
Show Answer
Answer:
Correct Answer: A
Solution:
$ n _{e}=8\times 10^{18}/m^{3},\ \ n _{h}=5\times 10^{18}/m^{3} $
$ {\mu _{e}}=2.3\frac{m^{2}}{volt-\sec },\ \ {\mu _{h}}=0.01\frac{m^{2}}{volt-\sec } $
$ \because n _{e}>n _{h} $ so semiconductor is N-type
Also conductivity $ \sigma =\frac{1}{\text{Resistivity}(\rho )}=e(n _{e}{\mu _{e}}+n _{h}{\mu _{h}}) $
Therefore $ \frac{1}{\rho }=1.6\times {{10}^{-19}}[8\times 10^{18}\times 2.3+5\times 10^{18}\times 0.01] $
Therefore r = 0.34 W-m.
