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Physics Question on Semiconductor electronics: materials, devices and simple circuits

An intrinsic semiconductor has a resistivity of 0.50Ωm0.50 \Omega\, m at room temperature. Find the intrinsic carrier concentration, if the mobilities of electrons and holes are 0.39m2/Volt0.39\, m^2/Volt , and 0.11m2/Voltsec0.11 m^2/Volt \,sec respectively

A

1.2×1018/m31.2 \times 10^{18}/m^3

B

2.5×1019/m32.5 \times 10^{19}/m^3

C

1.9×1020/m31.9 \times 10^{20}/m^3

D

3.1×1021/m33.1 \times 10^{21}/m^3

Answer

2.5×1019/m32.5 \times 10^{19}/m^3

Explanation

Solution

ρ=0.50Ωm\rho = 0.50 \,\Omega - m
σ=e(neue+nhuh)\sigma = e (n_e u_e + n_hu_h)
σ=1ρ\because \sigma = \frac{1}{\rho}
In intrinsic semi-conductor
ne=nh=nn_e = n_h = n
1ρ=e×n(μe+μh)\frac{1}{\rho} = e \times n(\mu_e + \mu_h)
10.50=1.6×1019×n(0.39+0.11)\frac{1}{0.50} = 1.6 \times 10^{-19} \times n(0.39 + 0.11)
1×10190.5×0.5×1.6=n\frac{1 \times 10^{19}}{0.5\times 0.5 \times 1.6 } = n
100×101925×1.6=n\frac{100\times 10^{19}}{25\times 1.6} = n
4016×1019=n\frac{40}{16} \times 10^{19} = n
n=2.5×1019/m2n = 2.5 \times 10^{19} /m^2