What will be the conductivity of a pure silic | vedclass

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(A) For an intrinsic semiconductor,the conductivity $\sigma$ is given by the formula: $\sigma = e(n_e \mu_e + n_h \mu_h)$.Since it is a pure crystal,t

Vedclass · Accepted Answer

$3.14 	imes 10^{-6} \ \Omega^{-1} \ cm^{-1}$

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(A) For an intrinsic semiconductor,the conductivity $\sigma$ is given by the formula: $\sigma = e(n_e \mu_e + n_h \mu_h)$.Since it is a pure crystal,the number of electrons equals the number of holes,so $n_e = n_h = n_i = 1.072 	imes 10^{10} \ cm^{-3}$.Substituting the given values:$\sigma = (1.6 	imes 10^{-19} \ C) 	imes (1.072 	imes 10^{10} \ cm^{-3}) 	imes (1350 \ cm^2/V \cdot s + 480 \ cm^2/V \cdot s)$.$\sigma = (1.6 	imes 10^{-19}) 	imes (1.072 	imes 10^{10}) 	imes (1830)$.$\sigma = 1.7152 	imes 10^{-9} 	imes 1830 \approx 3.14 	imes 10^{-6} \ \Omega^{-1} \ cm^{-1}$ (or $mho \ cm^{-1}$).

What will be the conductivity of a pure silicon crystal at $300 \ K$ temperature,if the electron-hole pairs per $cm^3$ is $1.072 \times 10^{10}$ at this temperature,$\mu_e = 1350 \ cm^2/V \cdot s$ and $\mu_h = 480 \ cm^2/V \cdot s$?

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