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(A) The conductivity $\sigma$ of an intrinsic semiconductor is given by the formula: $\sigma = e(n_e \mu_e + n_h \mu_h)$.Since it is a pure silicon cr

Vedclass · Accepted Answer

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

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(A) The conductivity $\sigma$ of an intrinsic semiconductor is given by the formula: $\sigma = e(n_e \mu_e + n_h \mu_h)$.Since it is a pure silicon crystal,the number of electrons $(n_e)$ equals the number of holes $(n_h)$,which is the intrinsic carrier concentration $n_i = 1.072 	imes 10^{10} \ cm^{-3}$.Given values: $e = 1.6 	imes 10^{-19} \ C$,$\mu_e = 1350 \ cm^2/V \cdot s$,$\mu_h = 480 \ cm^2/V \cdot s$.Substituting these values into the formula:$\sigma = (1.6 	imes 10^{-19}) 	imes (1.072 	imes 10^{10}) 	imes (1350 + 480)$$\sigma = (1.6 	imes 1.072 	imes 10^{-9}) 	imes (1830)$$\sigma = 1.7152 	imes 10^{-9} 	imes 1830$$\sigma \approx 3.1388 	imes 10^{-6} \ \Omega^{-1} \ cm^{-1} \approx 3.14 	imes 10^{-6} \ \Omega^{-1} \ cm^{-1}$.

What is the conductivity of a pure silicon crystal at $300 \ K$? The number of electron-hole pairs per $cm^3$ is $1.072 \times 10^{10}$. Given $\mu_n = 1350 \ cm^2/V \cdot s$ and $\mu_p = 480 \ cm^2/V \cdot s$.

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