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(A) The conductivity $\sigma$ of a semiconductor is given by $\sigma = e(n\mu_{n} + p\mu_{p})$.From the law of mass action,$np = n_{i}^{2}$,so $n = \f

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$n_{i} \sqrt{\frac{\mu_{n}}{\mu_{p}}}$

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(A) The conductivity $\sigma$ of a semiconductor is given by $\sigma = e(n\mu_{n} + p\mu_{p})$.From the law of mass action,$np = n_{i}^{2}$,so $n = \frac{n_{i}^{2}}{p}$.Substituting this into the conductivity equation: $\sigma = e\left(\frac{n_{i}^{2}}{p}\mu_{n} + p\mu_{p}\right)$.To find the minimum conductivity,we differentiate $\sigma$ with respect to $p$ and set it to zero: $\frac{d\sigma}{dp} = e\left(-\frac{n_{i}^{2}}{p^{2}}\mu_{n} + \mu_{p}\right) = 0$.This implies $\frac{n_{i}^{2}}{p^{2}}\mu_{n} = \mu_{p}$.Solving for $p$,we get $p^{2} = n_{i}^{2} \frac{\mu_{n}}{\mu_{p}}$,which gives $p = n_{i} \sqrt{\frac{\mu_{n}}{\mu_{p}}}$.

$A$ semiconductor has electron and hole mobilities $\mu_{n}$ and $\mu_{p}$ respectively. If its intrinsic carrier density is $n_{i}$,what will be the value of hole concentration $p$ for which the conductivity will be minimum at a given temperature?

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