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(D) The conductivity $\sigma$ of the semiconductor is given by $\sigma = ne(\mu_e + \mu_h)$.Given $n = 2 	imes 10^{19}\,m^{-3}$,$e = 1.6 	imes 10^{-

Vedclass · Accepted Answer

$0.64$

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(D) The conductivity $\sigma$ of the semiconductor is given by $\sigma = ne(\mu_e + \mu_h)$.Given $n = 2 	imes 10^{19}\,m^{-3}$,$e = 1.6 	imes 10^{-19}\,C$,$\mu_e = 0.36\,m^{2}/(V\cdot s)$,and $\mu_h = 0.14\,m^{2}/(V\cdot s)$.$\sigma = (2 	imes 10^{19}) 	imes (1.6 	imes 10^{-19}) 	imes (0.36 + 0.14) = 3.2 	imes 0.5 = 1.6\,(\Omega\cdot m)^{-1}$.The resistance $R$ is given by $R = ho \frac{l}{A} = \frac{l}{\sigma A}$.Here $l = 0.5\,mm = 0.5 	imes 10^{-3}\,m$ and $A = 1\,cm^{2} = 10^{-4}\,m^{2}$.$R = \frac{0.5 	imes 10^{-3}}{1.6 	imes 10^{-4}} = \frac{5}{1.6} = 3.125\,\Omega$.The current $i$ is $i = \frac{V}{R} = \frac{2}{3.125} = 0.64\,A$.

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