In an N-type Ge semiconductor,the mobility of | vedclass

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(A) The conductivity $\sigma$ of an $N$-type semiconductor is given by $\sigma = e n_e \mu_e$,where $e$ is the electronic charge,$n_e$ is the electron

Vedclass · Accepted Answer

$6.25 	imes 10^{15} \ cm^{-3}$

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(A) The conductivity $\sigma$ of an $N$-type semiconductor is given by $\sigma = e n_e \mu_e$,where $e$ is the electronic charge,$n_e$ is the electron concentration,and $\mu_e$ is the electron mobility.Given: $\sigma = 5 \ mho/cm$,$\mu_e = 5000 \ cm^2/V \cdot s$,and $e = 1.6 	imes 10^{-19} \ C$.Rearranging the formula for $n_e$: $n_e = \frac{\sigma}{e \mu_e}$.Substituting the values: $n_e = \frac{5}{1.6 	imes 10^{-19} 	imes 5000}$.$n_e = \frac{5}{8000 	imes 10^{-19}} = \frac{5}{8 	imes 10^{-16}} = 0.625 	imes 10^{16} = 6.25 	imes 10^{15} \ cm^{-3}$.Since it is an $N$-type semiconductor,the concentration of impurity atoms is equal to the electron concentration $n_e$.

In an $N$-type $Ge$ semiconductor,the mobility of electrons is $5000 \ cm^2/V \cdot s$ and the conductivity is $5 \ mho/cm$. If the effect of holes is negligible,what is the concentration of impurity atoms?

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