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If the electron mobility in N-type germanium is 3900 ${cm}^{2} / V - s$ and its conductivity is 6.24 mho/cm, then impurity concentration, if the effect of cotters is negligible, will be

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$10^{15} {cm}^{3}$

$10^{13} / {cm}^{3}$

$10^{12} / {cm}^{3}$

$10^{16} / {cm}^{3}$

answer is D.

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Detailed Solution

$σ = {en}_{e} μ_{e} \Rightarrow n_{e} = \frac{σ}{{eμ}_{e}} = \frac{6.24}{1.6 \times 10^{- 19} \times 3900} = 10^{16} / {cm}^{3}$

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