The electron mobility in N-type germanium is $3900 {cm}^{2} / V.s$ and its conductivity is $6.24 mho/cm$ , then impurity concentration is
(Assume that the effect of cotters is negligible)

10^{15} {/cm}^{3}

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

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

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

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Solution

The correct option is D $10^{16} {/cm}^{3}$
Conductivity $σ = e n_{e} μ_{e}$
$n_{e} = \frac{σ}{e μ_{e}} = \frac{6.24}{1.6 \times 10^{- 19} \times 3900}$
$\Rightarrow n_{e} = 10^{16} {/cm}^{3}$ .

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The electron mobility in N-type germanium is 3900 cm2/V.s and its conductivity is 6.24 mho/cm, then impurity concentration is (Assume that the effect of cotters is negligible)

The correct option is D 1016/cm3Conductivity σ=eneμe ne=σeμe=6.241.6×10−19×3900 ⇒ne=1016/cm3.

The electron mobility in N-type germanium is $3900 {cm}^{2} / V.s$ and its conductivity is $6.24 mho/cm$ , then impurity concentration is
(Assume that the effect of cotters is negligible)

The correct option is D $10^{16} {/cm}^{3}$
Conductivity $σ = e n_{e} μ_{e}$
$n_{e} = \frac{σ}{e μ_{e}} = \frac{6.24}{1.6 \times 10^{- 19} \times 3900}$
$\Rightarrow n_{e} = 10^{16} {/cm}^{3}$ .