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Extrinsic Semiconductors

On doping ger...

Question

On doping germanium with donor atoms of density $10$ $^{17}$ $c m$ $^{- 3}$ its conductivity in $m h o / c m$ will be
[Given: $μ_{e}$ $= 3800 c m$ $^{2}$ $/ V - s$ and $n_{i}$ $= 2.5$ $\times$ $10$ $^{13}$ $c m$ $^{- 3}$ ]

30.4

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60.8

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91.2

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121.6

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Solution

The correct option is A $60.8$
Conductivity, $σ = n_{i} e μ_{e} = 10^{17} \times (1.6 \times 10^{- 19}) \times 3800 = 60.8$ $m h o / c m$ .

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Similar questions

10^{17} {c m}^{- 3}

μ_{e} = 3800 {c m}^{2} / V - s

n_{i} = 2.5 \times 10^{13} {c m}^{- 13}

3900 {cm}^{2} / v - s

6.24 mho / cm

8 \times 10^{13} c m^{- 3}

= 5 \times 10^{12} c m^{- 3}

= 2.3 \times 10^{4} c m^{2} V^{- 1} s^{- 1}

= 100 c m^{2} V^{- 1} s^{- 1}

300 K i s 2.5 \times 10^{13} / c m^{3}

5 \times 10^{16} / c m^{3}

= 5 \times 10^{12} / c m^{3}

= 8 \times 10^{13} / c m^{3} (μ_{e} = 2.3 m^{2} V^{- 1} s^{- 1}, μ_{h} = 0.01 m^{2} V^{- 1} s^{- 1}

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