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Question: An intrinsic semiconductor has a resistivity of \(0.50\Omega\) m at room temperature. Find the intri...

An intrinsic semiconductor has a resistivity of 0.50Ω0.50\Omega m at room temperature. Find the intrinsic carries concentration if the mobilities of electrons and holes are 0.39m2V−1s−10.39m^{2}V^{- 1}s^{- 1} and 0.11m2V−1s−10.11m^{2}V^{- 1}s^{- 1} respectively

A

1.2×1018m−31.2 \times 10^{18}m^{- 3}

B

2.5×1019m−32.5 \times 10^{19}m^{- 3}

C

1.9×1020m−31.9 \times 10^{20}m^{- 3}

D

3.1×1021m−33.1 \times 10^{21}m^{- 3}

Answer

2.5×1019m−32.5 \times 10^{19}m^{- 3}

Explanation

Solution

: here, ρ=0.50Ωm\rho = 0.50\Omega m

}{\mu_{h} = 0.11m^{2}V^{- 1}s^{- 1}}$$ The resistivity of intrinsic semiconductor is $$\frac{1}{\rho} = e(n_{i}\mu_{e} + n_{i}\mu_{h})$$ Where $n_{i}$is the intrinsic carrier concentration $$\therefore n_{i} = \frac{1}{\rho e(\mu_{e} + \mu_{h})}$$ Substituting the given values, we get $$n_{i} = \frac{1}{(0.5)(1.6 \times 10^{- 19})(0.39 + 0.11)} = 2.5 \times 10^{19}m^{- 3}$$