Question:

If the resistivity of pure silicon is $3000 \, \Omega m$ and the mobilities of electrons and holes are $0.12 \, m^{2}V^{- 1}s^{- 1}$ and $0.045 \, m^{2}V^{- 1}s^{- 1}$ respectively, then what will be the resistivity of a specimen of silicon when $10^{19}$ atoms of phosphorous are added per cubic metre?

Updated On: Jul 28, 2022
  • $2.21\Omega m$
  • $3.21\Omega m$
  • $4.21\Omega m$
  • $5.21\Omega m$
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The Correct Option is D

Solution and Explanation

The resistivity of pure $S$ is given by $\rho =\frac{1}{\sigma }=\frac{1}{e \left(n_{e} \left(\mu \right)_{e} + n_{h} \left(\mu \right)_{h}\right)}=\frac{1}{\left(en\right)_{1} \left(\left(\mu \right)_{e} + \left(\mu \right)_{h}\right)}$ or $n_{1}=\frac{1}{e? \left(\left(\mu \right)_{e} + \left(\mu \right)_{n}\right)}=\frac{1}{1 . 6 \times \left(10\right)^{- 19} \times 3000 \left(\right. 0 . 12 + 0 . 045 \left.\right)}$ $=1.26\times 10^{16}m^{- 3}$ When $10^{19}$ atoms of phosphorous (donor atoms of valence five) are added per $m^{3},$ the semiconductor becomes $n$ - type semiconductor. $\therefore n_{e}-n_{h}\approx n_{e}=N_{d}=10^{19}\because n_{n}=1.26\times 10^{16}$ Resitivity $\rho =\frac{1}{n_{e} \mu _{e} e}=\frac{1}{1 . 6 \times 10^{- 19} \times 10^{19} \times 0 . 12}=5.21\Omega m$
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