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Physics Question on Semiconductor electronics: materials, devices and simple circuits

If the resistivity of pure silicon is 3000 Ωm3000 \, \Omega m and the mobilities of electrons and holes are 0.12 m2V−1s−10.12 \, m^{2}V^{- 1}s^{- 1} and 0.045 m2V−1s−10.045 \, m^{2}V^{- 1}s^{- 1} respectively, then what will be the resistivity of a specimen of silicon when 101910^{19} atoms of phosphorous are added per cubic metre?

A

2.21Ωm2.21\Omega m

B

3.21Ωm3.21\Omega m

C

4.21Ωm4.21\Omega m

D

5.21Ωm5.21\Omega m

Answer

5.21Ωm5.21\Omega m

Explanation

Solution

The resistivity of pure SS is given by ρ=1σ=1e(ne(μ)e+nh(μ)h)=1(en)1((μ)e+(μ)h)\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 n1=1e?((μ)e+(μ)n)=11.6×(10)−19×3000(0.12+0.045)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)} =1.26×1016m−3=1.26\times 10^{16}m^{- 3} When 101910^{19} atoms of phosphorous (donor atoms of valence five) are added per m3,m^{3}, the semiconductor becomes nn - type semiconductor. ∴ne−nh≈ne=Nd=1019∵nn=1.26×1016\therefore n_{e}-n_{h}\approx n_{e}=N_{d}=10^{19}\because n_{n}=1.26\times 10^{16} Resitivity ρ=1neμee=11.6×10−19×1019×0.12=5.21Ωm\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