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?

• A. $2.21\Omega m$
• B. $3.21\Omega m$
• C. $4.21\Omega m$
• D. $5.21\Omega m$

Answer 1

Answer: (D)

The resistivity of pure $S$ is given by
$\rho=\frac{1}{\sigma}=\frac{1}{\mathrm{e}\left(\mathrm{n}_{\mathrm{e}} \mu_{\mathrm{e}}+\mathrm{n}_{\mathrm{h}} \mu_{\mathrm{h}}\right)}=\frac{1}{\mathrm{en}_{\mathrm{l}}\left(\mu_{\mathrm{e}}+\mu_{\mathrm{h}}\right)}$
or $\mathrm{n}_1=\frac{1}{\mathrm{e} \rho\left(\mu_{\mathrm{e}}+\mu_{\mathrm{n}}\right)}=\frac{1}{1.6 \times 10^{-19} \times 3000(0.12+0.045)}$
$=1.26\times 10^{16}m^{- 3}$
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 $\rho =\frac{1}{n_{e} \mu _{e} e}=\frac{1}{1 . 6 \times 10^{- 19} \times 10^{19} \times 0 . 12}=5.21\Omega m$