Question

If the resistivity of pure silicon is $3000\Omega m$ and the mobilities of electrons and holes are $0.12m^2{V}^{-1}{s}^{-1}$ and $0.045m^2{V}^{-1}{s}^{-1}$ 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 $\rho =\frac{1}{n_e{\mu }_{e}e}=\frac{1}{1.6\times 10^{-19}\times 10^{19}\times 0.12}=5.21\Omega m$