Question

If resistivity of pure silicon is 3000$\Omega$m and the electron and hole mobilities are 0.12 $m^2{V}^{-1}{s}^{-1}$ and 0.045 $m^2{V}^{-1}{s}^{-1}$ respectively. The resistivity of a specimen of the material when 10¹⁹ atoms of Phosphorous are added per $m^3$ is

• A. 2.21Ωm
• B. 3.21 Ωm
• C. 4.21 Ωm
• D. 5.21 Ωm

Answer 1

The correct option is D

5.21 Ωm

The resistivity of pure Si is given by

$P=\frac{1}{\sigma }=\frac{1}{e(n_e{\mu }_e+n_h{\mu }_h)}=\frac{1}{e{n}_h({\mu }_e+{\mu }_h)}Or{n}_h=\frac{1}{e\rho ({\mu }_e+{\mu }_h)}=\frac{1}{1.6\times {10}^{-19}\times 3000(0.12+0.045)}=1.26\times {10}^{16}{m}^{-3}$

When ${10}^{19}$ atoms of phosphorous (donor atoms of valence five) are added per $m^3$, the semiconductor becomes n – type semiconductor.

$\therefore n_e-n_h\approx n_e=N_d={10}^{19}\because n_h=1.26\times {10}^{16}Resitivity\rho =\frac{1}{ne{\mu }_e}=\frac{1}{1.6\times {10}^{-19}\times {10}^{19}\times 0.12}=5.21\Omega m$