Physics - Semiconductors Question with Solution | TestHub

The resistivity of pure $\mathrm{S}$ is given by

$\mathrm{\rho }=\frac{1}{\mathrm{\sigma }}=\frac{1}{\mathrm{e}\left({\mathrm{n}}_{\mathrm{e}}{\mathrm{\mu }}_{\mathrm{e}}+{\mathrm{n}}_{\mathrm{h}}{\mathrm{\mu }}_{\mathrm{h}}\right)}=\frac{1}{{\mathrm{en}}_1\left({\mathrm{\mu }}_{\mathrm{e}}+{\mathrm{\mu }}_{\mathrm{h}}\right)}$

or ${\mathrm{n}}_1=\frac{1}{\mathrm{e\rho }\left({\mathrm{\mu }}_{\mathrm{e}}+{\mathrm{\mu }}_{\mathrm{n}}\right)}=\frac{1}{1.6\times {10}^{-19}\times 3000(0.12+0.045)}$

$=1.26\times {10}^{16}{\mathrm{m}}^{-3}$

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

$\therefore {\mathrm{n}}_{\mathrm{e}}-{\mathrm{n}}_{\mathrm{h}}\approx {\mathrm{n}}_{\mathrm{e}}={\mathrm{N}}_{\mathrm{d}}={10}^{19} \because {\mathrm{n}}_{\mathrm{n}}=1.26\times {10}^{16}$

Resitivity $\mathrm{\rho }=\frac{1}{{\mathrm{n}}_{\mathrm{e}}{\mathrm{\mu }}_{\mathrm{e}}\mathrm{e}}=\frac{1}{1.6\times {10}^{-19}\times {10}^{19}\times 0.12}=5.21\mathrm{\Omega m}$