Density of electrons in an n-type semiconductor is \( 10^{19}\,\mathrm{m^{-3}} \). If the drift velocity is 0.02 mm/s, when an electric field \( 1.33\times10^{-5}\,\mathrm{V/m} \) is applied. Find the resistivity of semiconductor [contribution of holes is ignored]
Answer & explanation
Correct answer: option 4
The correct answer is Option (4) → 0.4 Ωm
The resistivity is the ratio of the electric field to the current density, \( \rho = E/J \), and the current density of electrons moving with drift velocity \( v_d \) is \( J = nev_d \).
\( v_d = 0.02\,\mathrm{mm\,s^{-1}} = 2\times10^{-5}\,\mathrm{m\,s^{-1}} \)
\( J = nev_d = 10^{19}\times1.6\times10^{-19}\times2\times10^{-5} = 3.2\times10^{-5}\,\mathrm{A\,m^{-2}} \)
\( \displaystyle \rho = \frac{E}{J} = \frac{1.33\times10^{-5}\,\mathrm{V\,m^{-1}}}{3.2\times10^{-5}\,\mathrm{A\,m^{-2}}} \approx 0.42\,\Omega\,\mathrm{m} \approx 0.4\,\Omega\,\mathrm{m} \)