Question

3.) At room temperature the electrical conductivity of PbTe is \(500(\Omega-m)^{-1}\) whereas the electron and hole mobilities are 0.16 and \(0.075 \mathrm{~m}^{2} / \mathrm{V}-s,\) respectively. Compute the intrinsic carrier concentration for PbTe at room temperature.

a) \(1.33 \times 10^{20} \mathrm{~m}^{-3}\)

b) \(1.33 \times 10^{23} \mathrm{~m}^{-3}\)

c) \(1.33 \times 10^{22} \mathrm{~m}^{-3}\)

d) \(2.13 \times 10^{3} \mathrm{~m}^{-3}\)

e) \(2.13 \times 10^{5} \mathrm{~m}^{-3}\)

Answer 1

$$ \begin{aligned} & n_{i}=\text { intrinsic concentration }=?\\ &\sigma=\text { electrical conductivity }=500(\Omega \mathrm{m})^{-1}\\ &e=\text { charge of an electron }=1.6 \times 10^{-19} \mathrm{C}\\ &\mathrm{u}_{\mathrm{e}}=\text { electron mobilities }=0.16\\ &\mathrm{u}_{\mathrm{h}}=\text { hole mobilities }=0.075\\ &n_{i}=\frac{\sigma}{|e|\left(u_{e}+u_{h}\right)}\\ &n_{i}=\frac{500(\Omega m)^{-1}}{\left|1.6 \times 10^{-19}\right|(0.16+0.075) m^{2} / V . s}\\ &n_{i}=1.33 \times 10^{22} \mathrm{~m}^{-3} \end{aligned} $$