In an intrinsic semiconductor the energy gap E_g is 1.2 eV. Its hole mobility is much smaller than electron mobility and independent of temperature. What is the ratio between conductivity at 600 K and 300 K? Assume that the temperature dependence of intrinsic carrier concentration n_i is given by
$n_{i} = n_{0} \exp (- \frac{E_{g}}{2 k_{B} T}),$ where n₀ is a constant and k_B = 8.62 x 10^–5 eV/K.

Solution

Given,
Energy gap in a semiconductor, E_g = 1.2 eV

Assuming, temperature dependance of intrisic semiconductor is,

n_{i} = n_{0} \exp (- \frac{E_{g}}{2 k_{B} T})

and here,

\frac{E_{g}}{2 k} [\frac{1}{T_{1}} - \frac{1}{T_{2}}]

\frac{1.2}{2 \times 8.62 \times 10^{- 5}} [\frac{1}{300} - \frac{1}{600}] = 11.59

Therefore,

\frac{n_{i_{1}}}{n_{i_{2}}} = \frac{1.2 eV}{\frac{e^{2 \times k_{B} \times 600}}{\frac{1.2 eV}{e^{2 \times k_{B} \times 300}}}}

= e^{\frac{1.2 eV}{^{2 \times k_{B}}}} (\frac{1}{300} - \frac{1}{600})

= e^{\frac{1.2 \times 1.6 \times 10^{- 19}}{2 \times 1.381 \times 10^{- 23} \times 600}}

\Rightarrow

\frac{n_{i_{1}}}{n_{i_{2}}} = e^{11.59} = 1.072 \times 10^{5}

is the required ratio of conductivity.

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