Question

The room temperature electrical conductivity of a semiconductor specimen is 2.8 x 10^4 (Ω-m)1. The electron concentration is known to be 2.9x 10^22 m^-3. Given that the electron and hole mobilities are 0.14 and 0.023 m^2/N-s, respectively, calculate the hole concentration (in m^-3)

Answer 1

Given:

$\sigma _{s} = 2.8\times 10^{4} \Omega-m$

electron concentration, n = $2.9\times 10^{22} m^{-3}$

$\mu _{h} = 0.14$

$\mu _{e} = 0.023$

Solution:

Let holes concentration be 'p'

$\sigma _{s}$ = ne$\mu _{e}$ +pe$\mu _{h}$     (1)

substituting all given values in eqn (1):

$2.8\times 10^{4} = 2.9\times 10^{22}\times 1.6 \times 10^{-19}\times 0.14 + p\times1.6 \times 10^{-19}\times 0.023$

The cocentration of holes is:

$p = 7.432\times 10^{24} m^{-3}$