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* 10^(4) \Omega-m`

electron concentration, n =
`2.9* 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* 10^(4) = 2.9* 10^(22)* 1.6 * 10^(-19)* 0.14 + p*1.6 * 10^(-19)* 0.023`

The cocentration of holes is:

`p = 7.432* 10^(24) m^(-3)`