Question

For an intrinsic semiconductor with gap widtheg=0.7ev calculate the concentration of intrinsic charge carriers at 300k assuming that m*e=m*h=m0

Answer 1

Explanation:

The intrinsic carrier concentration is assumed to be ni = 1.5 x 1010 cm-3. - Comment Nd >> ni, so that the thermal-equilibrium majority carrier electron concentration is essentially equal to the donor impurity concentration

Answer 2

the concentration of intrinsic charge carriers is 1.45×$10^{18}$ $m^{-3}$

given: an intrinsic semiconductor with gap width eg=0.7e and T=300K

to find: concentration  of intrinsic charge carriers

solution:

given condition is

$m^{*} _{e}=m^{*}_{h}=m_{o}=9.1 * 10^{-31}$

the concentration of intrinsic charge carrier

$n_{i}= 2[2\pi KT/h^{2}]^{3/2} (m^{*}_{e} .m^{*}_{h})^{3/4}exp[-E_{g}/2K_{b}T]$ .......(1)

on solving

$2[2 \pi KT/h^{2} ] ^{3/2}$= 2(1.4421 ×$10^{70}$)

= 2.884×$10^{70}$

now we are going to find

$(m^{*} _{e}*m^{*}_{h})^{3/4}=(9.1 * 10^{-31})^{3/2}$

=3.767×$10^{-47}$

now solve exp[$-E_{g}/2K_{b}T$}

=exp[-0.7/ (2×$K_{b}$×300)

=1.335×$10^{-6}$

on putting values in equation

we get

$n_{i}$ = 2.884× $10^{70}$× 3.767×$10^{-47} }$×1.335×$10^{-6}$

$n_{i}$=1.45×$10^{18}$ $m^{-3}$

hence required concentration is 1.45×$10^{18}$ $m^{-3}$

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