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Question: A semiconductor has equal electron and hole concentration of \(6 \times 10^{8}\) per \(m^{3}\) on do...

A semiconductor has equal electron and hole concentration of 6×1086 \times 10^{8} per m3m^{3} on doping with certain impurity, electron concentration increases to 9×1012perm39 \times 10^{12}perm^{3} The new hole concentration is

A

2×1042 \times 10^{4}per m3m^{3}

B

2×1022 \times 10^{2} per m2m^{2}

C

4×1044 \times 10^{4} per m3m^{3}

D

4×1024 \times 10^{2}per m3m^{3}

Answer

4×1044 \times 10^{4} per m3m^{3}

Explanation

Solution

: As nenh=ni2n_{e}n_{h} = n_{i}^{2}

Here, ni=6×108perm3n_{i} = 6 \times 10^{8}perm^{3}and ne=9×1012perm3n_{e} = 9 \times 10^{12}perm^{3}

nh=nine=(6×108)29×1012=4×104perm3\therefore n_{h} = \frac{n_{i}}{n_{e}} = \frac{(6 \times 10^{8})^{2}}{9 \times 10^{12}} = 4 \times 10^{4}perm^{3}