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Physics Question on P-N Junction

Pure SiSi at 500 K500\, K has equal number of electron (ne)(n_e) and hole (nh)(n_h) concentrations of 1.5×1016 m−31.5 \times 10^{16}\, m^{-3} Doping by indium increases nhn_h to 4.5 ×1022 m−3\times 10^{22}\, m^{-3} The doped semiconductor is of

A

p-type having electron concentration ne=5×109 m−3n_e=5\times 10^9\, m^{-3}

B

n-type with electron concentration ne=5×1022 m−3n_e=5\times 10^{22} \,m^{-3}

C

p-type with electron concentration ne=2.5×1010 m−3n_e=2.5\times 10^{10}\, m^{-3}

D

n-type with electron concentration ne=2.5×1023 m−3n_e=2.5\times 10^{23}\, m^{-3}

Answer

p-type having electron concentration ne=5×109 m−3n_e=5\times 10^9\, m^{-3}

Explanation

Solution

nenh=ni2n _{ e } n _{ h }= n _{ i }^{2}
neNA=ni2n _{ e } N _{ A }= n _{ i }^{2}
ne=ni2NA=(1.5×1016)24.5×1022n _{ e }=\frac{ n _{ i }^{2}}{ N _{ A }}=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.5 \times 10^{22}}
=5×109/m3=5 \times 10^{9} / m ^{3}