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  • Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 x 1016 m-3. Doping by In

Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentrations of 1.5 x 1016 m-3. Doping by Indium increases nh to 4.5 x 1022 m-3.

The doped semiconductor is of:

Option: 1

n-type with electron concentration ne = 2.5 x 1023 m-3


Option: 2

p-type having electron concentration ne = 5 x 109 m-3


Option: 3

n-type with electron concentration ne = 2.5 x 1022 m-3


Option: 4

p-type with electron concentration ne = 2.5 x 1010 m-3


Answers (1)

best_answer

\begin{aligned} &n_{i}^{2}=n_{e} n_{h}\\ &n _{ e }=\frac{\left( n _{ i }\right)^{2}}{ n _{ h }}\\ &n _{ e }=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.5 \times 10^{22}}\\ &n_{e}=5 \times 10^{9} m ^{-3}\\ &So , n _{ h }>> n _{ e } \text { semiconductor is } p -\text { type }\\ &n_{i}^{2}=n_{e} n_{h}\\ &n _{ e }=\frac{\left( n _{ i }\right)^{2}}{ n _{ h }}\\ &n _{ e }=\frac{\left(1.5 \times 10^{16}\right)^{2}}{4.5 \times 10^{22}}\\ &n_{e}=5 \times 10^{9} m ^{-3}\\ &So , n _{ h }>> n _{ e } \text { semiconductor is } p -\text { type } \end{aligned}

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himanshu.meshram

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