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Question

Pure Si at 500 K has equal number of electron (ne) and hole (nh) concentration of 1.5×1016 m3. Doping by indium increases nh to 4.5×1022 m3. The doped semiconductor is of
  1. n-type with electron concentration ne=5×1022 m3
  2. p-type with electron concentration ne=2.5×1010 m3
  3. p-type having electron concentrations ne=5×109 m3
  4. n-type with electron concentration ne=2.5×1023 m3

A
p-type with electron concentration ne=2.5×1010 m3
B
n-type with electron concentration ne=5×1022 m3
C
n-type with electron concentration ne=2.5×1023 m3
D
p-type having electron concentrations ne=5×109 m3
Solution
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The correct option is D p-type having electron concentrations ne=5×109 m3
n2i=nenh
(1.5×1016)2=ne(4.5×1022)
ne=0.5×1010
ne=5×109
nh=4.5×1022
nhne
Semiconductor is p-type and ne=5×109m3.

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