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

ni2=nenh⇒ne=1.5 x 101624.5 x 1022=5 x 109m−3

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

ni2=nenh⇒ne=1.5 x 101624.5 x 1022=5 x 109m−3

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Pure Si at 500 K has equal number of electron (n_e) and hole (n_h) concentrations is 1.5 x 10¹⁶ m^-3. Doping by indium increases n_h to 4.5 x 10²² m^-3. The doped semiconductor is of :

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

n-type with electron concentration n_e = 5 x 10²² m^-3

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

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

answer is A.

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Detailed Solution

$n_{i}^{2} = n_{e} n_{h} \Rightarrow n_{e} = \frac{{(1 .5 x 10^{16})}^{2}}{4 .5 x 10^{22}} = 5 x 10^{9} m^{- 3}$

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

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Pure Si at 500 K has equal number of electron (n_e) and hole (n_h) concentrations is 1.5 x 10¹⁶ m^-3. Doping by indium increases n_h to 4.5 x 10²² m^-3. The doped semiconductor is of :