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

Question

Pure Si at 500 K has equal number of electron ( n e ) and hole ( n h ) concentrations to 1.5 × 1016 m -3. Doping by indium increases n h to 4.5 × 1022 m -3. The doped semiconductor is of :- (A) p-type having elecron concentrations n e =5 × 109 m -3 (B) n-type with electron concentration n e =5 × 1022 m -3 (C) P-type with electron concentration n e =2.5 × 1010 m -3 (D) n-type with electron concentration n e =2.5 × 1023 m -3

Tardigrade Admin · Accepted Answer

n e =( n i 2/ n h )=((1.5 × 1016)2/4.5 × 1022) =5 × 109 m -3

Q. Pure Si at $500 K$ has equal number of electron $(n_{e})$ and hole $(n_{h})$ concentrations to $1.5 \times 1 0^{16} m^{- 3}$ . Doping by indium increases $n_{h}$ to $4.5 \times 1 0^{22} m^{- 3}$ . The doped semiconductor is of :-

$p$ -type having elecron concentrations $n_{e} = 5 \times 1 0^{9} m^{- 3}$

$n$ -type with electron concentration $n_{e} = 5 \times 1 0^{22} m^{- 3}$

$P$ -type with electron concentration $n_{e} = 2.5 \times 1 0^{10} m^{- 3}$

$n$ -type with electron concentration $n_{e} = 2.5 \times 1 0^{23} m^{- 3}$

$n_{e} = \frac{n _{i}^{2}}{n _{h}} = \frac{( 1.5 \times 1 0 ^{16} ) ^{2}}{4.5 \times 1 0 ^{22}}$
$= 5 \times 1 0^{9} m^{- 3}$

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

p-type having elecron concentrations ne​=5×109m−3

n-type with electron concentration ne​=5×1022m−3

P-type with electron concentration ne​=2.5×1010m−3

n-type with electron concentration ne​=2.5×1023m−3