Pure Si at 500 ~K has equal number of electron (n_e) and hole (n_h) concentrations of 1.5 × 10^ 16 ~m^ -3 . Doping by indium increases n_h to 4.5 × 10^ 22 ~m^ -3 . The doped semiconductor is of

Correct answer: D. p-type having electron concentration n_e=5 × 10^9 ~m^ -3 aligned & n_i^2=n_e n_h & n_e=((n_i)^2)/(n_h) & n_e= (1.5 × 10^ 16 )^2 (4.5 × 10^ 22 ) & n_e=5 ×10^9 ~m^ -3 aligned So, n_ f_1 n_e semiconductor is p -type.

NEET 2011

Exam: NEET
Year: 2011
Subject: Physics
Chapter: Semiconductors

NEET 2011 · Physics · Semiconductors

Pure Si at 500 ~K has equal number of electron (n_e) and hole (n_h)...

Pure Si at

500 \mathrm{~K}

has equal number of electron \left(n_e\right)

and hole \left(n_h\right)

concentrations of

1.5 \times 10^{16} \mathrm{~m}^{-3}

. Doping by indium increases n_h to

4.5 \times 10^{22} \mathrm{~m}^{-3}

. The doped semiconductor is of

Options

A $n$ -type with electron concentration $n_e=5 \times 10^{22} \mathrm{~m}^{-3}$
B $p$ -type with electron concentration $n_e=2.5 \times 10^{10} \mathrm{~m}^{-3}$
Cn-type with electron concentration $n_e=2.5 \times 10^{23} \mathrm{~m}^{-3}$
Dp-type having electron concentration $n_e=5 \times 10^9 \mathrm{~m}^{-3}$

Correct answer

D. p-type having electron concentration $n_e=5 \times 10^9 \mathrm{~m}^{-3}$

Step-by-step solution

\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}{\left(4.5 \times 10^{22}\right)}\\ & n_e=5 \times10^9 \mathrm{~m}^{-3} \end{aligned}

So,

n_{f_1} \gg n_e

semiconductor is

p

-type.

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