Q70 / 120 · NEET 2011 (Mains) · Physics

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

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

Solution

$$ \begin{aligned} & n_i^2=n_e n_h \\ & n_e=\frac{\left(n_i\right)^2}{n_h} \\ & n_e=\frac{\left(1.5 \times 10^{16}\right)^2}{\left(4.5 \times 10^{22}\right)} \\ & n_e=5 \times 10^9 \mathrm{~m}^{-3} \end{aligned} $$ So, $n_{f_1} \gg n_e$ semiconductor is $p$-type.

Hence, the correct answer is option: 'D'.

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