Q138 / 160 · AP EAMCET 2014 · Physics

A crystal of intrinsic silicon at room temperature has a carrier concentration of $1.6 \times 10^{16} / \mathrm{m}^3$. If the donor concentration level is $4.8 \times 10^{20} / \mathrm{m}^3$, then the concentration of holes in the semiconductor is

  • A
    $53 \times 10^{12} / \mathrm{m}^3$
  • B
    $4 \times 10^{11} / \mathrm{m}^3$
  • C
    $4 \times 10^{12} / \mathrm{m}^3$
  • Correct answer
    D
    $5.3 \times 10^{11} / \mathrm{m}^3$

Solution

Given $$ \begin{aligned} & n_1=1.6 \times 10^{16} / \mathrm{m}^3 \\ & n_e=4.8 \times 10^{20} / \mathrm{m}^3 \\ & n_h=? \end{aligned} $$ The concentration of holes in the semiconductor $$ \begin{aligned} n_1^2 & =n_e \times n_h \\ \left(1.6 \times 10^{16}\right)^2 & =4.8 \times 10^{20} \times n_h \\ n_h & =\frac{2.56 \times 10^{32}}{4.8 \times 10^{20}}=5.3 \times 10^{11} / \mathrm{m}^3 \end{aligned} $$

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

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