Q115 / 180 · AIIMS 2013 · Physics

Given that the mobility of electrons in Ge is $0.4 \mathrm{~m}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}$ and electronic charge is $1.6 \times 10^{-19} \mathrm{C}$. The number of donor atom $\left(\right.$ per $\mathrm{m}^3$ ) semiconductor of conductivity $500 \mathrm{mho} / \mathrm{m}$ is

  • Correct answer
    A
    $8 \times 10^{21}$
  • B
    $8 \times 10^{15}$
  • C
    $5 \times 10^{21}$
  • D
    $8 \times 10^{16}$

Solution

$\mathrm{U} \operatorname{sing}, \sigma=n_e q \mu_e$ Here, $\sigma=500 \mathrm{mho} / \mathrm{m}$ $e=1.6 \times 10^{-19} \mathrm{C}$ $\mu_e=0.4 \mathrm{~m}^2 \mathrm{~V}^{-1} \mathrm{~s}^{-1}$ $\therefore \quad n_e=\frac{500}{1.6 \times 10^{-19} \times 0.4}$ $=7.8 \times 10^{21} \simeq 8 \times 10^{21} \mathrm{~m}^{-3}$

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

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