Q118 / 120 · VITEEE 2010 · Physics

A potential difference of $2 \mathrm{~V}$ is applied between the opposite faces of a Ge crystal plate of area $1 \mathrm{~cm}^{2}$ and thickness $0.5 \mathrm{~mm}$. If the concentration of electrons in Ge is $2 \times 10^{19} / \mathrm{m}^{2}$ and mobilities of electrons and holes are $0.36 \mathrm{~m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$ and $0.14$ $\mathrm{m}^{2} \mathrm{~V}^{-1} \mathrm{~s}^{-1}$ respectively, then the current flowing through the plate will be

  • A
    $0.25 \mathrm{~A}$
  • B
    $0.45 \mathrm{~A}$
  • C
    $0.56 \mathrm{~A}$
  • Correct answer
    D
    $0.64 \mathrm{~A}$

Solution

As we know, conductivity $$ \begin{aligned} \sigma &=n e\left(\mu_{e}+\mu_{h}\right) \\ &=2 \times 10^{19} \times 1.6 \times 10^{-19}(0.36+0.14) \\ &=1.6(\Omega \mathrm{m})^{-1} \end{aligned} $$ $$ \begin{array}{l} R=\rho \frac{l}{A}=\frac{l}{\sigma A}=\frac{0.5 \times 10^{-3}}{1.6 \times 10^{-4}}=\frac{25}{8} \Omega \\ i=\frac{V}{R}=\frac{2}{25 / 8}=0.64 \mathrm{~A} \end{array} $$

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

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