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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
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2866 Upvotes
Verified Answer
The correct answer is:
$0.64 \mathrm{~A}$
As we know, conductivity
$$
\begin{aligned}
\sigma &=n e\left(\mu_{e}+\mu_{h}\right) \\
&=2 \times 10^{19} \times 1.6 \times 10^{-19}(0.36+0.14) \\
&=1.6(\Omega \mathrm{m})^{-1}
\end{aligned}
$$
$$
\begin{array}{l}
R=\rho \frac{l}{A}=\frac{l}{\sigma A}=\frac{0.5 \times 10^{-3}}{1.6 \times 10^{-4}}=\frac{25}{8} \Omega \\
i=\frac{V}{R}=\frac{2}{25 / 8}=0.64 \mathrm{~A}
\end{array}
$$
$$
\begin{aligned}
\sigma &=n e\left(\mu_{e}+\mu_{h}\right) \\
&=2 \times 10^{19} \times 1.6 \times 10^{-19}(0.36+0.14) \\
&=1.6(\Omega \mathrm{m})^{-1}
\end{aligned}
$$
$$
\begin{array}{l}
R=\rho \frac{l}{A}=\frac{l}{\sigma A}=\frac{0.5 \times 10^{-3}}{1.6 \times 10^{-4}}=\frac{25}{8} \Omega \\
i=\frac{V}{R}=\frac{2}{25 / 8}=0.64 \mathrm{~A}
\end{array}
$$
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