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A semiconductor has equal electron and hole concentration of $2 \times 10^8 \mathrm{~m}^{-3}$. On doping with a certain impurity, the electron concentration increases to $4 \times 10^{10} \mathrm{~m}^{-3}$, then the new hole concentration of the semiconductor is
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$10^6 \mathrm{~m}^3$
Electron concentration, $n_e=2 \times 10^8 \mathrm{~m}^{-3}$
Hole concentration, $n_h=2 \times 10^8 \mathrm{~m}^{-3}$
After doping with a impurity, the new electron concentration, $n_e^{\prime}=4 \times 10^{10} \mathrm{~m}^{-3}$
New hole concentration, $n_h^{\prime}=$ ?
We know that, $n_e \times n_h=n_i^2$
$$
\begin{array}{ll}
\text { and } & n_e^{\prime} \times n_h^{\prime}=n_i^2 \\
\therefore & n_e \times n_h=n_e^{\prime} \times n_h^{\prime} \\
& n_h^{\prime}=\frac{n_e \times n_h}{n_e^{\prime}}=\frac{2 \times 10^8 \times 2 \times 10^8}{4 \times 10^{10}} / \mathrm{m}^3=10^6 / \mathrm{m}^3 \\
\Rightarrow & n_h^{\prime}=10^6 \mathrm{~m}^{-3}
\end{array}
$$
Hole concentration, $n_h=2 \times 10^8 \mathrm{~m}^{-3}$
After doping with a impurity, the new electron concentration, $n_e^{\prime}=4 \times 10^{10} \mathrm{~m}^{-3}$
New hole concentration, $n_h^{\prime}=$ ?
We know that, $n_e \times n_h=n_i^2$
$$
\begin{array}{ll}
\text { and } & n_e^{\prime} \times n_h^{\prime}=n_i^2 \\
\therefore & n_e \times n_h=n_e^{\prime} \times n_h^{\prime} \\
& n_h^{\prime}=\frac{n_e \times n_h}{n_e^{\prime}}=\frac{2 \times 10^8 \times 2 \times 10^8}{4 \times 10^{10}} / \mathrm{m}^3=10^6 / \mathrm{m}^3 \\
\Rightarrow & n_h^{\prime}=10^6 \mathrm{~m}^{-3}
\end{array}
$$
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