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If the ratio of electron and hole currents in a semiconductor is $\frac{7}{4}$ and the ratio of drift velocities of electrons and holes is $\frac{5}{4}$, then ratio of concentrations of electrons and holes will be
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The correct answer is:
$7: 5$
Ratio of electron \& Hole current,
$$
\frac{\mathrm{I}_{\mathrm{e}}}{\mathrm{I}_{\mathrm{n}}}=\frac{7}{4}
$$
Ratio of velocities, $\frac{v_e}{v_n}=\frac{5}{4}$
Current is given by: $\mathrm{I}=\mathrm{ne}_{\mathrm{d}}$.
$$
\mathrm{n}=\frac{\mathrm{I}}{\mathrm{eAV}}
$$
Ratio of concentration of electrons \& holes:
$$
\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{n}}}=\frac{\mathrm{I}_{\mathrm{e}}}{\mathrm{ev}_{\mathrm{e}} \mathrm{A}} \times \frac{\mathrm{ev}_{\mathrm{n}} \mathrm{A}}{\mathrm{I}_{\mathrm{n}}}=\frac{7}{4} \times \frac{4}{5}=\frac{7}{5}
$$
$$
\frac{\mathrm{I}_{\mathrm{e}}}{\mathrm{I}_{\mathrm{n}}}=\frac{7}{4}
$$
Ratio of velocities, $\frac{v_e}{v_n}=\frac{5}{4}$
Current is given by: $\mathrm{I}=\mathrm{ne}_{\mathrm{d}}$.
$$
\mathrm{n}=\frac{\mathrm{I}}{\mathrm{eAV}}
$$
Ratio of concentration of electrons \& holes:
$$
\frac{\mathrm{n}_{\mathrm{e}}}{\mathrm{n}_{\mathrm{n}}}=\frac{\mathrm{I}_{\mathrm{e}}}{\mathrm{ev}_{\mathrm{e}} \mathrm{A}} \times \frac{\mathrm{ev}_{\mathrm{n}} \mathrm{A}}{\mathrm{I}_{\mathrm{n}}}=\frac{7}{4} \times \frac{4}{5}=\frac{7}{5}
$$
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