Question

The electron density of intrinsic semiconductor at room temperature is ${10^{16}}\;{{\rm{m}}^{ - 3}}$. When doped with a trivalent impurity, the electron density is decreased to ${10^{14}}\;{{\rm{m}}^{ - 3}}$ at the same temperature. The majority carrier density is:
A. ${10^{16}}\;{{\rm{m}}^{ - 3}}$
B. ${10^{18}}\;{{\rm{m}}^{ - 3}}$
C. ${10^{21}}\;{{\rm{m}}^{ - 3}}$
D. ${10^{20}}\;{{\rm{m}}^{ - 3}}$
E. ${10^{19}}\;{{\rm{m}}^{ - 3}}$

Answer 1

Hint:The electrons and holes are the charge carriers in semiconductors. The charge carrier density varies with the temperature of the semiconductor. The product of carrier density of majority carrier and minority carriers is equal to the square of electron density at room temperature.

Complete Step by Step Answer:
The electron density of the semiconductor is ${10^{16}}\;{{\rm{m}}^{ - 3}}$, reduced electron density is ${10^{14}}\;{{\rm{m}}^{ - 3}}$.

Write the equation to calculate the majority carrier density.
${n_e} \times {n_h} = {n^2}$
Here, ${n_e}$ is the minority carrier density, ${n_h}$ is the majority carrier density, $n$ is the electron density.
Substitute $n$ as ${10^{16}}\;{{\rm{m}}^{ - 3}}$ and ${n_e}$ as ${10^{14}}\;{{\rm{m}}^{ - 3}}$ in the above equation.
$\begin{array}{l}
\left( {{{10}^{14}}\;{{\rm{m}}^{ - 3}}} \right){n_h} = {\left( {{{10}^{16}}\;{{\rm{m}}^{ - 3}}} \right)^2}\\
{n_h} = {10^{18}}\;{{\rm{m}}^{ - 3}}
\end{array}$

Therefore, the majority carrier density is ${10^{18}}\;{{\rm{m}}^{ - 3}}$and the option (B) is correct.

Note:Make sure to remember the majority or minority carrier by the type of the semiconductor and make sure that you know about carrier concentration in semiconductor physics.