The number of silicon atoms per $m^{3}$ is $5 \times 10^{28}$. This is doped simultaneously with $5 \times 10^{22}$ atoms per $m^{3}$ of Arsenic and $5 \times 10^{20}$ atoms per $m^{3}$ of Indium. Calculate the number of electrons and holes. Given that $n_{i} = 1.5 \times 10^{16} \; m^{-3}$. Is the material $n$-type or $p$-type?

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(A) Number of silicon atoms,$N = 5 \times 10^{28} \; \text{atoms}/m^{3}$.
Number of arsenic atoms (donor),$n_{D} = 5 \times 10^{22} \; \text{atoms}/m^{3}$.
Number of indium atoms (acceptor),$n_{A} = 5 \times 10^{20} \; \text{atoms}/m^{3}$.
Intrinsic carrier concentration,$n_{i} = 1.5 \times 10^{16} \; m^{-3}$.
Since $n_{D} > n_{A}$,the net donor concentration is $n_{e} \approx n_{D} - n_{A} = 5 \times 10^{22} - 0.05 \times 10^{22} = 4.95 \times 10^{22} \; \text{electrons}/m^{3}$.
Using the mass action law,$n_{e} n_{h} = n_{i}^{2}$.
$n_{h} = \frac{n_{i}^{2}}{n_{e}} = \frac{(1.5 \times 10^{16})^{2}}{4.95 \times 10^{22}} = \frac{2.25 \times 10^{32}}{4.95 \times 10^{22}} \approx 4.55 \times 10^{9} \; \text{holes}/m^{3}$.
Since $n_{e} > n_{h}$,the material is an $n$-type semiconductor.

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Similar Questions

Consider an $n$-type semiconductor in which $n_e$ and $n_h$ are the number of electrons and holes,respectively.
$(A)$ Holes are minority carriers.
$(B)$ The dopant is a pentavalent atom.
$(C)$ $n_e n_h \neq n_i^2$ (where $n_i$ is the number of electrons or holes in the semiconductor when it is in its intrinsic form).
$(D)$ $n_e n_h \geq n_i^2$.
$(E)$ The holes are not generated due to the donors.
Choose the correct answer from the options given below.

$P$-type semiconductors are made by adding which impurity element?

The process of adding impurities to a pure semiconductor is called

In an $n$-type semiconductor,the free electrons donated by the impurity atoms occupy energy levels in:

The charge carriers in a $p$-type semiconductor are

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