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 $5\times {10}^{20}$ per $m^3$ atoms 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?

Step 1: Find the number of electrons.
Hint: Arsenic is pentavalent dopant and Indium is trivalent dopant.
Given,
Number of silicon atoms $=5\times {10}^{28}\text{atoms per}$ $m^3$
Number of Arsenic atoms, $n_{As}=5\times {10}^{22}\text{atoms per}{m}^3$
Number of Indium atoms,
$n_{In}=5\times {10}^{20}\text{atoms}$ per $m^3$
As we know, Arsenic is pentavalent dopant and Indium is trivalent dopant.
So, holes of Indium atoms are balanced by electrons of Arsenic atoms.
No of electrons,
$n_e=n_{As}-n_{In}$
$n_e=(5\times {10}^{22})-(5\times {10}^{20})$
$n_e=4.95\times {10}^{22}$

Step 2: Find the number of holes.
Formula used: $n_e{n}_h=n_i^2$
Given,
intrinsic concentration,
$n_i=1.5\times {10}^{16}{m}^{-3}$
The electron and hole concentration in a semiconductor in thermal equilibrium is given by,
$n_e{n}_h=n_i^2$
Number of holes, $n_h=\frac{n_i^2}{n_e}$
$n_h=\frac{(1.5\times {10}^{16}{)}^2}{4.95\times {10}^{22}}$
$n_h=4.54\times {10}^9$
Here, $n_e>>n_h$
Therefore, material is n-type.
Final answer : $4.95\times {10}^{22},4.54\times {10}^9;$ n-type.