Question

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 holes. Given that $n_i=1.5\times {10}^{16}{m}^{-1}$.

• A. $4.54\times {10}^9{m}^{-3}$
  
• B. $4.54\times {10}^{10}{m}^{-3}$
• C. $4.95\times {10}^9{m}^{-3}$
• D. $4.95\times {10}^{10}{m}^{-3}$

Answer 1

The correct option is A $4.54\times {10}^9{m}^{-3}$

The Arsenic atoms density, $J_{Ar}=5\times {10}^{22}{m}^{-3}$

The Indium atoms density, $J_{In}=5\times {10}^{20}{m}^{-3}$

Therefore, the number of electrons, $n_e=\left[\left(5\times {10}^{22}\right)-\left(5\times {10}^{20}\right)\right]m^{-3}=4.95\times {10}^{22}{m}^{-3}$

The number of holes per $m^{-3}$ is-

$n_h=\frac{{\left(n_i\right)}^2}{n_e}$

$n_h=\frac{{\left(1.5\times {10}^{16}\right)}^2}{4.95\times {10}^{22}}=4.54\times {10}^9$