Question

Pure Si at $300K$ has equal electron and hole concentration of $1.5\times {10}^{16}/m^3$. Doping by indium increases $n_h$ to $4.5\times {10}^{22}/m^3$. What is $n_e$ in doped silicon?

• A. $5\times {10}^9/m^3$
• B. $1.5\times {10}^{16}/m^3$
• C. $4.5\times {10}^{22}/m^3$
• D. $4.5\times {10}^{16}/m^3$

Answer 1

The correct option is A $5\times {10}^9/m^3$

According to the mass action law, the product of hole concentration $n_h$ and free electron concentration $n_e$ remains constant at constant temperature.

Let $n_h^{{}^{\prime }}$ be the concentration of the holes before doping and $n_e^{{}^{\prime }}$ be the concentration of the free electrons before doping.
Let $n_h$ be the concentration of the holes after doping and $n_e$ be the concentration of the free electrons after doping

Then, $n_h{n}_e=n_h^{{}^{\prime }}{n}_e^{{}^{\prime }}$
$(1.5\times {10}^{16}{)}^2=4.5\times {10}^{22}\times n_e^{{}^{\prime }}$
$n_e^{{}^{\prime }}=5\times {10}^9/m^3$