The number densities of electrons and holes in pure silicon at 273 K are equal and their values are $1.5 \times 10^{16} m^{-3}$. On doping, the hole density increases to $4.5 \times 10^{27} m^{-3}$. The electron density in the doped silicon will be:

$5 \times 10^4 m^{-3}$

The correct answer is Option (1) → $5 \times 10^4 m^{-3}$

For intrinsic (pure) semiconductors, the product of the number densities of electron ($n_e$) and holes ($n_h$) is -

$n_e×n_h={n_i}^2$ [$n_i$ = Constant]

$∴n_e=\frac{{n_i}^2}{n_h}=\frac{(1.5×10^{16})^2}{4.5×10^{27}}$

$=5×10^4m^{-3}$