A intrinsic semiconductor has $5\times 10^{28}\,  atoms/m^3$. It is doped by 0.01 ppm. concentration of arsenic. If $ni=1.5 \times 10^{16} m^3$, then number of holes in the n-type semiconductor will be

$4.5 \times 10^{11}/m^3$

The correct answer is option (4) : $4.5 \times 10^{11}/m^3$

0.01 atom of Si doped out of $10^6$ atom 1 ppm.

In $5\times 10^{28}$ atom number of doped $=\frac{5\times 10^{28}}{\frac{10^6}{0.01}}$

$=5 \times 10^{20}$

1 as atom creates $1e^{-}$ excess so number of excess electron $=5\times 10^{20}=ne$

also $n_en_h=n^2_1$

$n_h=\frac{n^2_i}{n_e}=\frac{(1.5\times 10^{16})^2}{5\times 10^{20}}$

$=0.45 \times 10^{12}$

$=4.5 \times 10^{11}/m^3$