Question

Suppose a pure $Sicrystal$ has $5\times {10}^{28}atoms{m}^{-3}.$ It is doped by $1ppm$ concentration of pentavalent $As.$ Calculate the number of electrons and holes. Given that $n_i=1.5\times {10}^{16}m{}^{-}3.$

Answer 1

$n=5\times {10}^{28}$
$1PPM=\frac{1}{{10}^6}$
So no. of atom is pentavale $=\frac{5\times {10}^{28}}{{10}^6}$
$=5\times {10}^{22}$
Son no atom is $=5\times {10}^{22}\frac{1}{m^3}$
$h_n=\frac{h{i}^2}{he}$
$n_{nale}=\frac{(1.5\times {10}^{16})}{5\times {10}^{22}}$
$=0.3\times {10}^{10}$
$=3\times {10}^9$