Question

The product of the hole concentration and the conduction electron concentration turns out to be independent of the amount of any impurity doped. The concentration of conduction electrons in germanium is $6\times {10}^{19}$ per cubic metre. When some phosphorus impurity is doped into a germanium sample, the concentration of conduction electrons increases to $2\times {10}^{23}$ per cubic metre. Find the concentration of the holes in the doped germanium.

Answer 1

(Number of holes) (Number of conduction electrons) = Constant

At first,

Number of conduction electrons = $6\times {10}^{19}$

Number of holes = $6\times {10}^{19}$

After doping,

Number of conduction electrons = $2\times {10}^{23}$

Number of holes = x

$(6\times {10}^{19})(6\times {10}^{19})=(2\times {10}^{23})x$

$\frac{6\times 6\times {10}^{19+19}}{2\times {10}^{23}}=x$

$x=18\times {10}^{15}=1.8\times {10}^{16}$