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In Germanium it is found that concentration of conducting electron is $3 \times 10^{19}$ per cubic meter. When Phosphorus impurity is doped the concentration of conducting electron increases to $2 \times 10^{23}$ per cubic meter. If product of hole concentration and conducting electron is independent of amount of impurity find the hole concentration in doped Germanium.

$(a)\;2.8 \times 10^{15} \\ (b)\;4.5 \times 10^{16} \\ (c)\;1.8 \times 10^{16} \\ (d)\;4.5 \times 10^{15}$

The number of conducting electrons $\times$ number of holes = constant
Before doping, the number of conducting electrons = number of holes
After doping, the number of conducting electron is $2 \times 10^{23}$
$\therefore 3 \times 10^9 \times 3 \times 10^{19} =2 \times 10^{23} \times x$
$x= 4.5 \times 10^{15}$
Hence d is the correct answer.
edited Jul 13, 2014