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Numerical · Q32

Q.An intrinsic semiconductor has an electron (and hole) concentration of ni=1.5×1010 cm−3n_i = 1.5\times10^{10}\ \text{cm}^{-3} at room temperature. If it is doped with a pentavalent impurity so that the free-electron concentration rises to ne=5×1015 cm−3n_e = 5\times10^{15}\ \text{cm}^{-3}, find the new hole concentration nhn_h, using the law of mass action nenh=ni2n_e n_h = n_i^2.

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Given ni=1.5×1010 cm−3n_i=1.5\times10^{10}\ \text{cm}^{-3} and ne=5×1015 cm−3n_e=5\times10^{15}\ \text{cm}^{-3}, the law of mass action gives

nh=ni2ne=(1.5×1010)25×1015=2.25×10205×1015=4.5×104 cm−3n_h=\frac{n_i^2}{n_e}=\frac{(1.5\times10^{10})^2}{5\times10^{15}}=\frac{2.25\times10^{20}}{5\times10^{15}}=4.5\times10^{4}\ \text{cm}^{-3} …

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