Semiconductors

JEE Main level, numerical answer, medium

20 of 209

Question 20 of 209

In a semiconductor, the number density of intrinsic charge carriers at 27∘C27^{\circ}C is 1.5×1016m−31.5\times {10}^{16}{m}^{-3}. If the semiconductor is doped with an impurity atom, the hole density increases to 4.5×1022m−34.5\times {10}^{22}{m}^{-3}. The electron density in the doped semiconductor is _______×109m−3\times {10}^{9}{m}^{-3}

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Answer: 5

nenh=ni2{n}_{e}{n}_{h}={n}_{i}^{2}
ne=ni2nh=(1.5×1016)24.5×1022{n}_{e}=\dfrac{{n}_{i}^{2}}{{n}_{h}}=\dfrac{{\left(1.5\times {10}^{16}\right)}^{2}}{4.5\times {10}^{22}}
=1.5×1.5×10324.5×1022=\dfrac{1.5\times 1.5\times {10}^{32}}{4.5\times {10}^{22}}
5×109m−35\times {10}^{9}{m}^{-3}

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