Answer:
Since, we know, \[{{n}_{e}}{{n}_{h}}=n_{i}^{2}\] The number of holes, \[{{n}_{h}}=n_{i}^{2}/{{n}_{e}}\] \[=\frac{{{(1.5\times {{10}^{16}})}^{2}}}{5\times {{10}^{22}}}\cong 4.5\times {{10}^{9}}{{m}^{-3}}\] The number of electrons thermally generated \[({{n}_{i}}\tilde{\ }{{10}^{6}}{{m}^{-3}})\] are negligibly small as compared to those produced by doping. \[\therefore \] \[{{n}_{e}}\simeq {{N}_{D}}\] 1 ppm = 1 part per million \[{{N}_{D}}=\frac{5\times {{10}^{28}}}{{{10}^{6}}}=5\times {{10}^{22}}{{m}^{-3}}\] \[\therefore \] \[{{n}_{e}}\cong 5\times {{10}^{22}}{{m}^{-3}}\]
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