Answer:
\[{{n}_{e}}=4.95\times {{10}^{22}}{{m}^{-3}}\] \[{{n}_{h}}=4.5\times {{10}^{9}}{{m}^{-3}}\] material is n-type. Given, \[{{N}_{D}}=5\times {{10}^{22}}{{m}^{-3}},{{N}_{A}}=5\times {{10}^{20}}{{m}^{-3}}\] \[{{n}_{i}}=1.5\times {{10}^{16}}{{m}^{-3}}\] For the semiconductor to remain electrically neutral, \[{{N}_{D}}-{{N}_{A}}={{n}_{e}}-{{n}_{h}}\] ?(i) Also \[{{n}_{e}}{{n}_{h}}=n_{i}^{2}\] \[\therefore \] \[{{({{n}_{e}}+{{n}_{h}})}^{2}}={{({{n}_{e}}-{{n}_{h}})}^{2}}+2{{n}_{e}}{{n}_{h}}\] \[={{({{N}_{D}}-{{N}_{A}})}^{2}}=4n_{i}^{2}\] \[{{n}_{e}}+{{n}_{h}}=\sqrt{{{({{N}_{D}}-{{N}_{A}})}^{2}}=4n_{i}^{2}}\] ?(ii) On adding Eqs. (i) and (ii), we get \[{{n}_{e}}=\frac{1}{2}({{N}_{D}}-{{N}_{A}})+\sqrt{{{({{N}_{D}}-{{N}_{A}})}^{2}}=4n_{i}^{2}}\] \[=\frac{1}{2}\left[ (5\times {{10}^{22}}-0.05\times {{10}^{22}}) \right.\] \[\left. +\sqrt{{{(4.95\times {{10}^{22}})}^{2}}+{{(1.5\times {{10}^{16}})}^{2}}} \right]\] \[\approx \frac{1}{2}\left[ 4.95\times {{10}^{22}}+\sqrt{{{(4.95\times {{10}^{22}})}^{2}}} \right]\] \[=4.95\times {{10}^{22}}{{m}^{-3}}\] Also \[{{n}_{h}}=\frac{n_{i}^{2}}{{{n}_{e}}}=\frac{{{(1.5\times {{10}^{16}})}^{2}}}{4.95\times {{10}^{22}}}=\frac{2.25\times {{10}^{32}}}{4.95\times {{10}^{22}}}\] \[=4.5\times {{10}^{9}}{{m}^{-3}}\] Since \[{{n}_{e}}>{{n}_{h}},\] the material is of n-type.
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