Question

A pure semiconductor has equal electron and hole concentration of $ {{10}^{16}}{{m}^{-3}} $ . Doping by indium increases $ {{n}_{h}} $ to $ 5\times {{10}^{22}}{{m}^{-3}} $ . Then, the value of $ {{n}_{e}} $ in the doped semiconductor is

• A. $ {{10}^{6}}/{{m}^{3}} $
• B. $ {{10}^{22}}/{{m}^{3}} $
• C. $ 2\times {{10}^{6}}/{{m}^{3}} $
• D. $ {{10}^{19}}/{{m}^{3}} $
• E. $ 2\times {{10}^{9}}/{{m}^{3}} $

Answer 1

Answer: (E)

$ {{({{n}_{i}})}^{2}}={{n}_{e}}{{n}_{h}} $
$ {{({{10}^{16}})}^{2}}={{n}_{e}}\times 5\times {{10}^{22}} $
$ \therefore $ $ {{n}_{e}}=\frac{{{10}^{16}}\times {{10}^{16}}}{5\times {{10}^{22}}} $
$ {{n}_{e}}=2\times {{10}^{9}}/{{m}^{3}} $