question_answer

Suppose a pure Si crystal has \[5\times {{10}^{28}}\] atoms \[{{m}^{-3}}.\] It is doped by 1 ppm concentration of pentavalent As. Calculate the number of electrons and holes. Given that, \[{{n}_{i}}=1.5\times {{10}^{16}}{{m}^{-3}}\]

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 \[{{N}_{D}}=\frac{5\times {{10}^{28}}}{{{10}^{6}}}=5\times {{10}^{22}}{{m}^{-3}}\] \[\therefore \]      \[{{n}_{e}}\cong 5\times {{10}^{22}}{{m}^{-3}}\]