A) n-type with electron concentration \[{{n}_{e}}=5\times {{10}^{22}}{{m}^{-3}}\]
B) p-type with electron concentration \[{{n}_{e}}=2.5\times {{10}^{10}}{{m}^{-3}}\]
C) n-type with electron concentration \[{{n}_{e}}=2.5\times {{10}^{23}}{{m}^{-3}}\]
D) p-type having electron concentration \[{{n}_{e}}=5\times {{10}^{9}}{{m}^{-3}}\]
Correct Answer: D
Solution :
[d] \[{{n}_{i}}^{2}={{n}_{e}}{{n}_{h}}\] \[{{(1.5\times {{10}^{16}})}^{2}}={{n}_{e}}(4.5\times {{10}^{22}})\] \[\Rightarrow \,\,\,\,\,{{n}_{e}}=0.5\times {{10}^{10}}\] or \[{{n}_{e}}=5\times {{10}^{9}}\] Given \[{{n}_{h}}=4.5\times {{10}^{22}}\Rightarrow {{n}_{h}}>>{{n}_{e}}\] \[\therefore \] Semiconductor is p-type and \[{{n}_{e}}=5\times {{10}^{9}}\,{{m}^{-3}}.\]You need to login to perform this action.
You will be redirected in
3 sec