A) n-type
B) p-type
C) intrinsic
D) p-n type
Correct Answer: A
Solution :
In a doped semiconductor, the number density of electrons and holes is not equal. But it can be established that \[{{n}_{e}}{{n}_{h}}=n_{i}^{2}\] where \[{{n}_{e}},\,\,{{n}_{h}}\] are the number density of electrons and holes, respectively and \[{{n}_{i}}\] is tne number density of intrinsic carriers (i.e., electrons or holes) in a pure semiconductor. In n-type semiconductor, the number density of electrons is nearly equal to the number, density of donor atoms \[{{N}_{d}}\] and is very large as compared to number density of holes. Hence, \[{{n}_{e}}\approx {{N}_{d}}>>{{n}_{h}}\]. In p-type semiconductor, the number density of holes is nearly equal to the number density of acceptor atoms \[{{N}_{a}}\] and is very large as compared to number density of electrons. Hence, \[{{n}_{h}}\approx {{N}_{a}}>>{{n}_{e}}\] Since, electron concentration is \[7\times {{10}^{13}}\,c{{m}^{-3}}\] and hole concentration is \[5\times {{10}^{12}}\,c{{m}^{-3}}\], the semiconductor is n-type.
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