Question

If a semiconductor has an intrinsic carrier concentration of \[1.41\times {{10}^{16}}/{{\text{m}}^{3}}\], when doped with ${{10}^{21}}/{{\text{m}}^{3}}$ phosphorus atoms, then the concentration of holes$/{{\text{m}}^{3}}$ at room temperature will be:
A.$2\times {{10}^{21}}$
B.$2\times {{10}^{11}}$
C.$1.41\times {{10}^{10}}$
D.$1.41\times {{10}^{16}}$

Answer 1

Hint: Intrinsic carrier concentration refers to the number of electrons in the conduction band, or it can even be described as the number of holes in the valence band. Intrinsic semiconductors are those which do not have any impurities in them, i.e., they have not been doped.

Complete answer:
When the semiconductor is doped with phosphorus atoms then only the number of electrons will increase because phosphorus is a donor impurity and the doped semiconductor is a $n-$type semiconductor. In an $n-$type semiconductor, the major charge carriers are the electrons and thus, the number of holes will remain constant and will be equal to the intrinsic carrier concentration of the semiconductor i.e., \[1.41\times {{10}^{16}}/{{\text{m}}^{3}}\]. Therefore, the correct option is $D$.

Additional information:
The semiconductors for example silicon, germanium, etc. have a moderate conductivity, but to increase their conductivity they are doped with several elements like phosphorus, bismuth, boron, etc. There are two types of semiconductors formed after doping namely $p-$type semiconductors and $n-$type semiconductors. In a $p-$type semiconductor there is an excess of holes and the major charge carriers are the holes while in a $n-$type semiconductor there is an excess of electrons and the major charge carriers are these electrons.

Note:
If the semiconductor would have been doped with some other element other than phosphorus which is an acceptor, then a $p-$type semiconductor would have been formed in which the major charge carriers would have been holes. In that case the number of holes would have increased.