An n-type semiconductor has an impurity level \[20\,meV\] below the conduction band. In the thermal collision, transferable energy is \[kT\]. The value of T for which electrons jumps to conduction band is
A. 232 K
B. 348 K
C. 400 K
D. 600 K

Hint: The valence electrons in the n-type semiconductor are in impurity level. Since the difference in the conduction band and impurity level is 20 meV, the thermal energy \[kT\] will cause the electrons to jump into conduction level.

Complete step by step answer:
We have given that the impurity level is below the \[20\,meV\] conduction band. If the transferable energy is \[kT\], the electrons from the valence band will jump to the conduction band.

Therefore, we can say if the energy of the electron is \[kT\], the electrons start to jump into the conduction band.
\[kT = 20\,meV\]
\[ \Rightarrow kT = 20 \times {10^{ - 3}}eV\]

Here, T is temperature and k is Boltzmann constant and its value is \[8.62 \times {10^{ - 5}}eV/K\]. Therefore, if we substitute \[8.62 \times {10^{ - 5}}eV/K\] for k in the above equation, we will get the temperature at which the electrons jump into the conduction band.
\[T = \dfrac{{20 \times {{10}^{ - 3}}eV}}{{8.62 \times {{10}^{ - 5}}eV/K}}\]
\[ \Rightarrow T = 232\,K\]

So, the correct answer is “Option A”.

Additional information: In n-type semiconductor, pentavalent impurity is added into a pure semiconductor. Therefore, electrons are the majority carried in n-type semiconductors. The Fermi-level or impurity level is close to the conduction band and is an n-type semiconductor. By the application of thermal energy, the fifth electron of the pentavalent impurity from the Fermi level jumps into the conduction band.

Note:
The value of Boltzmann constant k should be taken in eV if the energy is given in eV. To answer such types of questions, the factors that should be on your fingertips is what causes semiconductor to become n-type and p-type, the Fermi level or impurity level.