Solveeit Logo

Question

Question: A semiconductor has an electron concentration of \(8 \times {10^{13}}c{m^{ - 3}}\) and a hole concen...

A semiconductor has an electron concentration of 8×1013cm38 \times {10^{13}}c{m^{ - 3}} and a hole concentration of 5×1013cm35 \times {10^{13}}c{m^{ - 3}}. The electron mobility is 25000cm2V1S125000c{m^2}{V^{ - 1}}{S^{ - 1}} and the hole mobility is 100cm2V1S1100c{m^2}{V^{ - 1}}{S^{ - 1}} Then,
A. The semiconductor is n-type
B. The conductivity is 320mmhocm1320mmhoc{m^{ - 1}}
C. Both A and B
D. None of these

Explanation

Solution

Since the concentration of electrons is greater than the concentration of hole so the semiconductor will be of n-type. So we have to calculate the conductivity of the semiconductor and also we know conductivity is the opposite of resistivity.

Formula used:
The conductivity of the semiconductor is.
neυe+nhυh\Rightarrow {n_e}{\upsilon _e} + {n_h}{\upsilon _h}
Where,
ne{n_e} is the concentration of electron
nh{n_h} is the concentration of hole
υe{\upsilon _e} is the mobility of electron
υh{\upsilon _h} is the mobility of the hole

Complete step by step Solution: In the question, it is given that the concentration of electron, concentration of hole, mobility of electron, and the mobility of hole all these terms are given and they are asking what will be the conductivity of the material.
Since from the question we can see that the ne>nh{n_e} > {n_h} means the concentration of electrons is greater than the concentration of hole.
So this will make us sure that It is a semiconductor of n-type.
Now we have to calculate the conductivity of the semiconductor.
By using the formula we will calculate it,
neυe+nhυh\Rightarrow {n_e}{\upsilon _e} + {n_h}{\upsilon _h}
Substituting the values, we get
1.6×1019((8×1013cm3)(25000cm2V1S1)+(5×1013cm3)(25000cm2V1S1)(100cm2V1S1))mhocm1\Rightarrow 1.6 \times {10^{19}}\left( {\left( {8 \times {{10}^{13}}c{m^{ - 3}}} \right)\left( {25000c{m^2}{V^{ - 1}}{S^{ - 1}}} \right) + \left( {5 \times {{10}^{13}}c{m^{ - 3}}} \right)\left( {25000c{m^2}{V^{ - 1}}{S^{ - 1}}} \right)\left( {100c{m^2}{V^{ - 1}}{S^{ - 1}}} \right)} \right)mhoc{m^{ - 1}}On solving the above solution we will get the following
0.32mhocm1\Rightarrow 0.32mhoc{m^{ - 1}}
On further converting the above solution, we get
320mmhocm1\Rightarrow 320mmhoc{m^{ - 1}}

Hence, the conductivity of the n-type semiconductor will be 320mmhocm1.320mmhoc{m^{ - 1}}.

Note: Properties of the semiconductors are as follows: Semiconductor acts like a dielectric at Zero Kelvin. By raising the temperature, it works as a conductor. Due to their exceptional electrical properties, semiconductors are changed by doping to form semiconductor devices appropriate for energy conversion, switches, and amplifiers. Lesser power losses. Their impedance is on top of conductors however lesser than insulators. The resistance of semiconductor materials decreases with the rise in temperature and vice-versa.