Question

What is the conductivity of a semiconductor if electron density = \[5 \times {10^{12}}/c{m^3}\], and hole density \[8 \times {10^{13}}/c{m^3}\], (\[{\mu _e} = 2.3{m^2}{V^{ - 1}}{s^{ - 1}}\],\[{\mu _h} = 0.01{m^2}{V^{ - 1}}{s^{ - 1}}\])?
A) 5.634
B) 1.968
C) 3.421
D) 8.964

Answer 1

Hint: First convert the values of electron density and hole density into standard metric form , then use the formula of conductivity of the semiconductor is given by
Conductivity =\[e({\mu _e}{n_e} + {\mu _h}{n_h})\] where we have provided (\[{\mu _e} = 2.3{m^2}{V^{ - 1}}{s^{ - 1}}\];\[{\mu _h} = 0.01{m^2}{V^{ - 1}}{s^{ - 1}}\]) solving this equation gives the value of conductivity of the semiconductor & this is our required answer.

Formula used:
The conductivity of the semiconductor is given by
Conductivity =\[e({\mu _e}{n_e} + {\mu _h}{n_h})\]
Where \[e = 1.6 \times {10^{ - 19}}\]Coulomb ; \[{n_e}\] is the electron density ; \[{n_h}\] is the hole density
\[{\mu _e}\] is the mobility of the electron ; \[{\mu _h}\] is the mobility of the hole .

Complete step by step solution:
Electron density =\[5 \times {10^{12}}/c{m^3}\]
Converting it to standard form
\[ \Rightarrow 1c{m^3} = {10^{ - 6}}{m^3}\]
Electron density =\[5 \times {10^{18}}{m^3}\]
\[\Rightarrow {n_e}\]=\[5 \times {10^{18}}{m^3}\];\[{\mu _e} = 2.3{m^2}{V^{ - 1}}{s^{ - 1}}\]
Hole density =\[8 \times {10^{13}}/c{m^3}\]
Converting it to standard form
\[ \Rightarrow 1c{m^3} = {10^{ - 6}}{m^3}\]
Hole density =\[8 \times {10^{19}}{m^3}\]
\[\Rightarrow {n_h}\]=\[8 \times {10^{19}}{m^3}\];\[{\mu _h} = 0.01{m^2}{V^{ - 1}}{s^{ - 1}}\]
The conductivity of the semiconductor is given by
Conductivity =\[e({\mu _e}{n_e} + {\mu _h}{n_h})\]
Putting all the values & \[e = 1.6 \times {10^{ - 19}}\]Coulomb
\[ \Rightarrow e({\mu _e}{n_e} + {\mu _h}{n_h})\]
\[ \Rightarrow 1.6 \times {10^{ - 19}}\left( {(2.3)(5 \times {{10}^{18}}) + (0.01)(8 \times {{10}^{19}})} \right)\]
Simplifying the equation we get ,
\[ \Rightarrow 1.6 \times {10^{ - 19}}(1.23 \times {10^{19}})\]
\[ \Rightarrow 1.6 \times 1.23\]
Further simplifying the value of conductivity is
\[ \Rightarrow 1.968\]
The conductivity of a semiconductor if electron density = \[5 \times {10^{12}}/c{m^3}\], and hole density \[8 \times {10^{13}}/c{m^3}\], (\[{\mu _e} = 2.3{m^2}{V^{ - 1}}{s^{ - 1}}\],\[{\mu _h} = 0.01{m^2}{V^{ - 1}}{s^{ - 1}}\]) is 1.968.

Hence option B is the correct option.

Note: The conductivity of a semiconductor material has an electrical conductivity falling between that of a conductor ( metals like copper , aluminium ) and an insulator .
The resistivity of the semiconductor falls as its temperature rises , it is just the opposite phenomenon as of the metals.