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Question: What is the conductivity of a semiconductor if electron density = \[5 \times {10^{12}}/c{m^3}\], and...

What is the conductivity of a semiconductor if electron density = 5×1012/cm35 \times {10^{12}}/c{m^3}, and hole density 8×1013/cm38 \times {10^{13}}/c{m^3}, (μe=2.3m2V−1s−1{\mu _e} = 2.3{m^2}{V^{ - 1}}{s^{ - 1}},μh=0.01m2V−1s−1{\mu _h} = 0.01{m^2}{V^{ - 1}}{s^{ - 1}})?
A) 5.634
B) 1.968
C) 3.421
D) 8.964

Explanation

Solution

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(μene+μhnh)e({\mu _e}{n_e} + {\mu _h}{n_h}) where we have provided (μe=2.3m2V−1s−1{\mu _e} = 2.3{m^2}{V^{ - 1}}{s^{ - 1}};μh=0.01m2V−1s−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(μene+μhnh)e({\mu _e}{n_e} + {\mu _h}{n_h})
Where e=1.6×10−19e = 1.6 \times {10^{ - 19}}Coulomb ; ne{n_e} is the electron density ; nh{n_h} is the hole density
μe{\mu _e} is the mobility of the electron ; μh{\mu _h} is the mobility of the hole .

Complete step by step solution:
Electron density =5×1012/cm35 \times {10^{12}}/c{m^3}
Converting it to standard form
⇒1cm3=10−6m3\Rightarrow 1c{m^3} = {10^{ - 6}}{m^3}
Electron density =5×1018m35 \times {10^{18}}{m^3}
⇒ne\Rightarrow {n_e}=5×1018m35 \times {10^{18}}{m^3};μe=2.3m2V−1s−1{\mu _e} = 2.3{m^2}{V^{ - 1}}{s^{ - 1}}
Hole density =8×1013/cm38 \times {10^{13}}/c{m^3}
Converting it to standard form
⇒1cm3=10−6m3\Rightarrow 1c{m^3} = {10^{ - 6}}{m^3}
Hole density =8×1019m38 \times {10^{19}}{m^3}
⇒nh\Rightarrow {n_h}=8×1019m38 \times {10^{19}}{m^3};μh=0.01m2V−1s−1{\mu _h} = 0.01{m^2}{V^{ - 1}}{s^{ - 1}}
The conductivity of the semiconductor is given by
Conductivity =e(μene+μhnh)e({\mu _e}{n_e} + {\mu _h}{n_h})
Putting all the values & e=1.6×10−19e = 1.6 \times {10^{ - 19}}Coulomb
⇒e(μene+μhnh)\Rightarrow e({\mu _e}{n_e} + {\mu _h}{n_h})
⇒1.6×10−19((2.3)(5×1018)+(0.01)(8×1019))\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 ,
⇒1.6×10−19(1.23×1019)\Rightarrow 1.6 \times {10^{ - 19}}(1.23 \times {10^{19}})
⇒1.6×1.23\Rightarrow 1.6 \times 1.23
Further simplifying the value of conductivity is
⇒1.968\Rightarrow 1.968
The conductivity of a semiconductor if electron density = 5×1012/cm35 \times {10^{12}}/c{m^3}, and hole density 8×1013/cm38 \times {10^{13}}/c{m^3}, (μe=2.3m2V−1s−1{\mu _e} = 2.3{m^2}{V^{ - 1}}{s^{ - 1}},μh=0.01m2V−1s−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.