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In a sample of semi conductor mobilities...

In a sample of semi conductor mobilities of electrons and holes are `24 xx 10^3 cm^2 V^-1 S^-1` and `0.2 xx 10^3 cm^2 V^-1 S^-1` respectively. If the density of electrons is `0.8 xx 10^14 cm^-3` and that of holes is `0.4 xx 10^14 cm^-3. Find the nature of semi-conductor and its conductivity.

Text Solution

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`n_(e)= 0.8 xx 10^(14) Cm^(-3) = 0.8 xx 10^(20) m^(-3)`
`n_(h)= 0.4 xx 10^(14) Cm^(-3) = 0.4 xx 10^(20) m^(-3)`
`mu_(e)= 24 xx 10^(3) Cm^(2) V^(-1) S^(-1) = 2.4 m^(2) V^(-1) S^(-1)`
`mu_(h)= 0.2 xx 10^(3) Cm^(2) V^(-1) S^(-1)= 0.02 m^(2) V^(-1) S^(-1)`
`because n e gt nh`
`therefore` It is a n-type semi Conductor.
Conductivity `= sigma = e (n_(e) mu_(3) + n_(h) mu_(h))`
`sigma = 1.6 xx 10^(-19) (0.8 xx 10^(20) xx 2.4 + 0.4 xx 10^(20) xx 0.02)`
`=1.6 xx 10^(-19) xx 10^(20) (1.92 + 0.008)`
`=1.6 xx 1.928`
`sigma = 30.848 Omega m`
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