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Show that for a room-temperature semiconductor with a band gap of 1eV, a temperature rise of 4K would raise the conductivity by about 30%.

Short Answer

Expert verified

The rise in conductivity of the semiconductor is30% .

Step by step solution

01

Given information

The band gap of the semiconductor is, Eg=1eV.

The initial room temperature is, Ti=300K.

The temperature rise isTf-Ti=4K,.

The final temperature is, Tf=304K.

02

Conductivity of semiconductors

For a semiconductor, the value of its conductivity relies upon the room temperature as well as the band gap energy of the material.

The formula showing the relationship between the conductivity and the temperature of the room is given below,

Nexcited=DkBTe-Eg2kBT

Here, D is the constant, T is the temperature of semiconductor, Egis the energy band gap of the semiconductor, andkB is the Boltzmann onstant.

03

Increase in conductivity

The ratio of the final conductivity to the initial conductivity of the semiconductor in terms of the ratio between the initial and final temperature can be calculated by using following relation,

NfNi=Tfe-Eg2kBTfTie-Eg2kBTi

Here,Nfis the final number of electrons in conduction band,Niis the initialof electrons in conduction band, and kBis the Boltzmann constant, its value is1.3810-23J/K.

Multiplying TiTfboth sides,

TiTfNfNi=TiTfTfTie-Eg2kBTf+Eg2kBTiTiTfNfNi=eEgTf-Ti2kBTiTf

Putting the values,

300K304KNfNi=e1.610-19J21.3810-23J/K4K300K304K0.987NfNi=e2.5410-51040.987NfNi=e0.254

Solving further,

NfNi=1.290.987Nf=1.3Ni

So, the rise in conductivity of the semiconductor is given by,

Nf-NiNi100=0.254Nf-NiNi100=1.3Ni-NiNi100Nf-NiNi100=0.3100Nf-NiNi100=30%

Hence proved, the rise in conductivity is30%

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