84 me b dc conductivity from electrons or holes are

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ivity metal covering the wavelengths that are not absorbed). 3. a) This is just straight application of the cyclotron frequency, easy. Effective mass of electron = 0.14 me Effective mass of hole = 0.84 me b) dc conductivity from electrons or holes are given by !!" = (!!,! )! ! !/!∗ . Have to calculate ne,h = Nc,v*Boltzmann factor for electrons or holes. Straight application of the equations I gave you relating the effective masses to the prefactors: Nc(200K) = 7.1*10^(17) cm^( ­3) Np(200K)=1.05*10^(19) cm^( ­3) The chemical potential is given by the equation above, which lies 0.046 eV closer to the conduction band from the middle of the gap. This results in two different Boltzmann factors: Boltzmann(electrons) = 0.186 Boltzmann(holes) = 0.000891 i.e. it makes a huge difference! ne = Nc*Boltzmann(electrons) = 1.32*10^(17) cm ­3 nh = Nv*Boltzmann(holes) = 0.093*10^(17) cm ­3 Conductivity for electrons = 8000 (Ohm cm) ­1 Conductivity for holes = 9.3 (Ohm cm) ­1 i.e. while both electrons and holes are present, the conductivity is predominantly determined by the electrons which is the lighter of the two!...
View Full Document

This document was uploaded on 02/04/2014.

Ask a homework question - tutors are online