Assignment5Solns

84 me b dc conductivity from electrons or holes are

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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!...
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