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Homework 6 { Solution - Michigan State University

PHY 491: Atomic, Molecular, and Condensed Matter PhysicsMichigan State University , Fall Semester 2012 Solve by: Wednesday, October 17, 2012 Homework 6 (i) Calculate the density of states of the electron gas in 2 and 1 dimensions.(ii) Derive expressions for the Fermi energy in atomic units, where the energy is expressed in Hartreeand the length is expressed in Bohr radius.(iii) Consider a 2D electron gas with the density of 1011cm 2. Express this density in atomic is the Fermi energy for this 2D electron gas? Solution :(i) In 2 dimensions (2D),D(E)dE= 2 2 k dk(2 L) ~k= h2k22m k dk=m h2dEwe getD(E) =A2 (2m h2),independent 1 dimension (1D),D(E)dE= 2 2dk(2 L)with one factor of 2 for spin and the other factor of 2 forkand k. Withk= 2m h2E1/2 dk=12 2m h2E 1/2dEwe getD(E) =L (2m h2)1/2E 1/2.(ii) In 2D,E F=N 1A , whereE F=EFHartreeandA = 1D,E F= 28(NL )2, whereE F=EFHartreeandL =LaB.

Homework 6 { Solution 6.1. (i) Calculate the density of states of the electron gas in 2 and 1 dimensions. (ii) Derive expressions for the Fermi energy in atomic units, where the energy is expressed in Hartree and the length is expressed in Bohr radius.

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Transcription of Homework 6 { Solution - Michigan State University

1 PHY 491: Atomic, Molecular, and Condensed Matter PhysicsMichigan State University , Fall Semester 2012 Solve by: Wednesday, October 17, 2012 Homework 6 (i) Calculate the density of states of the electron gas in 2 and 1 dimensions.(ii) Derive expressions for the Fermi energy in atomic units, where the energy is expressed in Hartreeand the length is expressed in Bohr radius.(iii) Consider a 2D electron gas with the density of 1011cm 2. Express this density in atomic is the Fermi energy for this 2D electron gas? Solution :(i) In 2 dimensions (2D),D(E)dE= 2 2 k dk(2 L) ~k= h2k22m k dk=m h2dEwe getD(E) =A2 (2m h2),independent 1 dimension (1D),D(E)dE= 2 2dk(2 L)with one factor of 2 for spin and the other factor of 2 forkand k. Withk= 2m h2E1/2 dk=12 2m h2E 1/2dEwe getD(E) =L (2m h2)1/2E 1/2.(ii) In 2D,E F=N 1A , whereE F=EFHartreeandA = 1D,E F= 28(NL )2, whereE F=EFHartreeandL =LaB.

2 (iii)NA= 1011cm 2,NA =NA/a2B=NAa2B= 1011cm 2 ( 10 8cm)2= 10 5,1E F= NA = 10 5,EF= 10 5 Hartree = 10 5 eV = 10 The3He atom is a fermion with spin 1/2 (why?). The density of liquid3He is g/cm3nearT= 0. Calculate the Fermi energyEFand the Fermi :Particles with a half-integer spin (here: nuclear spin) are fermions. The mass density of the liquid is =NVM(3He) = number density isn=NV= M(3He)= 10 24g= 1022cm 3= 1028m h22M(3He)(3 2NV)2/3= 10 23J = 10 4eV,TF= 10 4 104K = Assuming a free electron gas model for the valence electrons of the metals Li, Na, Cs, Cu, Mg, Al,In, Pb, calculate the Fermi energy (in eV) and the zero point pressure (in atmospheric pressure). UseTable 4( density and atomic concentration )in Chapter 1 of Kittel (p. 24 in 7th edition, p. 21 in 8thedition). Solution :The expression for the zero point pressure derived in Class isP=23(35EF)n ,wherenis the electron density andEFthe Fermi level that can be expressed in terms ofn.

3 Make sureyou get a result for a couple of elements ( Na, Mg, In).2


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