occupied by an electron is then given by the Fermi-Dirac distribution.This Fermi- Dirac distribution is a function of the temperature T and the Fermi energy 8F: 1 f(E)dE=- dE 1+exp (E-F a Draw the distribution function f(E)against the energy E,first for T=OK(absolute zero) and then for a finite temperature T>OK.Indicate on the diagram the Fermi energy sr. (3 points) b) Calculate the Fermi temperature Tr,i.e.the temperature corresponding to the Fermi energy sF for Silver (F(Ag)=5.48 ev).(2 points) R=8.3144 J/molK,k=8.6210-5eV/K=1.38-10-23J/K,NL=6.023-1023 Material Constants for Aluminium: G=27x10N/m2,a=4.04A,Tm=660℃，p=2.7g/cm3,A=27.0g/mol,=0.34 YSFE-18x10-2J/m2.YGB-0.6J/m2 Material Constants for Copper: G=48x109NWm2,a=3.61A,Tm=1083℃，p=87gcm3,A=63.5g/mol,V=0.35 YSFE-5x10-2J/m2,YGB-0.5J/m2.YSurface-1J/m2 88 occupied by an electron is then given by the Fermi-Dirac distribution. This Fermi￾Dirac distribution is a function of the temperature T and the Fermi energy εF: dE kT E f E dE F ⎟ ⎠ ⎞ ⎜ ⎝ ⎛ − + = ε 1 exp 1 ( ) a) Draw the distribution function f(E) against the energy E, first for T=0K (absolute zero) and then for a finite temperature T>0K. Indicate on the diagram the Fermi energy εF. (3 points) b) Calculate the Fermi temperature TF, i.e. the temperature corresponding to the Fermi energy εF for Silver (εF(Ag)= 5.48 eV). (2 points) R=8.3144 J/molK, k=8.62⋅10-5eV/K = 1.38⋅10-23J/K, NL=6.023⋅1023 Material Constants for Aluminium: G=27×109N/m2, a=4.04Å, Tm=660°C, ρ=2.7g/cm3, A=27.0g/mol, ν=0.34 γSFE=18×10-2J/m2, γGB=0.6J/m2 Material Constants for Copper: G=48×109 N/m2, a=3.61Å, Tm=1083°C, ρ=8.7g/cm3, A=63.5g/mol, ν=0.35 γSFE=5×10-2J/m2, γGB=0.5J/m2, γSurface=1J/m2