Problem 7.69. If you have a computer system that can do numerical integrals, it's not particularly difficult to evaluate μforT>Tc.

(a) As usual when solving a problem on a computer, it's best to start by putting everything in terms of dimensionless variables. So define t=T/Tc,c=μ/kTc,andx=ϵ/kTc. Express the integral that defines , equation 7.22, in terms of these variables. You should obtain the equation

2.315=0xdxe(x-c)/t-1

(b) According to Figure

the correct value of cwhen T=2Tcis approximately -0.8. Plug in these values and check that the equation above is approximately satisfied.

(c) Now vary μ, holding Tfixed, to find the precise value of μfor T=2Tc. Repeat for values of T/Tcranging from 1.2up to 3.0, in increments of 0.2. Plot a graph of μas a function of temperature.

Short Answer

Expert verified

(a) The equation was proved using the expression for density of states of fermi energy level Bose-Einstein condensation.

(b) The value of cis evaluated as 2.35which differs 1.6%from the accurate value.

(c) The graph is plotted as shown below.

Step by step solution

01

Step 1. Given Information

The equation for density of states of fermi energy level:
N=0g(ε)1e(ε-μ)kT-1dε (Equation-1)

For, critical temperature

kTc=nε

Here,

k= Boltzmann constant,

n=number of nearest neighboring dipole

ε=interaction energy.

Relation between fermi temperature and fermi energy

tεF=kT

Relation between chemical potential and fermi energy

μ=cεF

Substituting the values of tεF=kT,μ=cεFin Equation-1

N=0g(ε)1eε-εrur-1dε

=0g(ε)1eε-cptr-1dε

02

Step 2. (a) Solving the expression for density of state 

Substituting the values of tεF=kT,μ=cεFin Equation-1

N=0g(ε)1eε-cεFtεF-1dε

=0g(ε)1eε-cεFtεF-1dε

N=0g(ε)εFexεF-cεFtεF+1dx

N=0g(ε)εFe(x-c)t+1dx (Equation-2)

For bosons zero spin confined in a box of volume V, the function can be equated as,

N=2.6122πmh232V

Density of state =

g(ε)=2.6122πmh232Vεdε

Substituting the value of εF=xkTcanddε=kTcdx

g(ε)dε=2π2πmh232VxkTckTcdx

=2π2πmkTch232Vxdx

Substituting the value of N=2.6122πmh232V

N=02πN2.612xdxεFe(x-c)t-1

=2πN2.6120xe(x-c)t-1dx

2.315=0xe(x-c)t-1dx

Therefore, the equation is proved.

03

Step 3. (b)Checking the equation by putting values

The given mathematical function is used to evaluate the value of cwhen T=2Tc

NintegrateSqrt[x]Exp(x+0.8)2-1,{x,o,Infinity}

The value of cis evaluated as 2.35which differs 1.6% from the accurate value.

04

Step 4. (c) Plotting the graph

The integral equation we attained from part(a)

2.315=0xe(x-c)t-1dx

The following mathematical function is used to evaluate the value of cwhenT=2TcFind root2.315=NintegrateSqrt[x]Exp(x-c)2-1,{x,o,Infinity},{c,-0.8,-0.9}

This function gives the value of c=-0.820.

The following mathematical function generate a table for all values of c.

mutable=Tablet,Find root2.315=N integrateSqrt[x]Exp(x-c)2-1,{x,o,Infinity}{c,-0.1,-0.2},[1,2]{t,1.2,3,0.2}

The graph shows the chemical potential (μ) as the function of temperature.

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Most popular questions from this chapter

For a system of particles at room temperature, how large must ϵ-μbe before the Fermi-Dirac, Bose-Einstein, and Boltzmann distributions agree within 1%? Is this condition ever violated for the gases in our atmosphere? Explain.

Consider a system consisting of a single hydrogen atom/ion, which has two possible states: unoccupied (i.e., no electron present) and occupied (i.e., one electron present, in the ground state). Calculate the ratio of the probabilities of these two states, to obtain the Saha equation, already derived in Section 5.6 Treat the electrons as a monotonic ideal gas, for the purpose of determining μ. Neglect the fact that an electron has two independent spin states.


Fill in the steps to derive equations 7.112and7.117.

In analogy with the previous problem, consider a system of identical spin0bosonstrapped in a region where the energy levels are evenly spaced. Assume that Nis a large number, and again let qbe the number of energy units.

(a) Draw diagrams representing all allowed system states from q=0up to q=6.Instead of using dots as in the previous problem, use numbers to indicate the number of bosons occupying each level.

(b) Compute the occupancy of each energy level, for q=6. Draw a graph of the occupancy as a function of the energy at each level.

(c) Estimate values of μand Tthat you would have to plug into the Bose-Einstein distribution to best fit the graph of part(b).

(d) As in part (d) of the previous problem, draw a graph of entropy vs energy and estimate the temperature at q=6from this graph.

At the surface of the sun, the temperature is approximately 5800 K.

(a) How much energy is contained in the electromagnetic radiation filling a cubic meter of space at the sun's surface?

(b) Sketch the spectrum of this radiation as a function of photon energy. Mark the region of the spectrum that corresponds to visible wavelengths, between 400 nm and 700 nm.

(c) What fraction of the energy is in the visible portion of the spectrum? (Hint: Do the integral numerically.)

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