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Use a computer to sum the rotational partition function (equation 6.30) algebraically, keeping terms through j = 6. Then calculate the average energy and the heat capacity. Plot the heat capacity for values ofkT/ϵ ranging from 0 to 3. Have you kept enough terms in Z to give accurate results within this temperature range?

Short Answer

Expert verified

The average energy is E¯=ϵj(j+1)(2j+1)e-j(j+1)/t(2j+1)e-j(j+1)/tand

The average heat capacity is C=k(2j+1)e-j(j+1)/tj2(j+1)2(2j+1)e-j(j+1)/tt(2j+1)e-j(j+1)/t2-kj(j+1)(2j+1)e-j(j+1)/t2t(2j+1)e-j(j+1)/t2

Step by step solution

01

Given information

Heat capacity for values kT/ϵranging from 0 to 3.

j = 6

02

Explanation

Rotational partition function is:

Z=(2j+1)e-j(j+1)ϵ/kT

Let,

t=kTϵ

Therefore,

Z=(2j+1)e-j(j+1)/t(1)

The average energy is:

E¯=-1ZZβ(2)

By chain rule:

Zβ=ZttβZβ=Ztβt-1

But β=1/kT, henceβ=1/tϵhence,

Zβ=t(2j+1)e-j(j+1)/tt1tϵ-1Zβ=j(j+1)(2j+1)e-j(j+1)/t1t2-1t2ϵ-1Zβ=-ϵj(j+1)(2j+1)e-j(j+1)/t

Substitute into (2)

role="math" localid="1647453692824" E¯=ϵj(j+1)(2j+1)e-j(j+1)/t(2j+1)e-j(j+1)/t(3)

The partial derivative of total energy with respect to temperature equals the heat capacity, which is:

C=E¯TC=kϵE¯t

Substitute from (3):

role="math" localid="1647454207193" C=ϵktϵj(j+1)(2j+1)e-j(j+1)/t(2j+1)e-j(j+1)/tC=k(2j+1)e-j(j+1)/tj2(j+1)2(2j+1)e-j(j+1)/tt(2j+1)e-j(j+1)/t2-kj(j+1)(2j+1)e-j(j+1)/t2t(2j+1)e-j(j+1)/t2

03

Explanation

Using python to plot a function between t and C/k. The code is given below:

The graph is:

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Most popular questions from this chapter

This problem concerns a collection of N identical harmonic oscillators (perhaps an Einstein solid or the internal vibrations of gas molecules) at temperature T. As in Section 2.2, the allowed energies of each oscillator are 0, hf, 2hf, and so on. (

a) Prove by long division that

11-x=1+x+x2+x3+

For what values of x does this series have a finite sum?

(b) Evaluate the partition function for a single harmonic oscillator. Use the result of part (a) to simplify your answer as much as possible.

(c) Use formula 6.25 to find an expression for the average energy of a single oscillator at temperature T. Simplify your answer as much as possible.

(d) What is the total energy of the system of N oscillators at temperature T? Your result should agree with what you found in Problem 3.25.

(e) If you haven't already done so in Problem 3.25, compute the heat capacity of this system and check t hat it has the expected limits as T0 and T.

Each of the hydrogen atom states shown in Figure 6.2 is actually twofold degenerate, because the electron can be in two independent spin states, both with essentially the same energy. Repeat the calculation given in the text for the relative probability of being in a first excited state, taking spin degeneracy into account. Show that the results are unaffected.

A lithium nucleus has four independent spin orientations, conventionally labeled by the quantum number m = -3/2, -1/2, 1/2, 3/2. In a magnetic field B, the energies of these four states are E = mμB, where μ the constant is 1.03 x 10-7 eV/T. In the Purcell-Pound experiment described in Section 3.3, the maximum magnetic field strength was 0.63 T and the temperature was 300 K. Calculate the probability of a lithium nucleus being in each of its four spin states under these conditions. Then show that, if the field is suddenly reversed, the probabilities of the four states obey the Boltzmann distribution for T =-300 K.

The analysis of this section applies also to liner polyatomic molecules, for which no rotation about the axis of symmetry is possible. An example is CO2, with =0.000049eV. Estimate the rotational partition function for a CO2molecule at room temperature. (Note that the arrangement of the atoms isOCO, and the two oxygen atoms are identical.)

Prove that, for any system in equilibrium with a reservoir at temperature T, the average value of the energy is

E¯=-1ZZβ=-βlnZ

where β=1/kT. These formulas can be extremely useful when you have an explicit formula for the partition function.

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