In the real world, most oscillators are not perfectly harmonic. For a quantum oscillator, this means that the spacing between energy levels is not exactly uniform. The vibrational levels of an H2 molecule, for example, are more accurately described by the approximate formula

Enϵ1.03n-0.03n2,n=0,1,2,

where ϵ is the spacing between the two lowest levels. Thus, the levels get closer together with increasing energy. (This formula is reasonably accurate only up to about n = 15; for slightly higher n it would say that En decreases with increasing n. In fact, the molecule dissociates and there are no more discrete levels beyond n 15.) Use a computer to calculate the partition function, average energy, and heat capacity of a system with this set of energy levels. Include all levels through n = 15, but check to see how the results change when you include fewer levels Plot the heat capacity as a function of kT/ϵ. Compare to the case of a perfectly harmonic oscillator with evenly spaced levels, and also to the vibrational portion of the graph in Figure 1.13.

Short Answer

Expert verified

Therefore,

C=nNkβϵ1.03n-0.03n22eβϵ1.03n-0.03n2eβϵ1.03n-0.03n2-12

Step by step solution

01

Given information

In the real world, most oscillators are not perfectly harmonic. For a quantum oscillator, this means that the spacing between energy levels is not exactly uniform. The vibrational levels of an H2 molecule, for example, are more accurately described by the approximate formula

Enϵ1.03n-0.03n2,n=0,1,2,

where ϵ is the spacing between the two lowest levels. Thus, the levels get closer together with increasing energy. (This formula is reasonably accurate only up to about n = 15; for slightly higher n it would say that En decreases with increasing n. In fact, the molecule dissociates and there are no more discrete levels beyond n15.)

02

Explanation

In actual life, most oscillators are not perfect harmonic oscillators; the energy of real oscillators is provided by:

En=ϵ1.03n-0.03n2

Where

ϵis the energy difference between the first two levels.

The partition function is:

Z=ne-βEn(1)

Substitute with energy

localid="1647451366905">Z=ne-βϵ1.03n-0.03n2(2)

Using python the summation is found with the code:

03

Explanation

The average energy is given as:

E¯=-1ZZβ

Substitute from (1),

role="math" localid="1647451769495" E¯=-1-e-βEnβ11-e-βEnE¯=1-e-βEnEne-βEn1-e-βEn2E¯=Ene-βEn1-e-βEnE¯=EneβEn-1

Substituting the energy,

E¯=nϵ1.03n-0.03n2eβϵ1.03n-0.03n2-1

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

C=UT=βTUβC=βTβNEneβEn-1C=Tβ-1-NEn2eβEneβEn-12

But β=1/kTT=1/kβ

Tβ-1=-kβ2

Therefore,

C=NkβEn2eβEneβEn-12

Substitute the energy:

role="math" localid="1647452008304" C=nNkβϵ1.03n-0.03n22eβϵ1.03n-0.03n2eβϵ1.03n-0.03n2-12CNk=nβϵ1.03n-0.03n22eβϵ1.03n-0.03n2eβϵ1.03n-0.03n2-12

If x=1/βϵ, then

CNk=n1.03n-0.03n2/x2e1.03n-0.03n2/xe1.03n-0.03n2/x-12

04

Explanation

Using python to plot the function for different values of nmax. The code is given below:

In the graph below, the red curve is for harmonic oscillator:

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Most popular questions from this chapter

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?

Consider a classical "degree of freedom" that is linear rather than quadratic E=cqfor some constant c. (As example would be the kinetic energy of a highly relativistic particle in one dimension, written in terms of its momentum.) Repeat derivation of the equipartition theorem for this system, and show that the average energy isrole="math" localid="1646903677918" E-=kT.

2. Consider a classical particle moving in a one-dimensional potential well u(x), as shown The particle is in thermal equilibrium with a reservoir at temperature T, so the probabilities of its various states are determined by Boltzmann statistics.

{a) Show that the average position of the particle is given by

where each integral is over the entire xaxis.

A one-dimensional potential well. The higher the temperature, the farther the particle will stray from the equilibrium point.

(b) If the temperature is reasonably low (but still high enough for classical mechanics to apply), the particle will spend most of its time near the bottom of the potential well. In that case we can expand u(x)in a Taylor series about the equilibrium point x0: u(x)=ux0+x-x0dudxx0+12x-x02d2udx2x0

+13!x-x03d3udx3x0+

Show that the linear term must be zero, and that truncating the series after the quadratic term results in the trivial prediction x=x0.

(c) If we keep the cubic term in the Taylor series as well, the integrals in the formula for xbecome difficult. To simplify them, assume that the cubic term is small, so its exponential can be expanded in a Taylor series (leaving the quadratic term in the exponent). Keeping only the smallest temperature-dependent term, show that in this limit x differs from zo by a term proportional to kT. Express the coefficient of this term in terms of the coefficients of the Taylor series foru(x)

(d) The interaction of noble gas atoms can be modeled using the Lennard Jones potential,

u(x)=u0x0x12-2x0x6

Sketch this function, and show that the minimum of the potential well is at x=x0, with depth u0. For argon, x0=3.9Aand u0=0.010eV. Expand the Lennard-Jones potential in a Taylor series about the equilibrium point, and use the result of part ( c) to predict the linear thermal expansion coefficient of a noble gas crystal in terms of u0. Evaluate the result numerically for argon, and compare to the measured value

α=0.0007K-1(at80K)

Use Boltzmann factors to derive the exponential formula for the density of an isothermal atmosphere, already derived in Problems 1.16 and 3.37. (Hint: Let the system be a single air molecule, let s1 be a state with the molecule at sea level, and let s2 be a state with the molecule at height z.)

Verify from Maxwell speed distribution that the most likely speed of a molecule is2kTm.

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