Thermo and Intro to Stat Mech 2016

Homework assignment 3, Solution

Problem 1:Consider a harmonic oscillator coupled to a heat bath at temperature

T . The quantum mechanical treatment of a single harmonic oscillator showsthat it has equally spaced, non-degenerate energy levels εn = n~ω where ω isthe vibrational frequency and n is a quantum number n = 0, 1, 2, . . . . Here,the zero of energy has been chosen to be the ground state energy, but it is~ω/2 above the potential energy minimum (the so-called zero point energy).

(a) Write an expression for the partition function in terms of an infinite sum.

SolutionThe general expression for the partition function is

Z(T ) =∑n

Ω(n) e−εn/kBT

where the sum is over all energy levels, and Ω(n) is the multiplicity (de-generacy) and εn the energy of each energy level. For a one-dimensionalharmonic oscillator, Ω(n) = 1 for all n and εn = n~ω (since the zero ofenergy is chosen here to be the ground state energy) so the partition functi-on becomes

Z(T ) =∞∑n=0

e−n~ω/kBT .

(b) Carry out the summation to obtain a simple function [Hint:∑∞

n=0 xn =

1/(1− x)].

Solution

Z(T ) =∞∑n=0

(e−~ω/kBT

)n=

∞∑n=0

xn

where x = e−~ω/kBT . Using the geometric series (given in the hint), thepartition function can be written as

Z(T ) =1

(1− x)=

1

(1− e−~ω/kBT )

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(c) Give an expression for the relative probability of finding the oscillator inthe first excited state relative to the probability of finding it in the groundstate. Evaluate for an oscillator where ~ω = 0.1 eV and room temperature(recall the value of kBT at room temperature).

Solution

The relative probability is

P (n = 1)

P (n = 0)= e−(ε1−ε0)/kBT = e−~ω/kBT .

At room temperature, kBT = 0.025 eV , so if the vibrational frequency issuch that ~ω = 0.1 eV , then the relative probability is

P (n = 1)

P (n = 0)= e−0.1/0.025 = e−4 = 0.018

This shows that it is by far most likely that this harmonic oscillator is in theground state at room temperature.

(d) Calculate the average energy of the harmonic oscillator as a function ofT .

Solution

The average energy can be obtained from the summation

< ε > =∞∑n=0

εnP (n)

but the same result can also be obtained by differentiating partition functionwith respect to temperature

< ε > = −kBT 2 d

dTlnZ(T ) = −kBT 2

(1− e−~ω/kBT

) d

dT

[1

1− e−~ω/kBT

]

= −kBT 2(1− e−~ω/kBT

)(− ~ωkBT 2

)e−~ω/kBT

1− e−~ω/kBT=

~ωe−~ω/kBT

1− e−~ω/kBT

=~ω

e~ω/kBT − 1.

(e) Find the high temperature limit of the energy of the oscillator (use Taylorexpansions) and compare with the average energy associated with translationof a particle.

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SolutionAt high temperature, kBT >> ~ω, so e−~ω/kBT << 1 and can be

approximated with a first order Taylor expansion as e−~ω/kBT ≈ 1−~ω/kBT .Inserting this approximation in the results above gives

< ε > =~ω

1− ~ω/kBT − 1= kBT.

Note that this is twice as large as the average energy per degree of freedomof translation. The reason is that the harmonic oscillator can store energy inthe form of potential energy in addition to kinetic energy and thus has largerheat capacity.

(f) Derive an expression for the heat capacity as a function of temperature.

Solution

The heat capacity is

CV =

(d < ε >

dT

)V

=d

dT

[~ω

e~ω/kBT − 1

]=

~ωkBT 2

~ωe~ω/kBT

(e~ω/kBT − 1)2

=e~ω/kBT

kB(e~ω/kBT − 1)2

(~ωT

)2

(g) Sketch the heat capacity as a function of temperature and discuss how itcompares with the heat capacity of a two level system (discussed in problem 1of home work 2) as well as how it compares with the heat capacity associatedwith translation (why such a large difference?).

Solution

The heat capacity of the harmonic oscillator is shown in the figure below,as well as the heat capacity of a two level system. In each case, the unit oftemperature is kB/∆E where ∆E in the energy of the first excited state andthe ground state. In both cases, the heat capacity increases strongly whenkBT ≈ ∆E, but in the case of the two level system it then drops because theupper energy level gets saturated and the system cannot easily absorb moreenergy. In the case of the harmonic oscillator, however, there is an inifinitenumber of excited states, not just one, and the heat capacity does not drop,but rather reaches the classical, equipartition limit of kB as T →∞.

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0 0.5 1 1.5 2 2.5 30

0.1

0.2

0.3

0.4

0.5

0.6

0.7

0.8

0.9

1

T

CV

Harmonic oscillator

2−state system

Mynd 1: The heat capacity (in units of kB) of a harmonic oscillator and the heatcapacity of a two level system as a function of temperature (in units of kB/∆Ewhere ∆E in the energy of the first excited state and the ground state). The figureis taken from the homework solution by Gudbjartur.

Problem 2:

Consider a simple model for the opening of a double helix DNA molecule,consisting of N base pairs. Let ε be the binding energy of each conjugate basepair. Due to thermal energy, the double stranded DNA can partly dissociatein such a way that all adjacent pairs starting from one end of the moleculecan dissociate, i.e. n bonds in a row can open up starting from one of theends.

(a) What is the energy of the molecule when n bonds have broken?

SolutionThe energy is increased by ε when one bond is broken, and nε when n

bonds are broken.

(b) Give an expression for the partition function of the molecule consideringall the possible states of the N bonds. [Hint:

∑Nn=0 x

n = (1−xN+1)∑∞

n=0 xn].

Solution

Z =N∑n=0

e−nε/kBT =N∑n=0

(e−ε/kBT

)n=

N∑n=0

xn = (1−xN+1)∞∑n=0

xn =1− xN+1

1− x

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where x = e−ε/kBT . The result is

Z =1− e−(N+1)ε/kBT

1− e−ε/kBT

(c) Calculate the average number of broken bonds, < n >, as a function oftemperature.

Solution

< n >=N∑n=0

nP (n)

where P (n) is the probability that n bonds have broken,

P (n) =e−nε/kBT

Z

So,

< n >=N∑n=0

ne−nε/kBT

Z=

∑Nn=0 nxn∑Nn=0 xn

This expression can be obtained by differentiating Z with respect to x andthen multiplying by x, i.e.

< n >= xd logZ

dx= x

d

dx

(log

1− xN+1

1− x

)= x

d

dx

(log (1− xN+1)− log (1− x)

)= x

(−(N + 1)xN

1− xN+1− (−1)

1− x

)=

(x

1− x− (N + 1)xN+1

1− xN+1

)=

(e−ε/kBT

1− e−ε/kBT− (N + 1)e−(N+1)ε/kBT

1− e−(N+1)ε/kBT

)(d) Simplify your answer in part (c) for the case when ε >> kBT .

SolutionWhen ε >> kBT , x << 1 and 1/(1 − x) ≈ 1 + x + x2. So, < n >≈ x =

e−ε/kBT .

Problem 3:

Rare gas atoms adsorbed on the surface of a flat and smooth solid surfacecan in some cases be approximated as a two-dimensional gas. Assume theatoms can be described as classical particles with mass m, moving on a

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potential energy surface given by some two-dimensional function, V (x, y),and coupled to a heat bath (the surface) at temperature T .

(a) What is the normalized probability distribution function for a componentof the velocity of an atom, for example vx?

Solution

The probability distribution function for a component of the velocity ofan atom is the same in any dimension, so it is the same in a 2-dimensionalgas as in a 3-dimensional gas.

g(vx) =

√m

2πkBTe−mv

2x/2kBT .

(b) What is the normalized probability distribution function for the speedof an atom?

Solution

The area of the 2-dimensional velocity space corresponding to the circularribbon where the speed is between v and v + dv is

2πvdv

so the fraction of molecules with speeds between v and v + dv is given by

f(v)dv = Bve−mv2/2kBTdv

where B is a normalization factor which has absorbed the factor 2π. Thisnormalization factor can be found by setting

1 =

∫ ∞0

ve−mv2/2kBTdv

(c) Find the average kinetic energy of an atom.

Solution

The average kinetic energy of an atom is 12m < v2 > where

< v2 > =

∫ ∞0

v2f(v)dv = kBT.

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Problem 4:

Consider the isotope enrichment of uranium using effusion. Initially, thecontainer contains uranium hexafluoride, UF6, with natural abundance ofthe isotopes, 0.720% of 235U and 99.28% of 238U . The goal is to increase thefraction of 235U .

(a) What is the composition of the gas that escapes initially from the holeon the container?

Solution

The ratio of the probability that a 235U molecule escapes out of thecontainer divided by the probability that a 238U molecule escapes is given by

P235UF6

P238UF6

=

√M238UF6

M235UF6

The molecular weight of the molecules:

M238UF6= 238.05 + 6 · 18.998 = 352.038 g/mol

andM235UF6

= 235.04 + 6 · 18.998 = 349.028 g/mol

soP235UF6

P238UF6

=

√352.038

349.028= 1.0043

and the relative abundance of 235UF6 in the gas that escapes the box is1.0043 · 0.720% = 0.723%.

(b) How many such effusion steps would be needed to reach 99% purity of235U?

Solution

Assuming the same enrichment occurs in each step, the total enrichmentthat occurs in n steps is 1.0043n. In order to receive 0.99% purity of 235UF6,n needs to be such that

1.0043n · 0.0072 = 0.99

which means that n needs to be

n =ln (0.99/0.0072)

ln 1.0043= 1147.

So, it would require at least 1100 effusion steps to reach this level of purity.Not easy, but can be done!

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