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Homework #6 Solutions

1. Chap.4: prob. 3 -The partition function is

.0 1 1 1 2 2 2 2 2

,

exp( / ) 2.7803i Ballstates i

Z E k T e e e e e e e e e = = + + + + + + + + =

The probability of: i) the low energy state is e0/2.7803 = 0.3597, ii) one of the second energy states is e-1/2.7803 =

0.1323, and iii) one of the highest energy states is e-2/2.7803 = 0.0487. So the total probability of finding the

system with energy 100kB is 0.3969 and the total probability of finding the system with energy 200kB BB is 0.2434.The average energy is 0 + 100kB(0.3969) + 200kB BB(0.2434) = 88.37kB.B

2. Chap. 4: prob. 4 a) -

( )exp( )

exp( )ln( ) 1 1 1exp( )

ii i i

i i ii i i

Z ZP U

Z Z Z Z

= = = = =

= .

b) First, we have to notice that = but the averaging can be done on

the three terms separately so that = - 2 +

2= -

2. This is the

relation that we should show comes from the second partial of lnZ.

( )22 2

exp( ) exp( )ln( ) 1 1

exp( )i ii

i i iii iZZ Z

Z Z

= = +

22 2

2

1exp( ) exp( ) exp( ) ( )i i i i j j

i i j

E E E E Z Z

= =< > < > =< < > > 2

c) Mean square deviation of the thermal energy is 2(lnZ)/2 = -U/= -(U/T)(dT/d) = kBT2CV.

For your Monte Carlo simulations (whereEcan vary), this means that a calculation of the mean squaredeviation of the total energy, will give you the heat capacity of your system if you know the temperature.

3. Chap. 4: prob. 5 - Generally, you will find that the population of an energy level goes up as theenergy of the level goes down. Eventually, we want to find the contribution of these three energy levels

to the molar specific heat so lets assume that we have one mole of this molecule, i.e. Avogadros numberof molecules,N=NA. For a single molecule in thermal contact with the other molecules we know thatp3

= exp(-10)/Zone molecule = exp(-10)/[1 + exp(-) + exp(-10)] andp2 = exp(-)/Zone molecule = exp(-

)/[1 + exp(-) + exp(-10)]. So the second energy level will be occupiedp2/p3 = exp(9) times as

often as the highest energy level.Practically speaking we can probably ignore the highest energy level wheneverp2/p3 >> 1 (sayp2/p3 > 10

4

which would mean T< 9/kBln(104) ) and I would be satisfied if you gave this answer. However, we

could also argue that the number of molecules in the highest energy state is n3 =NAp3 and that if this is

less than 1 then, since we cant have less than one molecule in some state, we have to say that the highest

energy state is unoccupied. So we write n3 < 1 and plug in thep3. Doing this shows that e9 >NA/2 sothat T< 9/[kBln(NA/2)] (these two answers only differ by a factor of 6). This shows that whenever the

available thermal energy per degree of freedom, kBT, is less than about 0.2then 10energy level will beempty even though there are Avogadros number of molecules.

The average energy of one molecule is = 10p3 + p2 + 0p1 so just plug in the probability functionsabove to find the temperature dependence of .

The energy of one mole is U=NA and CV= U/T)Vso the molar specific heat is cV= U/T)Vif oursample size is one mole. If my math is correct you get

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0

0.1

0.2

0.3

0.4

0.5

0 2 4 6 8 10 12 14

3-level system

CV

T

( )( )2/10/

/11/10/

2

2

1

81100

TkTk

TkTkTk

B

AVBB

BBB

ee

eee

TkNc

++

++= .

This function is plotted in the figure to the right. The wholepoint of this problem was to let you see that the temperature

dependence of the heat capacity shows you the energy level

structure. At low Tyou see quantum effects very strongly.

Roughly speaking, as Tincreases, the heat capacity dropswhenever kBTpasses an energy level and increases again as kBT

approaches a new energy level. The heat capacity is really justtelling you about the density of energy levels per unit energy at

an energy of about kBT. At lowest Tonly the ground state has a

significant population. As Tis raised, eventually higher andhigher energy states have molecules in them. The heat capacity

increases every time new ways of having energy are added to

the system.

4. Chap. 4: prob. 12 - a) At low Tthe energy is as low as possible (all atoms in their ground state) and

the total entropy is as small as possible (only one state so zero entropy). This means that K0 =N.b) At high Tthe entropy increases as much as it can. Highest Twill have the highest entropy state which

is when each of the 10 states is equally occupied. This means that K0 =N/10.

c) You can factor this partition function so consider one ion. Probability of occupation of state i by this

ion is Pi(T) = e-i/kBT/ZwhereZ= [1 + e-

/kBT+ e-2/kBT+ + e-9

/kBT]. The low Tlimit is when /kBT> 9) you get e-i/kBT (1 i/kBT) etc. for all i soZ 10(1 4.5/kBT)

and K0(T) (N/10)(1 + 4.5/kBT) which increases (fromN/10) as Tdecreases.