Download G¨OTEBORGS UNIVERSITET Institutionen för fysik Solutions to

GÖTEBORGS UNIVERSITET
Institutionen för fysik
Solutions to exam in STATISTICAL PHYSICS FYP300, for the Physics
Programme.
Time:
Place:
Teacher:
Allowed Material:
Friday October 28 2011, kl 830 − 1330
V-building
Lennart Sjögren, tel. 786 9167 (work), 3365914 (home),
0730-61 38 04 (mobile)
Physics Handbook, Mathematical tables, Formularies,
optional calculator (not portable computer) and a handwritten A4 paper (two pages) with arbitraty content
Every correct solution gives 3 points, max 15p. The grading is: 7p -passed,
11p VG
1. The partition function of an ideal monatomic gas in two dimensions of
area A is given by
A2πm
ZN
Z = 1 , Z1 =
N!
βh2
Calculate the internal energy, the entropy, the specific heat at constant area
and the equation of state.
Solution: From the given expression and Stirlings formula ln N ! = N ln N −
N we obtain
A 2πm
+1
ln Z = N ln
N βh2
This gives the internal energy
U =−
N
∂ ln Z
=
= N kB T
∂β
β
The entropy is obtained from F = U − T S = −kB T ln Z
U
A 2πm
S = + kB ln Z = N kB 2 + ln
T
N βh2
The heat capacity is
CA =
∂U
∂T
= N kB
A
The equation of state or pressure is
P =
1 ∂ ln Z
1N
=
= nkB T
β ∂A
βA
1
2. In a crystal there is a magnetic Cr3+ ion on each site. Their electronic
ground state has spin angular momentum J = 3/2. The crystal field splits
this level into a doublet separated by an energy gap .
When a uniform magnetic field B is applied to the crystal, this doublet
splits into four levels of energy
m = 0m + mgµB B
where the magnetic quantum number m takes the values −3/2, −1/2, 1/2, 3/2,
and where 0m = 0 for m = ±3/2 and 0m = for m = ±1/2. µB = e~/2me c
is the Bohr magneton and g = 2.
Calculate the partition function for this 4-level system. Find also the
expression for the mean energy.
Solution: The four energy levels are given by −3µB B, −µB B, +µB B, 3µB B.
With x = βµB B this gives the partition function for one site
Z1 = e3x + e−3x + e−β+x + e−β−x = 2 cosh(3x) + 2e−β cosh(x)
This gives
ln Z = N ln 2 + ln cosh(3x) + e−β cosh(x)
The internal energy is then
µB B 3 sinh(3x) + e−β sinh(x) − e−β cosh(x)
∂ ln Z
= −N
U =−
∂β
cosh(3x) + e−β cosh(x)
3. Calculate the number of phonons in a crystalline solid in the Debye model
for T ΘD and T θD , where θD denotes the Debye temperature.
R∞
Hint: 0 xm /(ex − 1)dx = Γ(m + 1)ζ(m + 1)
Solution: The number of phonons is given by
Z ∞
X
X
1
D()d
Nph =
nr =
=
eβr − 1
eβ − 1
0
r
r
Here r = ~ωr and the density of states in the Debye approximation is
(
3V
2
< D = ~ωD
2π 2 (~cs )3
D() =
0
> D
The cutoff energy is choosen so that
Z
0
D
V
D()d = 2
2π
2
D
~cs
3
= 3N
where N is the number of atoms in the lattice. Then
3 Z D 2
3 Z θD /T 2
3V
1
3V kB T
x dx
d
Nph =
= 2
2
β
2π
~cs
e −1
2π
~cs
ex − 1
0
0
3 Z θD /T 2
T
x dx
= 9N
θD
ex − 1
0
Here D = ~ωD = kB θD . For T θD the upper limit in the integral can be
taken as infinity. This gives
3 Z ∞ 2
3
T
x dx
T
Nph = 9N
= 9N
Γ(3)ζ(3) (T θD )
x
θD
e −1
θD
0
In the other limit T θD we make the expansion
Z
0
θD /T
x2 dx
≈
ex − 1
and so
Nph =
Z
θD /T
0
9N T
,
2 θD
x2 dx
1
=
x
2
θD
T
2
(T θD )
4. The unwinding of a double-stranded DNA molecule is like unraveling a
zipper. The DNA has N links, each of which can be in one of two states: a
closed state with energy 0, and an open state with energy ∆. A link can be
opened only if all the links to its left are already open, as illustrated in the
figure. Due to thermal fluctuations, links will spontaneously open and close.
What is the average number of open links? Find also the limiting behavior
of open links for low (β∆ 1) and high (β∆ 1) temperatures.
Solution: Consider the state with n open links. The energy of this state
is En = n∆ and the number of configurations with this energy gn = 1. This
follows because in order for one link to be open all links to the left are already
open. The probability for n open links is therefore
1
pn = e−βn∆ ,
Z
Z=
N
X
n=0
3
gn e
−βEn
=
N
X
n=0
e−βn∆
The average number of open links is
n=
N
X
N
1 ∂ ln Z
1 X −βn∆
ne
=−
Z n=0
β ∂∆
npn =
n=0
For the partiton function we have
Z=
N
X
e
−βn∆
n=0
1 − e−β(N +1)∆
=
1 − e−β∆
since we just have a geometric series with N terms. This gives
n=
1
N +1
−
eβ∆ − 1 eβ∆(N +1) − 1
Then for β∆ 1, n ≈ e−β∆ and for β∆ 1 we have
N +1
β∆ +
β∆(N + 1) + 12 (β∆(N + 1))2
1
N
1
1
≈
1 − β∆ − 1 + β∆(N + 1) =
β∆
2
2
2
n =
1
1
(β∆)2
2
−
5. Molecules of a gas can be trapped at the surface of a solid. This is the
phenomenon of adsorption. In chemisorption the adsorbed molecules form a
chemical bond with the atoms of the surface at a finite number of sites and
form a monomolecular layer.
In the Langmuir model of chemisorption, it is assumed that the adsorbing
solid possesses N0 sites at which the gas molecules can be trapped with an
energy −. To each site is attributed a number ni (i = 1, . . . N0 ) equal to 0
if the site is unoccupied, and to 1 if the site contains one molecule.
a) Calculate the mean number NA of molecules in the adsorbed phase as a
function of T and the chemical potential µ of the adsorbed phase.
b) Use the condition of equilibrium between the gas phase and the adsorbed
phase to express NA as a function of the temperature T and the pressure P
of the gas.
Hint: The chemical potential for the gas is given by
n
,
µgas = kB T ln
gnQ
where n is the density of the gas, nQ = 1/λ3D the quantum density with
λD = (2π~2 /mkB T )1/2 the deBroglie thermal wavelength and g is the spindegeneracy.
4
c) The figure shows experimental data (dots)
and theoretical results (full curve) for an
isotherm NA (P ) ∝ m of nitrogen adsorbed
on wood charcoal. The experimental data
are usually presented in the form
P
= a(T ) + b(T )P
x
where x ∝ m is the ratio of the mass of adsorbed gas to that of the adsorbent. Show
that this form is compatible with the Langmuir equation.
Solution: The number of adsorbed molecules in a certain state r together
with their energy can be written
Nr =
N0
X
ni ,
i=0
N0
X
Er = −
ni = 0, 1
ni i=0
The grand partition function is then
Z =
X
e−β(Er −µNr ) =
r
=
N0
Y
X
{ni }
1
X
exp β
N0
X
!
( + µ)ni
=
i=0
N0
XY
exp (β( + µ)ni )
{ni } i=0
N0
exp (β( + µ)ni ) = 1 + eβ(+µ)
i=0 ni =0
Then we find the mean number of adsorbed molecules
eβ(+µ)
N0
1 ∂ ln Z
= N0
=
NA =
β(+µ)
β ∂µ
1+e
1 + e−β(+µ)
In equilibrium with the gas molecules we have µ = µgas and so
n
1
P
µ = kB T ln
= ln
gnQ
β
P0
where P = nkB T and P0 = gnQ kB T is a reference pressure. This gives
NA =
N0 P
N0
=
P + e−β P0
1 + e−β PP0
From this we obtain
P N0
= e−β P0 + P
NA
which is an expression of the form above.
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