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```Physics 2203, Fall 2011 Modern Physics .   Monday, Nov. 5th , 2012 ‐‐‐Sta=s=cal Physics: Quantum sta=s=cs: Ch. 15 in our book. Notes from Ch. 10 in Serway ‐‐Quantum Sta=s=cs: BE and FD ‐‐Example of BE‐Blackbody radia=on: Speciﬁc heat   Announcements ‐‐‐Term papers due Nov. 19th our book. ‐‐‐New Syllabus ‐‐‐First‐ever physics‐major ice cream social: Wed. Nov. 14th 4:30 pm library. Lets ﬁnd the average number of par=cles with some energy‐‐Ej n j = n j1 p1 + n j 2 p2 + i i
n j1 is the number of particles found in the j level in 1
p1 is the probablility of observing arrangement 1
Basic postulate of sta=s=cal mechanics is that there is equal probability of ﬁnd any microstate! Lets calculated the average number of par=cles with energy zero: Total number of microstates is 1287 n0 = n01 p1 + n02 p2 + n03 p3 − − − − −
n01 p1 = (5)(6 / 1287)
n02 p2 = (4)(30 / 1287)
find n1
n0 = 2.307
Lets ﬁnd the average number of par=cles with some energy‐‐Ej n j = n j1 p1 + n j 2 p2 + i i
n j1 is the number of particles found in the j level in 1
p1 is the probablility of observing arrangement 1
Lets calculated the average number of par=cles with energy zero: Total number of microstates is 1287 n0 = n01 p1 + n02 p2 + n03 p3 − − − − −
n0 = (5)(6 / 1287) + (4)(30 / 1287) + (4)(30 / 1287) +
(3)(60 / 1287) + (4)(30 / 1287) + (3)(120 / 1287) +
(2)(60 / 1287) + (4)(15 / 1287) + (3)(120 / 1287) +
(3)(60 / 1287) + (2)(180 / 1287) + (1)(30 / 1287) +
(3)(60 / 1287) + (2)(90 / 1287) + (2)(180 / 1287) +
(1)(120 / 1287) + (0)(60 / 1287) + (2)(15 / 1287) +
(1)(60 / 1287) + (0)(15 / 1287) = 2.307
Probability of ﬁnding a par=cle with energy 0, if we reach randomly into any of the boxes represen=ng 6 par=cles with total energy 8E. p(0) =
n0 2.307
=
= 0.385
6
6
Prove that p(1E)=0.256
fMB = Ae
−E / kB T
g(E) is the Density of States N
N = ∑ ni → =
V
Determine A.
∞
∫ g(E) f
MB
(E)dE
0
Velocity Distribu=on 4π N  m 
n(v)dv =


V  2 π k BT 
3/2
2 −mv 2 /2 kB t
ve
dv
Equipar;;on of Energy –a classical molecule in thermal equilibrium at temperature T has an average energy of kBT/2 for each independent mode of mo;on—degrees of freedom Three dimensional motion
1 2 1 2
1
1
3k T
mv = mvx + + mvy2 + mvz2 = B
2
2
2
2
2m
MB sta;s;cs works when the average distance d between par;cles is greater than the quantum uncertainty. Δx << d
N
3
<< 1
 
3/2
 V  8(mkBT )

ΔxΔpx ≤
2
Low density, high mass, high Temperature Fermions: ψab
(
 
r1 , r 2 = ψa (r1 )ψb (r2 ) -ψa (r2 )ψb (r1 )
)
Lets try to put two electrons into the same state, a=b
 
ψab r 2 , r1 = ψa (r2 )ψa (r1 ) -ψa (r1 )ψa (r2 )
 
ψab r 2 , r1 = ψa (r2 )ψa (r1 ) -ψa (r2 )ψa (r1 ) ≡ 0
(
(
)
)
Bosons: ψab
(
 
r1 , r 2 = ψa (r1 )ψb (r2 ) +ψa (r2 )ψb (r1 )
)
1945 Nobel Prize for discovery of the Exclusion Principle.” Lets try to put two bosons into the same state, a=b
 
ψab r 2 , r1 = ψa (r2 )ψa (r1 ) +ψa (r1 )ψa (r2 )
 
ψab r 2 , r1 = ψa (r2 )ψa (r1 ) +ψa (r2 )ψa (r1 ) ≡ 2ψa (r2 )ψa (r1 )
(
(
)
)
Bose Condensation
Bosons: ψab
(
 
r1 , r 2 = ψa (r1 )ψb (r2 ) +ψa (r2 )ψb (r1 )
)
Bose − Einstein statistics
Integral Spin Systems: Alpha par=cle (S=0), Photons (S=1), deuteron (S=1) Fermions: ψab
(
 
r1 , r 2 = ψa (r1 )ψb (r2 ) -ψa (r2 )ψb (r1 )
)
Fermi-Dirac Statistics
Half Integral Spin Systems: Electrons (S=1/2), neutron (S=1/2) Fermi Dirac Important: par;cles are iden;cal so each microstate has iden;cal count. Now the total number is 20, not 1287. Lets find n 0
1  5 + 4 + 4 + 3 + 4 + 3 + 2 + 4 + 3 + 3 +
n0 = 

2
+1+
3
+
2
+
2
+1+
0
+
2
+1+
0
20 

n 0 = 2.45
n0
p(0) = = 0.408
6
You should be able to determine
p(1E), p(2E), etc.
MB distribu=on BE distribu=on Important: par;cles are iden;cal so each microstate has iden;cal count. Fermions: Only two electrons in each state Lets find n 0
1
n 0 = ( 2 + 2 + 2 ) = 2.00
3
Lets find n1
1
n1 = ( 2 + 2 +1) = 5 / 3
3
n0
n1
p(0) = = 0.333 : p(1E) = = 0.2.78
6
6
What is n 8 ?
X X X X X X X X X X X X X X X X X MB distribu=on FD distribu=on Ground rules are a ﬁxed number of par=cles at a ﬁxed temperature T. f(E) is the probability of ﬁnding a par=cle in an energy state E at temperature T. B and H determined 1
1
by ﬁxing the fBE (E) = E / k T
fFD (E) =
B
Be
−1 number of par=cles HeE / kB T +1
n(E)dE = g(E) fBE (E)dE
N
=
 
 V bosons
fBE (E) =
1
e
E / kB T
∞
∫ Be
0
dE
E / kB T
−1
n(E)dE = g(E) fFD (E)dE
N
=
 
 V Fermions
H = e−EF / kB T
−1 EF → Fermi Energy
∞
∫ He
0
dE
E / kB T
+1
fFD (E) =
1
e
( E −EF ) / kB T
+1
This deﬁnes the Fermi energy! It is the energy between ﬁlled and empty states—keeps the system neutral. Fermi‐Dirac Distribu=on at T=0 n(E)dE = g(E) fFD (E)dE
Fermi Energy can be a function of T
Fermi‐Dirac Distribu=on at T≠0 1
High T fFD ( E ) = ( E − E ) / kT
e F
+1
1
fFD ( E << EF ) = −∞
=1
e +1
1
fFD ( E >> EF ) = ∞
=0
e +1
The Fermi‐Dirac func=on is symmetric. What you lose below EF you gain above EF. kT > EF
fFD ( E ) =
1
eE / kT + 1
= e− E / kT
BE at high temperature
fBE (E) =
1
e E / kB T
e−E / kB T
−E / kB T
=
≈
e
−1 1 − e−E / kB T
FD at high temperature
E>>E F : k B T>>E F
fFD (E) =
1
( E −EF ) / kB T
=
e
− ( E −EF ) / kB T
e
+1 1+ e
fFD (E >> EF ) ≈ e−E / kB T
− ( E −EF ) / kB T
Fermions ψ (1, 2 ) = −ψ ( 2,1)
Bosons ψ (1, 2 ) = +ψ ( 2,1)
In the Quantum region we use the Pauli principle for and put everything in the lowest state for Bosons. At an elevated temperature we should go over to a classical Maxwell‐Boltzmann distribui=on Today we will talk about heat capacity in solids. The electrons are and the laece vibra=ons are Bosons. Photons in a box at Temperature T. BE sta=s=cs n(E)dE = g(E) fBE (E)eE
Energy density u(E)dE is
g(E)EdE
u(E)dE=En(E)dE= e/ k T −1
e B
g(E) =
Need to find g(E)
We have done this
Standing waves in box.
8π E 2
(hc)
3
u(E)dE=
8π h
f3
u( f ,T ) = 3 hf / k T
c e B −1
E = hf
8π
(hc)
3
E 3dE
ee/ kB T −1
Just BE (a) Find an expression for the number of photons/unit volume with energies between E and E+dE in a cavity of temperature T n(E)dE=
8π
(hc)
3
E 2 dE
ee/ kB T −1
(b) Find an expression for the total number of photons per unit volume (all energies). N
=
V
∞
∞
∫ n(E)dE = ∫
0
N 8 π (kBT )3
=
3
V
(hc)
0
∞
∫
0
8π
(hc)
3
E 2 dE
ee/ kB T −1
( E / k BT )
2
dE / kBT
N
(kBT )3
= 8π
3
V
hc
( )
∞
∫
0
z 2 dz
ez −1
ee/ kB T −1
(c) Find the number of photons/cm3 inside a cavity whose walls are T=3000 K ∞
∫
0
z 2 dz
≈ 2.40
ez−1
3
 (8l62x10 −5 eV / K )(3000K ) 
N
11
3
= 8 π (2.40)
=
5.47x10
photons
/
cm

V
1.24x10 −4 eV icm


The molar speciﬁc heat C, is determined by the change in thermal energy U dU
added to a mole of substance, as a func=on of T. C=
dT
Model our solid as a collec=on of independent harmonic oscillators. A one dimensional harmonic oscillator has two degrees of freedom: poten=al energy and kine=c energy Equipartition Theorem:
k BT
2
U = 3N A kBT = 3RT
R universal gas constant
R=N A kB = 8.31J / moliK
1-D HO is 2
k BT
for each degree of freedom
2
Low temperature wrong: Einstein solved this in 1907 d(3RT )
C=
= 3R = 5.97cal / moliK
dT
Three dimensional
3kBT
Einstein assumed as we did that a solid can be represented by independent harmonic oscillators. He showed that in 1‐D system the average energy is: E=
ω
Atoms independent so U=3N A E =
eω / kB T −1
2
 ω 
dU
e ω / k B T
C=
= 3R 
 ω / k B T
dT
−1)2
 kBT  (e
when ω <<k BT
ω
E = ω / k T
≈ k BT
B
e
−1
One paramater ω
ω = k BTE : Einstein Temp.
3N A ω
eω / kB T −1
d(3RT )
C=
= 3R Dulong-Petit Law
dT
2
 ω 
dU
e ω / k B T
C=
= 3R 
 ω / k B T
2
dT
k
T
(e
−1)
 B 
Equipartition Theorem:
k BT
for each degree of freedom
2
3kBT
for three translational degrees of freedom.
2
E=
3kBT 1 2 1 2 1
1
+ Iω x + Iω y + µ (dr / dt)2 + k(r − r0 )2
2
2
2
2
2
(a) Calculate the vibra=on frequency of the carbon atoms in diamond if the Einstein temperature is 1300K. Find the energy‐level spacing. k T
ω = k BTE : ω = B E

ω = 0.112eV
8.62x10
(
=
−5
eV / K ) (1300K )
6.58x10 −16 eV is
= 1.70x1014 Hz
(b) Calculate the average oscillator energy at room temperature and at 1500K, compare to level spacing. E(300) =
ω
e
ω / kB T
−1
E(300)
= 0.01
ω
= 0.00149eV
E(1500) =
ω
e
ω / kB T
−1
E(1500)
=1
ω
= 0.0813eV
```
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