Download Physics 2203, Fall 2011 Modern Physics Monday, Nov. 5th , 2012

Survey
yes no Was this document useful for you?
   Thank you for your participation!

* Your assessment is very important for improving the work of artificial intelligence, which forms the content of this project

Document related concepts
no text concepts found
Transcript
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
Related documents