Lecture 1 - HMJ Pendidikan Fisika | Universitas Islam
Download
Report
Transcript Lecture 1 - HMJ Pendidikan Fisika | Universitas Islam
CURICULUM VITAE
A. DATA DIRI
01. N a m a
02. Tempat/Tanggal Lahir
03. Jenis Kelamin
04. Fakultas/Jurusan
05. Pangkat/Golongan/NIP
06. Bidang Keahlian
07. Alamat Rumah
:
:
:
:
:
:
:
09. e-mail
Dr. H. Muris, M.Si
Tinggas, 1965
Laki-laki
FMIPA/Fisika
Lektor Kepala/IV/a/131925820
Fisika Material
BTN Minasa Upa G20/14 Makassar.
90224.
Telp. (0411) 886307
HP. 081342403676
: Jurusan Fisika FMIPA UNM
Kampus Parangtambung Makassar
Tlp/Fax. (0411)840622, HP. 081342403676
: [email protected]
10. Riwayat Pendidikan Tinggi
:
08. Alamat Kantor
Jenis Pendidikan
Sarjana (S1)
Pra Magister (Pra S2)
Magister (S2)
Doktor (S3)
Tempat
IKIP Ujung Pandang
ITB Bandung
ITB Bandung
Université de la Méditerranée
Marseille, Prancis
Tahun lulus
1989
1992
1994
2001
Spesialisasi
Pendidikan Fisika
Fisika
Fisika Material
Fisika Material
B. Riwayat Pekerjaan
1.Dosen Tetap Jurusan Fisika FMIPA Universitas Negeri Makassar, 1990 - sekarang.
2.Ketua Program Studi Fisika FMIPA Universitas Negeri Makassar, 2003 - 2004.
3.Pembantu Dekan Bidang Akademik FMIPA Universitas Negeri Makassar, 2004 - sekarang.
4.Dosen Program Pascasarjana UNM Makassar, 2006 - sekarang
Fisika Statistik
Rujukan Utama :
Introdution to Statistical Physics for Students
by
Pointon
Longman, England
Rujukan Tambahan :
Buku Buku Fisika Zat Padat, Fisika Kuantum dan Fisika
Modern yang relevan
Pokok Bahasan
1. Pengantar
2. Statistik Maxwell Boltzmann
3. Aplikasi Statistik Maxwell Boltzmann
4. Statistik Bose Einstein
5. Statistik Fermi Dirac
6. Temperatur dan Entropy
7. Aplikasi Statistik Termodinamika
8. Ensemble Kanonik
9. Grand Ensemble Kanonik
Pokok Bahasan
1. Pengantar
2. Statistik Maxwell Boltzmann
3. Aplikasi Statistik Maxwell Boltzmann
4. Statistik Bose Einstein
5. Statistik Fermi Dirac
6. Temperatur dan Entropy
7. Aplikasi Statistik Termodinamika
8. Ensemble Kanonik
9. Grand Ensemble Kanonik
Sistim Termodinamika, Parameter Makroskopik
Sistim terbuka dimana dimungkinkan
terjadi pertukanan energi dan materi
dengan lingkungan.
Sistim tertutup terjadi pertukaran
energi maupun materi dengan
lingkungannya
Isolated systems tidak
memungkinkan terjadinya pertukaran
energi maupu materi dengan
lingkungannya
Paramater internal dan external : temperatur, volume, tekanan, energi,
medan magnet, dll. (nilai rata-rata, fluktuasi diabaikan).
Pengertian Dasar Statistik
Mean : Rata-rata
Mode : yang paling mungkin
Median : Titik tengah
Varians : Ragam, Lebar Distribusi
Pengertian Dasar Statistik
Misalkan suatu variabel yang diselidiki : 3,4,4,3,5,3,4
X
3 4 43635
7
X
28
4
7
x1 x 2 x 3 x 4 x 5 x 6 x 7
7
x
X
i
N
i
Pengertian Dasar Statistik
Rata-rata dengan fungsi probabilitas
xi
f
f(xi)
xi f(xi)
3
3
3/7
9/7
4
3
3/7
12/7
5
1
1/7
5/7
7
1
28/7 = 4
Ternyata diperoleh hasil rata-rata yang sama yakni 4
Pengertian Dasar Statistik
Hasil ini diperoleh dari pengembangan bentuk
X
f .( x i ). x i
i
f ( xi )
f ( x i ). x i
f ( xi ) 1
i
i
Jika fungsinya kontinyu maka :
X
x . f ( x ) dx
Bagaimana anda mengartikan parameter statistik berikut ?
kontinyu
diskrit
Pengertian Dasar Statistik
Fungsi Gaussian
Fungsi seperti akan banyak dijumpai dalam pembahasan statistik
partikel
Ruang Euclid dan Ruang Fase
Ruang Euclid
dV dxdydz
dV
z
x
y
dz
dx
dy
Ruang Euclid dan Ruang Fase
px py pz
2
2
2
2m
d dxdydzdp
x
dp y dp z
Ruang fase Ruang momentum
d 6 N dx 1 dy 1 dz 1 dp
........ dx i dy i dz i dp
........ dx
N
dy
N
dz
N
xi
dp
dp
dp
x1
xN
yi
dx
i
dy i dz i dp xi dp yi dp zi
i 1
N
d
i 1
i
dp
dp
N
y1
dp
z1
zi
yN
dp
zn
Rata Rata Sifat Assembly
Misalkan dalam assembly terdapat sejumlah N molekul dengan
energi total E dan berada dalam volume V.
p(N) menyatakan koordinat momentum
x(N) menyatakan koordinat posisi
p(N)
x(N)
Rata Rata Sifat Assembly
Jika X adalah perilaku yang ingin dicari rata-ratanya dalam ruang
fase tersebut
X
X x ( N ), p ( N )P x ( N ), p ( N )d
6 N
Normalisasi terhadap ruang
X x ( N ), p ( N )P x ( N ), p ( N )d
X
6 N
P x ( N ), p ( N )d
6 N
6N
6N
6N
Rata Rata Sifat Assembly
Jika X merupakan fungsi yang diskrit, maka perata-rataan fungsi X
dapat dinyatakan dengan :
X
pi X i
i
pi
i
Normalisasi probabilitas menghasilkan
pi 1
i
X
i
pi X i
Assembli Klasik dan Kuantum
a.
Klasik
- Terbedakan antara satu dengan lainnya (distinguishable)
- Energi kontinu
- Tak memenuhi prinsip larangan Pauli
b.
Kuantum : Terdapat dua tipe
Tipe I (fermion) :
- Tak terbedakan antara satu dengan lainnya (indistinguishable)
- Energi disktrit
- Memenuhi prinsip larangan Pauli
Misalnya : elektron dalam zat padat
Assembli Klasik dan Kuantum
b.
Kuantum : Terdapat dua tipe
Tipe II (boson) :
- Tak terbedakan antara satu dengan lainnya (indistinguishable)
- Energi disktrit
- Tidak memenuhi prinsip larangan Pauli
Misalnya : foton atau partikel alpha
Statistik Maxwell Boltzmann
Distribusi Energi
Misalkan dalam sistim yang ditinjau terdapat N sistim :
Sistem 1 dengan energi ε1
Sistem 2 dengan energi ε2
…………………….
Sistem i dengan energi εi
…………………….
Sistem N dengan energi εN
Statistik Maxwell Boltzmann
Distribusi Energi
Misalkan dalam sistim yang ditinjau terdapat N sistim :
Sistem 1 dengan energi ε1
Sistem 2 dengan energi ε2
…………………….
Sistem i dengan energi εi
…………………….
Sistem N dengan energi εN
Statistik Maxwell Boltzmann
Prinsip Kekekalan
Statistik Maxwell Boltzmann
Jumlah pilihan jika memilih sejumlah N1 di antara N partikel
Jika g1 menyatakan bobot, maka jumlah pilihan yang ada adalah :
Statistik Maxwell Boltzmann
Perluas lagi dengan mengambil sejumlah N2 dari N-N1
Perluas lagi dengan mengambil sampai n kali
Statistik Maxwell Boltzmann
Secara umum dapat ditulis :
Contoh Pemakaian
Empat partikel dengan notasi a,b,c dan d didistribusi pada dua pita energi 2 pada pita 1 dan 2
pada sistim 2. Bobot masing-masing adalah 3 dan 4.
Jadi : N1 = N2 = 2
g1 = 3 , g2 = 4
W
N!
2
N 1 !. N 2 !
W
4!
2 !.2!
g 1 .g 2
3 .4
2
2
2
864
Contoh Pemakaian
a
b
a
c
d
c
c,a
b
d
d
Ini hanyalah 3 contoh gambar dari 864 kemungkinan yang ada.
Sekarang adalah giliran anda untuk melengkapinya.
b
Statistik Maxwell Boltzmann
Peluang terbesar diperoleh dengan mengambil dw/dn = 0
Rumus Stirling
Distribusi Maxwell Boltzmann
n ( ) d
kT
g()
exp
k
T
B
0
2 N
3/2
e
/ kT
1/ 2
d
P()
g C
=
0
0
Aplikasi Statistik Maxwell Boltzmann
ky
2D
Untuk partikel kuantum dalam kotak 2D (e.g., electron pd FET):
n y
n x
2
2
k
kx
ky
Lx
k
k
k area
N k
4
4
kx
2
1
kx ky
Ly
2
G k
Lx
Lx
3D
kz
Ly
# states within ¼ of
a circle of radius k
N k
1
4 / 3
k
3
8
Lx Ly Lz
kx
g
3D
g
k
3
volume
6
2D
G k
2
k
k
4
2
g()
2 s 1 2 m
4
3/2
2
2
1/ 2
1 2m
G
2 s 1 m
3
6
2
2
2
G
4
2
- Tak
bergantung pd
1
6
2
2m
2
3/2
3D
2D
1D
ky
Thus, for 3D electrons
(2s+1=2):
g
3D
1
2
2
2m
2
3/2
1/ 2
Distribusi Kecepatan Maxwell
m
f v fv
2 k B T
3/2
vy
mv 2
4 v 2 dv
exp
2 k BT
" volume"
f v dv
m
C
2 k B T
1
0
v dv
4 v dv
2
Nampak bahwa persamaan ini merupakan perkalian
antara faktor Boltzmann dengan sebuah tetapan.
Tetapan tersebut dapat diperoleh dari normalisasi
v
v
vx
3/2
vz
P(v)
d
dN NP d N exp
k BT
Distribusi energi, N – the total # of particles
m
dN v NP v dv N
2 k B T
3/2
mv 2
dv
4 v exp
2
k
T
B
2
speed distribution (distribusi kecepatan)
m
dN v x NP v x dv N
2 k B T
1/ 2
v
P(vx)
mv 2
dv
exp
2
k
T
B
Distrbusi kecepatan dalam arah x, vx
vx
Karakteristik Nilai Kecepatan
m
P v
2 k B T
3/2
mv 2
4 v exp
2 k BT
2
P(v)
Lihat bahwa distribusi ini tidak simetrik, sehingga
perlu dicari perata-rataan sebagai berikut
The root-mean-square speed is proportional to
the square root of the average energy:
E
1
2
v max
v
2
v rms
2E
3k BT
m
m
v
v rms
dP v
0
dv
v v max
Harga kec.maksimum :
Kelajuan rata-rata :
m v rms
m
v v P v dv
2 k B T
0
v max v v rms
2
8 /
v max
2 k BT
m
mv 2
3
4 v exp
2 k T dv
B
0
3 1 1 . 13 1 . 22
8k BT
m
Soal (Maxwell distr.)
Consider a mixture of Hydrogen and Helium at T=300 K. Find the speed at which
the Maxwell distributions for these gases have the same value.
m
P v , T , m
2 k B T
m1
2 k T
B
3/2
2
2
ln
m1
m2
v
2
2 k BT
3/2
mv 2
4 v exp
2
k
T
B
2
m1v 2 m 2
2
4 v exp
2 k B T 2 k B T
3
3
m1 m 2
ln m 1
m1v
2
2 k BT
3
2
3 k B T ln
v
ln m 2
3/2
m 2v
m2v 2
4 v exp
2 k BT
2
2
2 k BT
m1
m2
m1 m 2
3 1 . 38 10
23
300 ln 2
2 1 . 7 10
27
1 . 6 km/s
Soal (Maxwell distr.)
Find the temperature at which the number of molecules in an ideal Boltzmann gas
with the values of speed within the range v - v+dv is a maximum.
m
P v , T , m
2 k B T
3
m
2 2 k B T
1/ 2
3/2
mv 2
4 v exp
2
k
T
B
2
m
2 k T 2 exp
B
mv 2
m
2 k T 2 k T
B
B
mv 2
0
2
2 k BT
3
maximum:
T
3/2
mv
P v , T
T
mv 2 mv 2
exp
2
2 k BT 2 k BT
0
0
2
3k B
At home:
Find the temperature T at which the rms speed of Hydrogen molecules exceeds their
most probable speed by 400 m/s.
Answer: 380K
o
Pelebaran Garis Spektrum Doppler
Bagian ini adalah salah satu contoh penerapan distribusi laju dari
o
statistik Maxwell Boltzmann, yakni pelebaran spektrum akibat efek
Doppler.
Misalkan molekul gas melakukan radiasi dengan panjang
gelombang
dalam arah x dengan kecepatan vx menuju kepada
seorang pengamat. Pengamat akan menerima radiasi dengan
panjang gelombang.
o
Pelebaran Garis Spektrum Doppler
Karena efek Doppler, maka panjang gelombang yang diamati
pengamat adalah :
o 1
v
vx
c
c o
o
dv x
c
o
d
o
Pelebaran Garis Spektrum Doppler
Dari distribusi Maxwell Boltzamann
m
dN v Nf v dv N
2 k B T
3/2
mv 2
dv
4 v exp
2
k
T
B
2
Ubah sebagai fungsi panjang gelombang
m
f d
2 k B T
3/2
mc 2 o 2
exp
2
2k T
o
B
c
d
o
o
Pelebaran Garis Spektrum Doppler
Intensitas radiasi :
mc 2 o 2
I d Cf ( ) d I o exp
2
2k T
o
B
I (o )
I ( )
d
Dengan mengukur intensitas
radiasi maka dapat ditentukan
temperatur gas emisi
o
o
p
2
x
/ 2 me
e / KT
d
e
e / kT
dT
Prinsip Ekipartisi Energi
Jika energi sistem dinyatakan dalam bentuk kuadrat posisi dan momentum maka tiap
bentuk kuadrat tersebut akan memberikan energi rata-rata ½ kT
Contoh molekul gas dengan massa m, energinya dapat dinyatakan dengan
2
x
px
2m
Maka energi rata-ratanya adalah :
2
px / 2m e
e / KT
e
e / kT
dT
d
Prinsip Ekipartisi Energi
2
px
2m
Nyatakan energi sebagai
x
exp (
Misalkan
2
px
2m
exp (
2
px
dan
x
) / kT dxdydzdp
2
px
2m
dp z
y
2m
y
dp x x
maka
2 mkT
x
e
u
2
2
u du
e
u
2
exp( p x / 2 mkT ) dp x
2
kT
exp( p x / 2 mkT ) dp x
) / kT dxdydzdp
= u2
2
px
2
du
Prinsip Ekipartisi Energi
Hasilnya memberikan :
e
u
2
u du
2
1
2
e
u
2
du
2
Maka :
x
1
px
kT
2 mkT
2
Karena ada satu bentuk kuadrat maka memberikan energi rata-rata ½ kT
Contoh 2 : Osilator harmonik dengan dua jenis energi
2
x
px
2m
1
2
x
2
u
Prinsip Ekipartisi Energi
Maka :
x
1 2
2
p
/
2
m
x
2
x
e
e / kT
e e / kT d
d
dp
xdx p x2
1
2 1
2
2
p x 2 x exp 2 m 2 x / kT dxdp
x
p x2
1
2
exp 2 m 2 x kT dxdp x
Ubah ke koordinat polar :
2
px
2m
r sin
2
2
,
1
x r cos
2
2
2
1
2
dp x dp y 2 ( m / ) rdrd
2
x
Prinsip Ekipartisi Energi
Maka :
2x
d
0
2x
0
e
r
2
/ kT
3
r dr
kT
0
d
e
r
2
/ kT
rdr
0
Karena terdiri dari dua bentuk kuadrat maka energinya adalah
2 x ½ kT = kT
Untuk osilator harmonik 3D maka :
2
2
p2
py
p x 1
1
1
3
x
kT
kT
kT
kT
2
m
2
m
2
m
2
2
2
2
3
2
kT
Prinsip Ekipartisi Energi
Energi rata-rata untuk osilator harmonik 3 D.
2
2
p 2
p
px
1
1
1
y
2
2
2
x
1x
2y
3z
2
2m
2
2m
2
2 m
6.
1
kT
2
3 kT
Jadi dalam hal ini ada 6 derajat kebebasan ( f = 6) dimana tiap
derajat kebebasan memberikan kontribusi energi sebesar ½ kT
Prinsip Ekipartisi Energi
Jika terdapat NA (bil. Avogadro) molekul gas dan berlaku sebagai
osilator harmonik 3D, maka, terdapat 6 derajat kebebasan,maka :
E 6N A
1
kT 3 RT
2
Panas jenis per gram atom zat padat :
E
o
3 R 5 , 94 kal/ K/gr.atom
T v
Panas jenis gas
Jika terdapat NA (bil. Avogadro) molekul gas dan berlaku sebagai
osilator harmonik 3D, maka, terdapat 6 derajat kebebasan,maka :
E 6N A
1
kT 3 RT
2
Panas jenis per gram atom zat padat :
E
o
3 R 5 , 94 kal/ K/gr.atom
T v
STATISTIK BOSE-EINSTEIN
g s g s 1 n s !
g s g s 1 n s !
ws
g s
g
s
1 n s !
s
1 ! n s !
g s
1 ! n s !
g s 1 n s !
g s 1 ! n s !
w
w
s
s
STATISTIK BOSE-EINSTEIN
g s 1 n s !
g 1! n !
s
s
s
w
w
s
s
STATISTIK BOSE-EINSTEIN
s
log w
dn s 0
x
s
n
s
log w
n s
log w
log
x
s
0
ws
s
g
s
s
1 n s log g s 1 n s g s 1 log g s 1 n s log n s
STATISTIK BOSE-EINSTEIN
log w
n s
log w
n s
log g s 1 n s log n s
g ns
log s
ns
g ns
x
log s
ns
gs
ns
e
x
0
s
s
1
STATISTIK BOSE-EINSTEIN
ns
gs
e
x
ns
s
1
gs
1
e
s / kT 1
A
ws
g s!
n s ! g s n s !
STATISTIK BOSE-EINSTEIN
ns
gs
e
x
ns
s
1
gs
1
e
s / kT 1
A
ws
g s!
n s ! g s n s !
STATISTIK BOSE-EINSTEIN
STATISTIK FERMI-DIRAC
W
w
s
s
ws
W
g s!
n s ! g s n s !
s
g s!
n s ! g s n s !
Jumlah untuk semua kemungkinan susunan
yang berbeda untuk satu tingkatan energi
Jumlah untuk semua kemungkinan susunan
yang berbeda
STATISTIK FERMI-DIRAC
log W
log
s
g
s
g s!
n s ! g s n s !
log g s n s log n s g s n s log g s n s
s
s
log W
s
n s
log W
n s
s 0
dn s 0
Gunakan rumus Stirling
STATISTIK FERMI-DIRAC
log W
log
n s
log
ns
g s ns
ns
gs
ns
g s ns
s 0
e
s
1
STATISTIK FERMI-DIRAC
~ kBT
1
f
e
F kT
1
n d f g d
T=0
(with respect to )
=
F
0 ,
f
F
0 ,
f
1
e
1
1
e
1
1
0
STATISTIK FERMI-DIRAC
ns
gs
e
s
1
1
f
e
F kT
1
Distribusi jumlah partikel partikel
Melalui normalisasi gs = 1 diperoleh
fungsi distribusi. Maka f(e) merupakan
probabilitas sebagai fungsi energi
Sebagai fungsi probabilitas maka harga fungsi ini maksimum 1
dan minimum 0
Radiasi Benda Hitam
Two types of bosons:
(a)
Composite particles which contain an even
number of fermions. These number of these
particles is conserved if the energy does not
exceed the dissociation energy (~ MeV in the
case of the nucleus).
(b) particles associated with a field, of which the
most important example is the photon. These
particles are not conserved: if the total
energy of the field changes, particles appear
and disappear. We’ll see that the chemical
potential of such particles is zero in
equilibrium, regardless of density.
Radiation in Equilibrium with Matter
Typically, radiation emitted by a hot body, or from a laser is not in equilibrium: energy
is flowing outwards and must be replenished from some source. The first step towards
understanding of radiation being in equilibrium with matter was made by Kirchhoff,
who considered a cavity filled with radiation, the walls can be regarded as a heat
bath for radiation.
The walls emit and absorb e.-m. waves. In equilibrium, the walls and radiation must
have the same temperature T. The energy of radiation is spread over a range of
frequencies, and we define uS (,T) d as the energy density (per unit volume) of the
radiation with frequencies between and +d. uS(,T) is the spectral energy density.
The internal energy of the photon gas:
u T u S , T d
0
In equilibrium, uS (,T) is the same everywhere in the cavity, and is a function of
frequency and temperature only. If the cavity volume increases at T=const, the
internal energy U = u (T) V also increases. The essential difference between the
photon gas and the ideal gas of molecules: for an ideal gas, an isothermal expansion
would conserve the gas energy, whereas for the photon gas, it is the energy density
which is unchanged, the number of photons is not conserved, but proportional to
volume in an isothermal change.
A real surface absorbs only a fraction of the radiation falling on it. The absorptivity
is a function of and T; a surface for which ( ) =1 for all frequencies is called a
black body.
Photons Apa Itu ?
The electromagnetic field has an infinite number of modes (standing
waves) in the cavity. Any radiation field is a superposition of plane
waves of different frequencies. The characteristic feature of the
radiation is that a mode may be excited only in units of the quantum
of energy hf (similar to a harmonic oscillators) :
i n i 1 / 2 h
T
This fact leads to the concept of photons as quanta of the electromagnetic field. The
state of the el.-mag. field is specified by the number n for each of the modes, or, in other
words, by enumerating the number of photons with each frequency.
According to the quantum theory of radiation, photons are massless
bosons of spin 1 (in units ħ). They move with the speed of light :
The linearity of Maxwell equations implies that the photons do not
interact with each other. (Non-linear optical phenomena are
observed when a large-intensity radiation interacts with matter).
E ph h
E ph cp ph
p ph
E ph
c
h
c
Presence of a small amount of matter is essential for establishing equilibrium in the
photon gas. We’ll treat a system of photons as an ideal photon gas, and, in particular,
we’ll apply the BE statistics to this system.
The mechanism of establishing equilibrium in a photon gas is absorption and emission
of photons by matter.
Potensial Kimia Foton = 0
F
N
The mechanism of establishing equilibrium in a photon gas is absorption
and emission of photons by matter. The textbook suggests that N can be
found from the equilibrium condition:
On the other hand,
F
N
ph
T ,V
Thus, in equilibrium, the chemical
potential for a photon gas is zero:
0
T ,V
ph 0
However, we cannot use the usual expression for the chemical potential, because one
cannot increase N (i.e., add photons to the system) at constant volume and at the same
time keep the temperature constant:
F
- does not exist for the photon gas
N T ,V
G N
Instead, we can use
G F PV
F T , V
F
P
V
V T
- by increasing the volume at T=const, we proportionally scale F
Thus,
G F
F
V 0
V
- the Gibbs free energy of an
equilibrium photon gas is 0 !
ph
G
0
N
For = 0, the BE distribution reduces to the Planck’s distribution:
n ph f ph , T
1
1
exp
k
T
B
1
h
1
exp
k
T
B
Planck’s distribution provides the average
number of photons in a single mode of
frequency = /h.
n h
The average energy in the mode:
k BT
In the classical (high temperature) limit:
h
h
1
exp
k
T
B
In order to calculate the average number of photons per small energy interval d, the
average energy of photons per small energy interval d, etc., as well as the total
average number of photons in a photon gas and its total energy, we need to know the
density of states for photons as a function of photon energy.
Rapat Keadaan Foton
kz
N k
kx
ky
g
extra factor of 2:
two polarizations
g
3D
ph
dG
d
g
3D
ph
1
4 / 3
3
8
Lx L y Lz
cp c k
k
d
d
h
k
volume
6
G
h 2
3
2
c
3
6
8
c
3
2
G k
2
3
g
c 3
3D
ph
2
g
3D
ph
k
6
3
2
2
8
c
3
2
2
2
c 3
Spektrum Radiasi Benda Hitam
Rata-rata jumlah foton per satuan volume denga frekwensi dan +d:
g f d u S , T d
u s , T h g f
8 h
c
3
3
exp h 1
- Rapat Spektrum (hukum Radiasi Planck)
u adalahfungsi energi:
u , T d u , T d
Radiasi spektrum
benda hitam
u , T u , T
u , T
8
hc 3
d
d
u h , T h
3
1
exp
k BT
u(,T) - the energy density per unit photon
energy for a photon gas in equilibrium with
a blackbody at temperature T.
Pendekatan Klasik (f kecil , besar), Hkm Rayleigh-Jeans
Pd frekwensi rendah dan temp. tinggi
u s , T
8 h
c
3
3
exp h 1
8
c
3
2
k BT
Hukum Rayleigh-Jeans
This equation predicts the so-called
ultraviolet catastrophe – an infinite
amount of energy being radiated at
high
frequencies
or
short
wavelengths.
h 1
exp h 1 h
- purely classical result (no h), can be
obtained directly from equipartition
Hukum Rayleigh-Jeans
u sebagai fungsi dari panjang gelombang
hc
d
2
d
u , T d u , T d
u , T
8
hc 3
c
h
3
hc
exp
k BT
1
hc 8 hc
2
5
In the limit of large :
u , T
large
8 k B T
4
1
4
1
hc
exp
k BT
1
frekwensi tinggi , Hukum Pergeseran Wien’s
At high frequencies:
u s , T
8 h
c
3
h 1
exp h
3
exp h 1 exp h
- Ditemukan secara eksperimen oleh Wien
Wien
Nobel 1911
Maksimum u() berfeser ke frekwensi tinggi ketika temperatur naik.
du
max 2 . 8
d
k BT
h
const
d
h
d
k
T
B
3
h
k T
3 x
3x2
x e
B
const x
x
e
1
h
e 1
exp
k T 1
B
3 x e x
3
h max
u(,T)
k BT
2
0
x 2 .8
2 .8
Hukum
Pergeseran
Wien
- the “most likely” frequency of a photon in a
blackbody radiation with temperature T
Numerous applications
(e.g., non-contact radiation thermometry)
max max
u , T
u , T
h max
2 .8
k BT
hc
max
u , T d u , T d
k B T max
max
hc
d
2
d
u , T
8
hc 3
c
h
- does this mean that
2 .8
Wrong!
?
3
hc
exp
k BT
1
hc 8 hc
2
5
1
hc
exp
k BT
1
2
d
1
5
x exp 1 / x
const
5
0
5
const 6
2
df
dx x exp 1 / x 1
x exp 1 / x 1
x exp 1 / x 1
du
5 x exp 1 / x 1 exp 1 / x
T = 300 K
“night vision” devices
max
hc
5 k BT
max 10 m
Radiasi Sinar Matahari
max
Temperatur permukaan- 5800K
hc
0 .5 m
5 k BT
As a function of energy, the spectrum of sunlight peaks at a photon energy of
u max h max 2 . 8 k B T 1 . 4 eV
(umax) 0.88 m, near infrared
- close to the energy gap in Si, 1.2 eV,
which has been so far the best material
for photovoltaic devices (solar cells)
Spectral sensitivity of the eye:
Hukum Radiasi Stefan-Boltzmann
Jumlah total foton persatuan volume :
n
N
V
n g d
0
8
c
3
0
2
h
1
exp
k BT
8 k T
d 3 B
c h
5
3
15 h c
4
u T
Tetapan Stefan-Boltzmann
2
0
3
2
kB 3
8
T 2 .4
x
e 1
hc
x dx
- increases as T 3
Energi total foton per satuan volume : (apat
energi gas foton)
2 k B
3
U
u T
g
V
exp 1d
k B T 4
3
15 hc
8
0
4
T
5
Hukum StefanBoltzmann
4
c
Energi rata-rata per foton :
u T
N
k B T 4 hc 3
3
3
15 hc 8 k B T 2 . 4
8
5
4
15 2 . 4
k B T 2 .7 k B T
(just slightly less than the “most”
probable energy)
Daya yang dipancarkan oleh Benda Hitam
For the “uni-directional” motion, the flux of energy per unit area c u
energy density u
1m2
c 1s
Integration over all angles provides a factor of ¼:
power emitted
by unit area
4
(the hole size must be >> the wavelength)
Thus, the power emitted by a unit-area
surface at temperature T in all directions:
power
c
4
u T
c
4
4
T
4
T
c
The total power emitted
by a sphere of radius R:
total power emitted
by a sphere 4 R T
2
1
4
T
4
cu
Beberapa Contoh
u T
The value of the Stefan-Boltzmann constant:
4
T
4
c
5 . 76 10
8
W / K m
4
2
T 500 W / m
4
Consider a human body at 310K. The power emitted by the body:
2
While the emissivity of skin is considerably less than 1, it emits sufficient infrared
radiation to be easily detectable by modern techniques (night vision).
Radiative transfer:
Liquid nitrogen is stored in a vacuum or Dewar flask, a container surrounded by a thin
evacuated jacket. While the thermal conductivity of gas at very low pressure is small, energy
can still be transferred by radiation. Both surfaces, cold and warm, radiate at a rate:
J rad 1 r T i
4
W /m
2
i=a for the outer (hot) wall, i=b for the inner (cold) wall,
r – the coefficient of reflection, (1-r) – the coefficient of emission
Let the total ingoing flux be J, and the total outgoing flux be J’:
Dewar
J 1 r T a r J
4
The net ingoing flux:
J J
J 1 r Tb rJ
4
1 r
1 r
T a Tb
4
4
If r=0.98 (walls are covered with silver mirror), the net flux is reduced to
1% of the value it would have if the surfaces were black bodies (r=0).
Efek Rumah Kaca
Absorption:
2
4 R Sun
P ower in R E T Sun
R orbit
the flux of the solar radiation energy
received by the Earth ~ 1370 W/m2
Emission:
2
Power out 4 R E T E
2
TE
4
R Sun
R
orbit
2
4
1/ 4
T Sun
Rorbit = 1.5·1011 m
Transmittance of the Earth atmosphere
= 1 – TEarth = 280K
However, in reality
RSun = 7·108 m
= 0.7 – TEarth = 256K
To maintain a comfortable temperature on the Earth, we need the Greenhouse Effect !
The complicated issue of global worming: adding CO2 (and other “greenhouse” gases)
to the atmosphere tends in itself to raise the earth’s average temperature, but also may
increase cloudiness, which lowers it. One thing is clear: since climate is largely
determined by the heat balance in the atmosphere, anything that changes the
atmospheric absorption is bound to have climatic consequences.
Pengurangan Massa Matahari
The spectrum of the Sun radiation is close to the black body spectrum with the
maximum at a wavelength = 0.5 m. Find the mass loss for the Sun in one second.
How long it takes for the Sun to loose 1% of its mass due to radiation? Radius of the
Sun: 7·108 m, mass - 2 ·1030 kg.
max = 0.5 m
P power emitted
max
hc
T
5 k BT
by a sphere
34
8
6 . 6 10
3 10
K
5
,
740
K
5 1 . 38 10 23 0 . 5 10 6
hc
5 k B max
2 k B
5
4 R 2 T 4
3
15 h c
4
2
5 . 7 10
8
W
2
m K
4
This result is consistent with the flux of the solar radiation energy received by the Earth
(1370 W/m2) being multiplied by the area of a sphere with radius 1.5·1011 m (Sun-Earth
distance).
P 4 R Sun
hc
2
2 . 8 k B max
4
4 7 10 8 m
the mass loss per one second
dm
dt
1% of Sun’s mass will be lost in
t
P
c
2
2
5 . 7 10
W
8
2
m K
3 . 8 10
26
3 10
m
0 . 01 M
dm / dt
8
W
2
2 10
4
5 ,740K
4
3 . 8 10
26
W
s 1.5 10
11
yr
4 . 2 10 kg/s
9
28
kg
4 . 2 10 kg/s
9
4 . 7 10
18
Fungsi Distribusi untuk gas Fermi Ideal
The probability of the i-state with energy i to be occupied
by ni particles (the total energy of this state ni i) :
The grand partition function for all particles in the ith singleparticle state (the sum is taken over all possible values of ni) :
P i , n i
Zi
n ni
exp i i
Z
k BT
1
exp
ni
n i i
k BT
Z i 1 exp
k
T
B
The mean number of particles in this state:
exp
k B T
1
n i n i P n i 0 P 0 1 P 1
ni
1 exp
1 exp
k T
k B T
B
1
- the Fermi-Dirac
n
~ kBT
distribution
1
exp
k BT
If the particles are fermions, n can only be 0 or 1:
At T = 0, all the states with < have the average #
of particles 1 (i.e., they are occupied with 100%
probability), all the states with > have the average
# of particles 0 (i.e., they are unoccupied). With
increasing T, the step-like function is “smeared” over
the energy range ~ kBT.
T=0
(with respect to )
=
Fungsi Distribusi Gas Bose Ideal
The grand partition function for all particles in the
ith single-particle state:
(the sum is taken over the possible values of ni)
If the particles are bosons,
n can any integer 0:
Z i 1 exp
exp
k BT
ni
The mean
number of
particles in this
state:
ni
ni
Z x
1 e
x
exp
ni
n i i
k BT
2
3
Z i 1 exp
exp
exp
....
k
T
k
T
k
T
B
B
B
2
exp
k BT
3
....
k BT
1 exp
1
k BT
n i P n i 0 P 0 1 P 1 2 P 2 ... x
k BT
ni
1 Z
Zi
n
exp i
k BT
1
ni
Z
Z
x
1
e
1
x
x
x
x 1 e 1 e
e 1
n i
ni
x
exp n i x
1 Z
Z
x
1
exp
k BT
1
Distribusi Bose
Einstein
The mean number of particles in a given state for the BEG can exceed unity, it diverges as
, and is nonexistent for > .
Probabilitas, Fungsi Distribusi, Rapat Keadaan ….
The probability that the system is in state s
with energy E and N particles
U(x)
P i
x
T n i
exp i
1
Z
k
T
B
1
P E 1
s
The macrostate of such system is completely defined if
we know the mean occupancy for all energy levels,
which is often called the distribution function:
f E n E
While f(E) is often less than unity (much less in the case of an ideal gas), it is not a
probability. (e.g., it can exceed unity in a Bose gas).
f E n
where n=N/V – the density of particles
i
If we can neglect the
spectrum discreteness:
n
0
g f d
where g() is the
density of states
Kaitan Termodinamika, Potensial Kimia
Consider the grand potential
k B T ln Z
which is a generalization of F=-kBT lnZ
d SdT PdV Nd
- the appearance of μ as a variable, while computationally very convenient for the grand
canonical ensemble, is not natural. Thermodynamic properties of systems are
eventually measured with a given density of particles. However, in the grand canonical
ensemble, quantities like pressure or N are given as functions of the “natural” variables
T,V and μ. Thus, we need to use
n=N/V.
/ T ,V
N
to eliminate μ in terms of T and
S
U
F
N U ,V N S ,V N T ,V
n , T T
Boltzmann
Boltzmann
Gas
MB < 0:
nQ
0
k B T ln
n
μ for an ideal gas is negative: when you add a
particle to a system and want to keep S fixed, you
typically have to remove some energy from the
system.
- the occupancy n B exp
k
T
B
cannot be negative for any
Potensial Kimia untuk Gas Fermi
Fermi
Gas
n F f F
1
exp
k BT
n
1
0
g f d
g
0
T , n
1
exp
k
T
B
d
T , V , N n N / V T , n
When the average number of fermions in a system (their density) is known, this equation
can be considered as an implicit integral equation for (T,n). It also shows that
determines the mean number of particles in the system just as T determines the mean
energy. However, solving the eq. is a non-trivial task.
/EF
2
2
depending on n and T, for 1
EF
1
k BT
....
12 E F
fermions may be either
positive or negative.
1
kBT/EF
The limit T0: adding one fermion to the system at T=0 increases its energy U by EF. In
this situation F = U-TS = U (S is also 0: all the fermions are packed into the lowest-energy
states), so that the chemical potential, which is the change in F produced by the addition
of one particle, is EF:
T 0 E
F
The change of sign of (n,T) indicates the crossover from the
n
4 EF
degenerate Fermi system (low T, high n) to the Boltzmann statistics.
nQ
3 k BT
The condition kBT << EF is equivalent to n >> nQ:
The crossover occurs at n~nQ When n<<nQ the chemical potential
nQ
becomes negative:
0
Boltzmann k B T ln
n
3/2
Potensial Kimia untuk Gas Bose
Bose
Gas
n BE
1
exp
k BT
1
n
g f d
0
g
0
exp
k BT
1
The occupancy cannot be negative for any , thus, for bosons,
0 ( varies from 0 to ). Also, as T0, 0
n BE T 0
1
exp 0 / 0 1
n BE T 0
d
T
0, 0
1, 0
For bosons, the chemical potential is a non-trivial function of the density and temperature
(for details, see the lecture on BE condensation).
Pendekatan Klasik
The Fermi-Dirac and Bose-Einstein distributions must reduce to the MaxwellBoltzmann distribution in the classical limit, n i 1 for all i. Hence,
exp
1
k
T
B
ni
exp
k BT
exp
k BT
1
and
the MaxwellBoltzmann
distribution
The same result, of course, we would get if we start from the equation for the
average nk in Boltzmann statistics:
n i NP s
Z
Z
k B T ln 1 1 exp
exp
exp
k T
k T
Z1
N
N
B
k BT
B
N
exp
exp
k T
k BT
B
Comparison of the MB, FD, and BE
distributions plotted for the same
value of . Note that the MB
distribution makes no sense when the
average # of particle in a given state
becomes comparable to 1 (violation of
the dilute limit).
=
Pendekatan Klasik (cont.)
2 mk B T
n n Q
2
h
In terms of the density, the classical limit
corresponds to n << the quantum density:
3/2
We can also rewrite this condition as T>>TC where TC is the so-called degeneracy
temperature of the gas, which corresponds to the condition n~ nQ. More accurately:
TC
h
2
2 mk B
n
2 .6
2/3
For the FD gas, TC ~ EF/kB where EF is the Fermi energy (Lect. 24) , for the BE gas
TC is the temperature of BE condensation (Lect. 26).
Critical density for bosons:
n
0
g
3/2
3/2
1/ 2
2s 1 2m
2
s
1
2
mk
T
x
1/ 2
B
d g
dx
2
2
2
2
exp 1
4
4
exp
x
1
0
Since 0, the maximum possible value of
n is obtained when = 0, and
x
1/ 2
exp x 1 dx
1 .3
0
n cr
2 s 1 2 mk B T
1 .3
2
2
4
3/2
2 .6 n Q
where nQ is the quantum concentration,
which varies as T 3/2
Pendekatan Ketiga Distribusi
S
Fermi-Dirac
Maxwell-Boltzmann
Nk B
3
Bose-Einstein
U
2
k B TC
3
1
2
T
1
zero-point
energy,
Pauli
principle
2
3 TC
1
T
1
2
3
TC
TC
h
2
2 mk B
n
2 .6
2/3
Comparison between Distributions
CV /NkB
Fermi-Dirac
Maxwell-Boltzmann
Bose-Einstein
2
1.5
0
1
T/TC
Comparison between Distributions
Maxwell
Boltzmann
nk
Bose
Einstein
1
exp
k BT
nk
Fermi
Dirac
1
exp
k BT
1
nk
1
exp
k BT
1
distinguishable
Z=(Z1)N/N!
nK<<1
indistinguishable
integer spin 0,1,2 …
indistinguishable
half-integer spin 1/2,3/2,5/2 …
spin doesn’t matter
bosons
fermions
localized particles
don’t overlap
wavefunctions overlap
total symmetric
wavefunctions overlap
total anti-symmetric
photons
atoms
free electrons in metals
electrons in white dwarfs
unlimited number of
particles per state
never more than 1
particle per state
gas molecules
at low densities
“unlimited” number of
particles per state
nK<<1
4He
Aplikasi Statistik Termodinamika
Paramagnetism
Fungsi Partisi
Aplikasi Statistik Termodinamika
Momen magnet rata-rata
Fungsi Partisi
Aplikasi Statistik Termodinamika
Kapasitas panas magnetik
Aplikasi Statistik Termodinamika
Untuk temperatur rendah
Aplikasi Statistik Termodinamika
Jika dideferensial terhadap B
Aplikasi Statistik Termodinamika