CURICULUM VITAE A. DATA DATA DIRI DI RI
01. N a m a 02. Tempat/Tanggal Tempat/Tanggal Lahir 03. Jenis Kelamin 04. Fakultas/Jurusan 05. Pangkat/Golongan/NIP Pangkat/Golongan/NIP 06. Bidang Keahlian 07. Alamat Rumah
: : : : : : :
09. e-mail
Dr. H. Muris, M.Si Tinggas, 1965 Laki-laki FMIPA/Fisika FMIPA/Fisika Lektor Kepala/IV/a/131925820 Kepala/IV/a/131925820 Fisika Material BTN Minasa Upa G20/14 Makassar. Makassar. 90224. Telp. (0411) 886307 HP. 081342403676 : Jurusan Jurus an Fisika FMIPA UNM Kampus Parangtambung Makassar Tlp/Fax. (0411)840622, HP. 081342403676 :
[email protected] [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 Spesiali sasi 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 R u j u k a n U t am am a : Introdution to Statistical Physics for Students by Pointon Longman, England
R u j u k a n Ta Ta m b a h a n : Buku Buku Fisika Zat Padat, Fisika Kuantum dan Fisika Modern yang relevan
Pokok Bahasan 1.
Pengantar
2. Stat Statis isti tik k Max Maxwel well Bolt Boltzm zman ann n 3. Apli Aplika kasi si Stat Statis isti tik k Maxw Maxwel elll Bolt Boltzm zman ann n 4. Stat Statis isti tik k Bose Bose Eins Einsttein ein 5.
Statistik Fe Fermi Di Dirac
6. Tempe empera rattur dan Entro ntropy py 7. Apli Aplika kasi si Stat Statis isti tik k Termo ermodi dina nami mika ka 8.
Ensemble Kanonik
9.
Grand En Ensemble Ka 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 43 635
X
7
28 7
4
x1 x 2 x3 x 4 x5 x6 x7 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 kontinyu
Hasil ini diperoleh dari pengembangan bentuk
f .( x ). x X f ( x ). x f ( x ) i
i
i
i
i
i
f ( x ) 1 i
i
i
Jika fungsinya kontinyu maka :
X x. f ( x)dx
Bagaimana anda mengartikan parameter statistik berikut ?
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
p x2 p y2 p z 2 2m
d dxdydzdp x dp y dp z Ruang fase Ruang momentum
d 6 N dx1 dy1 dz 1 dp x1 dp y1 dp z 1 ........dxi dyi dz i dp xi dp yi dp zi ........dx N dy N dz N dp xN dp yN dp zn N
dxi dyi dz i dp xi dp yi dp zi i 1
N
d i i 1
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
6 N
Normalisasi terhadap ruang
X x( N ), p( N ) P x( N ), p( N )d
6 N
X
6 N
P x( N ), p( N )d
6 N
6 N
Rata Rata Sifat Assembly
Jika X merupakan fungsi yang diskrit, maka perata-rataan fungsi X dapat dinyatakan dengan :
p X X p i
i
i
i
i
Normalisasi probabilitas menghasilkan
p
i
1
i
X
p X i
i
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 ! N 1!. N 2 !
W
4!
g 12 . g 22
3 2.4 2 864
2 !.2!
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
exp k BT
0
2 N
kT 3 / 2 g ( )
e
/ kT
1 / 2 d
g C
P ( )
= 0
0
Aplikasi Statistik Maxwell Boltzmann Untuk partikel kuantum dalam kotak 2D (e.g., electron pd FET):
k y
2D
k
k x
L x
3D k z
k y
L x
n y
k k x k y 2
L y
1 k 2 k 2 area N k 4 4 L x L y
k x
n x
g 2 D
# states within ¼ of a circle of radius k
N k
1
4 / 3 k 3
8 L x L y L z
Gk
k 3 volume 6
2
Gk
k 2 4
2
G
2 s 1 m
k 3 6 2
2 2
1 2m 4
2
- Tak bergantung pd
1 2m G 6 2 2
3/ 2
g ( ) 3D
3D
k x
g
2 s 1 2m 3 / 2 4 2
2
1/ 2
2D
1D
k y
Thus, for 3D electrons (2s +1=2):
1 2m g 2 2 2 3D
3/ 2
1/ 2
Distribusi Kecepatan Maxwell m f v f v 2 k T B
3/ 2
mv 2 4 v 2 dv exp 2k BT
v y
" volume"v v dv
4 v 2dv
Nampak bahwa persamaan ini merupakan perkalian antara faktor Boltzmann dengan sebuah tetapan . Tetapan tersebut dapat diperoleh dari normalisasi
m C 2 k BT
f v dv 1
0
v v x
3/ 2
v z P(v)
dN NP d N exp
d
k BT
Distrib usi energi, N – the total # of particles
m dN v NP v dv N 2 k BT
3/ 2
mv 2 dv 4 v exp 2k BT 2
s p e e d d i s t r i b u t i o n (d i s t r i b u s i k e c e p a t an ) 1/ 2
m dN v x NP v x dv N 2 k BT
mv 2 dv exp 2k BT
v P (v x )
Karakteristik Nilai Kecepatan m P v 2 k BT
3/ 2
mv 2 4 v exp 2k BT 2
P (v )
Lihat bahwa distribusi ini tidak simetrik, sehingga perlu dicari perata-rataan sebagai berikut
The roo t-m ean-sq uare speed is proportional to the square root of the average energy:
E vm a x
v
1 2
mvrms
2
vrms
m
3k BT
m
v
v rms
H a r g a k e c .m a k s i m u m :
dP v dv v v
0 vmax ma x
Kelaju an rata-rata :
2 E
2k BT m
m mv2 8k BT 3 4 v exp dv v v P v dv m 2 k BT 0 2k BT 0
vmax v vrms 2 8 / 3 1 1.13 1.22
Soal (Maxwell distr.) Consider a mixture of Hydrogen and Helium at T=300 K. Fi nd the speed at which the Maxwell distributions for these gases have the same value.
m P v, T , m 2 k BT m1 2 k BT
3/ 2
2
2
ln
m1 m2
v2 2k BT
mv 2 4 v exp 2k BT 2
m1v 2 m2 2 4 v exp 2k BT 2 k BT 3
3
3/ 2
ln m1
m1v 2 2k BT
v
m2v 2 4 v exp 2k BT
3
m2v 2
2
2k BT
ln m2
3k BT ln
m1 m2
3/ 2
m1 m2
m1 m2
2
3 1.38 1023 300 ln 2 2 1.7 1027
1.6 km/s
Soal (Maxwell distr.) Find the temperature at which the number of molecules m olecules 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 BT 1/ 2
3 m 2 2 k BT
3/ 2
mv 2 4 v exp 2k BT 2
maximum:
mv 2 m m exp 2 2 k BT 2k BT 2 k BT
mv2 0 2 2k BT 3
T
3/ 2
P v, T 0 T
mv 2 mv 2 0 exp 2 2k BT 2k BT
mv 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, Doppler, maka panjang gelombang yang diamati pengamat adalah :
o 1
v
c o
v x
c
o
dv x
c
o
d
o
Pelebaran Garis Spektrum Doppler
Dari distribusi Maxwell Boltzamann m dN v Nf v dv N 2 k T B
3/ 2
mv 2 dv 4 v exp 2 k T B 2
Ubah sebagai fungsi panjang gelombang m f d 2 k BT
3/ 2
mc 2 o 2 c exp exp 2k T 2 d B o o
o
Pelebaran Garis Spektrum Doppler Intensitas radiasi : mc 2 o 2 I d Cf ( )d I o exp 2k T 2 d B o I ( o ) I ( )
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 x
p x2
2m
Maka energi rata-ratanya adalah :
p x2 / 2 m e e / KT d
e e / kT dT
Prinsip Ekipartisi Energi 2
p x
2m
Nyatakan energi sebagai
x
Misalkan
dan x
p x2 p x2 exp( ) / kT dxdydzdp y dp z exp( p x2 / 2mkT )dp x 2m 2m
exp(
p x2 2m
) / kT dxdydzdp y dp x x
exp( p x2 / 2mkT )dp x
p x2
= u 2
maka
2mkT
k T e u u 2 du x
2
e u du 2
Prinsip Ekipartisi Energi
e
Hasilnya memberikan :
u
2
u du 2
Maka :
x
1
1 2
e
u 2
du
p x2 k T
2 mkT
2
Karena ada satu bentuk kuadrat maka memberikan energi rata-rata ½ kT Contoh 2 : Osilator harmonik dengan dua jenis energi
x
p x2 2m
1 2
x 2
u
Prinsip Ekipartisi Energi
Maka :
1 2 2 p / 2 m x e e / kT d x 2 x e / kT e d dp xdx p x2 1 2 2 1 2 p x exp x / kT dxdp x x 2 2 m 2 x p x2 1 2 exp 2m 2 x kT dxdp x
Ubah ke koordinat polar : 2
p x
2m
r 2 sin 2 ,
1 2
x 2 r 2 cos2 1 2
dp x dp y 2(m / ) rdrd
Prinsip Ekipartisi Energi
Maka : 2 x
d e r
0 2 x
2
/ kT
r 3 dr
k T
0
d e r
0
0
2
/ kT
rdr
Karena terdiri dari dua bentuk kuadrat maka energinya adalah 2 x ½ kT = kT Untuk osilator harmonik 3D maka : p 2 p y2 p 2 1 1 1 3 x kT kT kT kT x 2 2 2 2m 2m 2m 2
3 2
kT
Prinsip Ekipartisi Energi Energi rata-rata untuk osilator harmonik 3 D. 2 2 p x2 1 p p 1 1 y 2 2 2 x 1 x 2 y 3 z 2m 2 2m 2 2m 2
1
6. kT 2 3kT
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 N A (bil. Avogadro) molekul gas dan berlaku sebagai osilator harmonik 3D, maka, terdapat 6 derajat kebebasan,maka :
E 6 N A
1 2
kT 3 RT
Panas jenis per gram atom zat padat :
E 3 R 5,94 kal/ o K/gr.atom T v
Panas jenis gas
Jika terdapat N A (bil. Avogadro) molekul gas dan berlaku sebagai osilator harmonik 3D, maka, terdapat 6 derajat kebebasan,maka :
E 6 N A
1 2
kT 3 RT
Panas jenis per gram atom zat padat :
E 3 R 5,94 kal/ o K/gr.atom T v
STATISTIK BOSE-EINSTEIN
g s g s 1 n s ! w s
s
g s 1 n s ! g s 1! n s !
g s g s
g s
1 n s ! 1! n s !
g s 1 n s ! g s 1!n s ! w
w
s
s
STATISTIK BOSE-EINSTEIN
s
w
g s 1 n s ! g s 1! n s !
w
s
s
STATISTIK BOSE-EINSTEIN
log w x s dns 0 n s s
log w x s 0 n s
log w
log w
s
s
g s 1 n s log g s 1 n s g s 1log g s 1 n s log n s s
STATISTIK BOSE-EINSTEIN
log w log g s 1 n s log n s n s g n log w log s s n s n s g s n s x s 0 n s
log
g s n s
e x 1 s
STATISTIK BOSE-EINSTEIN
n s
g s e
n s
w s
x s
1
g s
1 s / kT 1 e A
g s ! n s ! g s
n s !
STATISTIK BOSE-EINSTEIN
n s
g s e
n s
w s
x s
1
g s
1 s / kT 1 e A
g s ! n s ! g s
n s !
STATISTIK BOSE-EINSTEIN
STATISTIK FERMI-DIRAC
W
w
s
s
w s
W
g s ! n s ! g s
n s ! g s !
n ! g n ! s
s
s
s
Jumlah untuk semua kemungkinan susunan yang berbeda untuk satu tingkatan energi
Jumlah untuk semua kemungkinan susunan yang berbeda
STATISTIK FERMI-DIRAC
log W
g s !
log n ! g s
s
s
ns !
g s log g s n s log n s g s n s log g s n s s
log W n s s
s
log W 0 n s s
dn s 0
Gunakan rumus Stirling
STATISTIK FERMI-DIRAC
g s n s log W log n s n s log
g s
n s n s
g s n s
0 s
e
s
1
STATISTIK FERMI-DIRAC ~ k B T
f
1 e
F kT
1
nd f g d
T =0
(with respect to )
=
1
F 0,
f
F 0,
f
e 1
1 e 1
1
0
STATISTIK FERMI-DIRAC
n s
g s e
f
s
1
Distribusi jumlah partikel partikel
1
Melalui normalisasi gs = 1 diperoleh fungsi distribusi. Maka f(e) merupakan probabilitas sebagai fungsi energi
1 F kT
e
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 c a v i t y f i l l e d w i t h r a d i at i o n , 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 u S ( ,T ) d as the energy density (per unit volume) of the radiation with frequencies between and +d . u S ( ,T ) is the spectral energy density. The internal energy of the photon gas:
u T uS , T d 0
In equilibrium, u S ( ,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 e n e r g y d e n s i t y 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 m o d e m a y b e ex c i t ed o n l y i n u n i t s o f t h e q u an t u m o f e n e r g y h f (similar to a harmonic oscillators) :
T
i ni 1 / 2 h
This fact leads to the concept of p h o t o n s a s q u a n t a o f t h e e l ec t r o m a g n e t i c f i el d . 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 m a s s l e s s b o s o n s o f s p i n 1 (in units ħ). They move with the speed of light :
E ph h
The linearity of Maxwell equations implies that t h e p h o t o n s d o n o t i n t e r a c t w i t h e a c h o t h e r . (Non-linear optical phenomena are observed when a large-intensity radiation interacts with matter).
E ph
E ph cp ph p 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 a n i d e al p h o t o n g a s , and, in particular, we’ll apply the BE statistics to this system. The mechanism of establishing equilibrium in a photon gas is a b s o r p t i o n a n d e m i s s i o n of photons by matter.
Potensial Kimia Foton = 0 The mechanism of establishing equilibrium in a photon gas is a b s o r p t i o n a n d e m i s s i o n of photons by matter. The textbook suggests that N can be found from the equilibrium condition: On the other hand,
F ph N T ,V
F 0 N T ,V
Thus, in equilibrium, the c h e m i c a l p o t e n t i a l for a photon gas is zero:
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 V V T
P
- by increasing the volume at T =const, we proportionally scale F Thus,
G F
F V
V 0
- the Gibbs free energy of an equilibrium photon gas is 0 !
For = 0, the BE distribution reduces to the
n ph f ph , T
1
1 k T
exp
1
h 1 k T
exp
ph
G N
0
Planck’s distribution:
Planck’s distribution provides the average number of photons in a single mode of frequency = /h .
The average energy in the mode:
h 1 k BT
exp
k BT
In the classical (high temperature) limit:
h
n h
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 d e n s i t y o f s t a t es f o r p h o t o n s as a function of photon energy.
Rapat Keadaan Foton
k z
N k k x k y
g
extra factor of 2: two polarizations
3 D ph
g
dG d
g
3D ph
1
4 / 3 k 3
8 L x L y L z
k 3 volume
cp ck G
2
6
3 6 c
h 2 8 2 h 2 3 3 d c c d
2
3
Gk
3D ph
g
3D ph
g
k 3 6 2
2 2 2 c
3
8 2 c
3
Spektrum Radiasi Benda Hitam Rata-rata jumlah foton per satuan volume denga frekwensi dan +d:
g f d uS , T d
u s , T h g f
8 h
3
c3 exp h 1
- R a p at S p e k t r u m ( h u k u m R ad i a s i P l an c k ) u adalahfungsi energi:
u , T d u , T d u , T u , T
R a d i as i s p e k t r u m b e n d a h i t am
u , T
d d
3
8
hc3
u h , T h
1 k BT
exp
u ( ,T ) - the energy density per unit photon
energy for a photon gas in equilibrium with a blackbody at temperature T .
Pendekatan Klasik (f k e c i l , Pd frekwensi rendah dan temp. tinggi
u s , T
8 h
3
c3 exp h 1
8 2 c3
k BT
H u k u m R a y l ei g h - J e an s
This equation predicts the so-called u l t r a v i o l e t c a t as t r o p h e – an infinite amount of energy being radiated at high frequencies or short wavelengths.
besar), Hkm Rayleigh-Jeans 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 3
c h hc 8 1 d hc 8 hc u , T d u , T d 2 u , T 5 hc 2 hc hc 3 d 1 1 exp exp k T k T B B In the limit of large :
u , T large
8 k BT
4
1 4
frekwensi tinggi , Hukum Pergeseran Wien’s At high frequencies:
u s , T
8 h c
3
h 1 exp h 1 exp h
3 exp h
- Ditemukan secara eksperimen oleh Wien
Wien Nobel 1911
Maksimum u ( ) berfeser ke frekwensi tinggi ketika temperatur naik.
m ax 2.8
k BT h
h 3 3 x 2 k T du d x 3e x B const const x 0 2 x d e 1 h h e 1 exp 1 d k BT k BT
3 x e x 3 h m ax )
T ,
(
u
k BT
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
u , T
h m ax
u , T
k BT
2 .8
hc m ax
m a x
k BT m ax
max
- does this mean that
2.8 ?
Wrong!
3
c h d hc 8 1 hc 8 hc u , T d u , T d 2 u , T 5 hc 2 hc hc 3 d 1 1 exp exp k BT k BT
5 x 2 exp1 / x const 5 0 const x 6 exp1 / x 1 5 2 df dx x exp1 / x 1 x exp1 / x 1
du
d
1
5 xexp 1 / x 1 exp 1 / x
T = 300 K
m ax
m ax
hc 5 k BT
10 m
Radiasi Sinar Matahari m ax
Temperatur permukaan- 5800K
hc 5 k BT
0.5 m
As a f u n c t i o n o f en e r g y , the spectrum of sunlight peaks at a photon energy of
um ax h m ax 2.8k BT 1.4 eV
(u m ax )
- 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:
0.88 m, near infrared
Hukum Radiasi Stefan-Boltzmann Jumlah total foton persatuan volume :
n
N V
n g d 0
8 c3
0
8 k BT
4
3
x 2 dx
h 1 k BT
exp
- increases as T 3
Energi total foton per satuan volume : (apat energi gas foton)
2 5 k B
3
k d 3 8 B T 3 2.4 x c h 0 e 1 hc
2
u T
Tetapan Stefan-Boltzman n
15 h 3c 2
U V
u T
g
c
4
d 3 exp 1 15hc
0
4
8 5 k BT
H u k u m S t ef a n - Boltzmann
4
T
Energi rata-rata per foton :
u T N
8 5 k BT hc 4
3
15hc 8 k BT 2.4 3
3
4 15 2.4
(just slightly less than the “most” probable energy)
k BT 2.7 k BT
Daya yang dipancarkan oleh Benda Hitam For the “uni-directional” motion, the flux of energy per unit area c u e n e r g y d e n s i t y u
1m2
c 1s
Integration over all angles provides a factor of ¼:
power emitted by unit area (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 c
T 4 T 4
The total power emitted by a sphere of radius R :
total power emitted by a sphere 4 R 2 T 4
T
1 4
cu
Beberapa Contoh
u T
The value of the Stefan-Boltzmann constant:
4 c
T 4
5.76 10 8 W / K 4 m 2
Consider a human body at 310K. The power emitted by the body:
T 4 500 W / m 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 / m2
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
Dewar
J’ :
J 1 r T a4 r J J 1 r T b4 rJ The net ingoing flux:
J J
1 r 1 r
T a4 T b4
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:
RSun Rorbit
2
P ower in R E T Sun 2
4
the flux of the solar radiation energy received by the Earth ~ 1370 W/m 2 Emission:
Power out 4 R E T E 2
4
1/ 4
R 2 T E Sun T Sun 4 Rorbit 11 m R orbit = 1.5·10
Transmittance of the Earth atmosphere = 1 – T Earth = 280K
However, in reality
8 R Sun = 7·10 m
= 0.7 – T Earth = 256K
To maintain a comfortable temperature on the Earth, we need the Greenhouse Effect ! The complicated issue of g l o b a l w o r m i n g : adding CO 2 (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: 7·10 8 m, mass - 2 ·10 30 kg. max = 0.5 m
ma x
hc 5 k BT
T
hc 5 k B ma x
6.6 1034 3 108 K 5 , 740 K 23 6 5 1 . 38 10 0 . 5 10
P power emitted by a sphere 4 R T 2
4
2 5 k B
4
15 h 3c 2
5.7 10 8
W m 2 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 1.5·10 11 m (Sun-Earth distance). 4
hc 2 W 2 4 7 108 m 5.7 108 2 4 5,740K 4 3.8 1026 W P 4 RSun m K 2.8 k B max the mass loss per one second
1% of Sun’s mass will be lost in
dm dt
t
P c2
3.8 1026 W
3 10 m
0.01 M dm / dt
8
2
4.2 109 kg/s
2 10 28 kg 4.2 10 9 kg/s
4.7 10 18 s 1.5 10 11 yr
Fungsi Distribusi untuk gas Fermi Ideal The probability of the i -state with energy i to be occupied by n i particles (the total energy of this state n i i) :
P i , ni
The grand partition function for all particles in the i t h singleparticle state (the sum is taken over all possible values of n i ) :
1 Z
ni i ni k BT
exp
ni i Z i exp k BT n i
k BT
If the particles are f e r m i o n s , n can only be 0 or 1:
Z i 1 exp
The mean number of particles in this state:
1 k BT ni ni P ni 0 P 0 1 P 1 n 1 exp 1 exp k BT k BT exp
i
n
1
1 exp k T B
- the Ferm i-Dirac
~ k B T
distribution
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 ~ k B T .
T =0
Fungsi Distribusi Gas Bose Ideal The grand partition function for all particles in the i th single-particle state: (the sum is taken over the possible values of n i ) If the particles are b o s o n s , n can any integer 0:
Z i
ni i k T B
exp
ni
2 3 exp exp k T .... k T k T B B B
Z i 1 exp 2
3
1 Z i 1 exp exp exp .... k BT k BT k BT 1 exp k T B ni
The mean number of particles in this state:
ni P ni 0 P 0 1 P 1 2 P 2 ... x
ni
ni ni
ni k BT 1
exp
x
1 Z exp n x i Z n x Z x
Z
1 1 e x 1 e ni x 1 e x 1 e x e x 1 Z x 1 Z
k BT
i
ni
1
1 exp k T B
D i s tr i b u s i B o s e 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 p r o b a b i l i t y that the system is in state s with energy E and N particles
U (x )
P i x
1 Z
i T ni 1 k T B
exp
P E 1 s
The macrostate of such system is completely defined if we know the m e a n o c c u p a n c y for all energy levels, which is often called the d i s t r ib u t i o n f u n c t i o n :
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 g f d 0
where g ( ) is the d e n s i t y o f s t a t es
Kaitan Termodinamika, Potensial Kimia Consider the g r a n d p o t e n t i al
k BT ln Z
which is a generalization of F=-k B T 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:
n k BT ln Q 0 n
μ for an ideal gas is n e g a t i v e : when you add a particle to a system and want to keep S fixed, you typically have to r e m o v e some energy from the system.
cannot be negative for any k BT
- the occupancy n B exp
Potensial Kimia untuk Gas Fermi F e r m i n F f F Gas
1
n g f d
1 exp k T B
0
g
T , n 0 1 exp k BT
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. / E F 2 depending on n and T , for 1 2 k BT .... f e r m i o n s may be either 1 E F 12 E F 1 p o s i t i v e or n e g a t i v e . k B T/E F
The limit T 0: adding one fermion to the system at T= 0 increases its energy U by E F . 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 E F : T 0 E F
The change of sign of (n ,T ) indicates the crossover from the degenerate Fermi system (low T , high n ) to the Boltzmann statistics. The condition k B T > n Q : The crossover occurs at n ~n Q When n
