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# Fisika Modern (10) statistical physics

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### Fisika Modern (10) statistical physics

1. 1. Fisika Modern Pertemuan 10-11 Statistical Physics Hadi Nasbey, M.Si <ul><li>Jurusan Fisika </li></ul><ul><li>Fakultas Matematika dan Ilmu Pengetahuan Alam </li></ul>01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
2. 2. Outline <ul><li>The Boltzmann Distribution </li></ul><ul><li>The Maxwell Distribution </li></ul>01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
3. 3. Introduction We believe we now have the basic laws that, in principle, can be used to predict the detailed behavior of an arbitrarily large assembly of atoms and molecules But even a tiny piece of matter consists of millions of atoms In practice, the complexity of the calculation is far beyond the capability of any conceivable computer and we need a different approach 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
4. 4. The Boltzmann Distribution Ludwig Boltzmann 1844 - 1906 The Austrian physicist Boltzmann asked the following question: in an assembly of atoms, what is the probability that an atom has total energy between E and E+dE ? His answer: where 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
5. 5. The Boltzmann Distribution is called the Boltzmann distribution , e -E/kT is the Boltzmann factor and k = 8.617 x 10 -5 eV/K is the Boltzmann constant The Boltzmann distribution applies to identical , but distinguishable particles 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
6. 6. The Boltzmann Distribution The number of particles with energy E is given by where g(E) is the statistical weight , i.e., the number of states with energy E . However, in classical physics the energy is continuous so we must replace g(E) by g(E)dE , which is the number of states with energy between E and E + dE . g(E) is then referred to as the density of states . 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
7. 7. The Boltzmann Distribution Classically, the total energy of a gas molecule of mass m , near the Earth’s surface, is where z is the vertical distance above the ground Example: The Law of Atmospheres z So we can write Wanted: the fraction of particles between z and z+dz 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
8. 8. The Boltzmann Distribution A basic rule of probability is: sum, or integrate, over quantities whose values are either unknown or not of interest. We are interested only in z . After integrating the Boltzmann distribution with respect to p we get Example: The Law of Atmospheres z 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
9. 9. The Boltzmann Distribution The fraction of molecules between z and z + dz is Example: The Law of Atmospheres z At T = 300K, the ratio of the fraction at z = 1000 m to that at z = 0 m is just f B (1000) / f B (0) = 0.893 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
10. 10. The Boltzmann Distribution At temperature T, the atoms of a gas will occupy different energy levels. For hydrogen, the energy difference E 2 - E 1 between the 1 st excited state and the ground state is 10.2 eV . What is the ratio of the number of atoms in the 1 st excited state to the number in the ground state at T= 5800 K (the temperature of the Sun’s “surface”)? Example: H Atoms in First Excited State 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
11. 11. The Boltzmann Distribution <ul><li>Example : </li></ul><ul><li>Number of atoms in state E </li></ul><ul><li>Ratio of number of atoms in E 2 and E 1 </li></ul>01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
12. 12. The Boltzmann Distribution <ul><li>Example : </li></ul><ul><li>Ratio of statistical weights. The degeneracy for each orbital quantum number l is given by 2l+1 . For the ground state of hydrogen, l = 0 , which gives 1 . For the 1 st excited state l = 0 and l = 1 , which gives 4 . But for each of these states the electron has 2 spin states. So we have g 1 = 2 and g 2 = 8 . So </li></ul>01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
13. 13. The Boltzmann Distribution Example : 4. For T = 5800 K (kT ≈ 0.5 and  E = E 2 -E 1 = 10.2 eV) we have Even at the Sun’s surface there are relatively few atoms in the 1 st excited state. This is because the energy gap  E >> kT 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
14. 14. The Maxwell Distribution An important application of the Boltzmann distribution is the distribution of molecular speeds v in a gas of N molecules: This distribution, in fact, was derived by James Clerk Maxwell before Boltzmann’s work. But it is an important (and famous) special case. 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
15. 15. The Maxwell Distribution Different summaries of the molecular speed computed from the Maxwell distribution Most probable Average RMS 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
16. 16. The Maxwell Distribution Example The average speed of a nitrogen molecule at T = 300 K is given by With k = 1.39 x 10 -23 J/K and m = 4.68 x 10 -26 kg one gets <v> = 475 m/s = 1700 km/h 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
17. 17. Summary <ul><li>Statistical physics is the study of the collective behavior of large assemblies of particles </li></ul><ul><li>Ludwig Boltzmann derived the following energy distribution for identical, but distinguishable , particles </li></ul><ul><li>The Maxwell distribution of molecular speeds is a famous application of Boltzmann’s general formula </li></ul>01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |
18. 18. TERIMA KASIH 01/02/11 © 2010 Universitas Negeri Jakarta | www.unj.ac.id |