Physics 2203, Fall 2011 Modern Physics

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1 Physics 2203, Fall 2011 Modern Physics. Friday, Nov. 2 nd, Energy levels in Nitrogen molecule Sta@s@cal Physics: Quantum sta@s@cs: Ch. 15 in our book. Notes from Ch. 10 in Serway Announcements Second midterm: Average 71, high 92, low 51 Term papers due Nov. 19 th our book. New Syllabus

4 1: (10pts). The figure to the right is a schema@c of the bonding of the outer electrons in the N 2 molecule. With this picture in mind answer the following ques@ons. (a) What does the molecular orbital picture look like for the bonding in the 1s 2 N 2 atomic states draw a picture and label the molecular states. (b) Relabel the molecular states star@ng with the ones you found in part (a) as a func@on of binding energy. Combine your bonding an@ bonding picture from (a) with the diagram above.

5 First we must make sure we have accounted for all electrons. 4 electrons 14 Electrons 2 electrons 2 electrons 4 electrons 2 electrons ~18eV

6 ~100 mev splizng Today s measurement is 97 mev What are these small peaks? These are electronic excita@ons.

Einstein (BE) and Fermi Dirac (FD) Distribu@ons Applica@ons of Bose Einstein Sta@s@cs An

7 Goal is to use a microscopic approach to explain thermodynamic bulk proper@es. Maxwell Boltzmann (MB) Distribu@on When do we use MB Sta@s@cs Quantum Sta@s@cs Bose Einstein (BE) and Fermi Dirac (FD) Distribu@ons Applica@ons of Bose Einstein Sta@s@cs An applica@on of Fermi Dirac Sta@s@cs. Maxwell Boltzmann Depressed because his theory was not accepted Suicide in 1908.

8 1) The are in terms of physical but in trajectory, etc. not quantum. 2) The equilibrium is the most probable way of the among various allowed energy states subject to constraints of fixed number of and fixed energy. 3) No on number of but low enough density and high enough T that no more than on par@cle at in a state. Example, 6 par/cles with energy 8E. Figure shows 20 ways to do this. But these are dis@nguishable par@cles, so we need to learn how many configura@ons are there for each box see ( ) above.

9 Each of the 20 arrangements can be decomposed into many substates, or Microstates. Look at the simplest one with one electron with energy 8E. As shown below there are 6 ways to have this arrangement as shown below. If these were quantum par/cles they would be indis/nguishable!!

10 1) An example, 2 par@cles with E, 1 with 6E and 3 with 0. Lets count: N (total)=6, n 1 (0E)=3, n 2 (1E)=2, n 3 (2E)=0, n 4 (3E)=0, n 5 (4E)=0, n 6 (5E)=0, n 7 (6E)=1 n 8 (7E)=0, n 9 (8E)=0 The MB sta/s/cs should be able to tell us how many of these microstates exist for a given configura@on. N MB = N! n 1!n 2!n 3!i i You do this one. Check : N MB = 6! = 60 : only occupied states in denominator. 3!2!1!

Lets find the average number of par@cles with some energy E j n j = n j1 p 1 + n j 2 p 2 + i i n j1 is the number of particles found in the j level in 1 p 1 is the probablility of observing

11 Lets find the average number of with some energy E j n j = n j1 p 1 + n j 2 p 2 + i i n j1 is the number of particles found in the j level in 1 p 1 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 n 0 = n 01 p 1 + n 02 p 2 + n 03 p 3 n 01 p 1 = (5)(6 /1287) n 02 p 2 = (4)(30 /1287) find n 1 n 0 = 2.307

12 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) = n 0 6 = = Prove that p(1)=0.256

13 The M B distribu/on for a system in thermal equilibrium can be calculate assuming that the total number of par@cles N is fixed and that the total system energy is fixed at a temperature T. f MB = Ae E i /k B T f MB is the probability that a state E i is occupied at a temperature T If the number of states with energy E i is given by g i degeneracy then n i = g i f MB = g i Ae E i /kt A is determined by requiring N= n i When N is very large sum replaced g i g(e) : f MB Ae E /k BT

14 f MB = Ae E /k BT When N is very large sum replaced g i g(e) : f MB Ae E /k BT g(e) is the Density of States n i = g i f MB n(e)de = g(e) f MB (E)dE N = n i N V = n(e)de = g(e) f MB (E)dE 0 0

15 (a) Find the of the first and second excited states rela%ve to the ground state for atomic hydrogen at room temperature use MB Find energies and degeneracy Ground state: E 1 = 13.6eV : g 1 = 2 1 st excited state: E 2 = 3.4eV : g 2 = 8 2 nd excited state: E 3 = 1.51eV : g 3 = 18 n i = g i f MB n 2 n 1 = g 2 Ae E2 /KBT g 1 Ae E 1 /K B T = g 2 g 1 e (E 1 E 2 )/k BT n 2 n 1 = 8 2 exp{( 10.2eV ) / (8.16x10 5 ev / K)(300K)} n 2 n 1 = 4 exp( 395) 0

16 (b) Find the of the first and second excited states rela%ve to the ground state for atomic hydrogen at 20,000K in a star. Find energies and degeneracy Ground state: E 1 = 13.6eV : g 1 = 2 k B T = 1.72eV n 2 n 1 = g 2 g 1 e (E 1 E 2 )/k BT = 4e 10.2/1.72 =.0107 n 3 n 1 = g 3 g 1 e (E 1 E 3 )/k BT = 9e 12.1/1.72 = st excited state: E 2 = 3.4eV : g 2 = 8 2 nd excited state: E 3 = 1.51eV : g 3 = 18 Emission intensity in this simple picture depends upon the occupancy of the states: Show that the 3 to 1 and 3 to 2 are 75% of the 2 to 1

17 Lets show that the MB can be converted into and for the is speed of a gas. n(v)dv = 4πN V m 2πk B T 3/2 v 2 e mv2 /2k B t dv n(v)dv is the average number of par@cles per unit volume with velocity between v and v+dv. E = mv2 2 n(e)de = g(e) f MB (E)dE = g(e)ae mv2 /2k B T de To evaluate this we need to look at the density of states in velocity space.

18 The number of states in v space f(v) between v and v+dv is f (v)dv = C4πv 2 dv : Volume in shell Each v corresponds to an E= mv2 2 g(e)de = f (v)dv = C4πv 2 dv n(e)de = A4πv 2 e mv2 /2k B T dv : C absorbed into A Finally: n(e)de=n(v)dv=a4πv 2 e mv2 /2k B T dv Find A: N V = n(v)dv = A4πv 2 e mv2 /2k T B dv = 0 0 N V ( m 2πk B T )3/2 n(v)dv = 4πN V m 2πk B T 3/2 v 2 e mv2 /2k B t dv

19 Average speed v v = 0 vn(v)dv N / V = 4πN V m 2πk B T 3/2 0 v 3 e mv2 /2k B T dv v = 8k B T πm RMS v 2 v 2 = 0 v 2 n(v)dv N / V = 4πN V m 2πk B T 3/2 0 v 4 e mv2 /2k B T dv = 3k B T m v RMS = v 2 = 3k B T m

average energy of k B T/2 for each independent mode of mo/on degrees of freedom Three dimensional

20 RMS v 2 v 2 = 0 v 2 n(v)dv N / V = 4πN V m 2πk B T 3/2 0 v 4 e mv2 /2k B T dv = 3k BT m v RMS = v 2 = 3k B T m Equipar//on of Energy a classical molecule in thermal equilibrium at temperature T has an average energy of k B T/2 for each independent mode of mo/on degrees of freedom Three dimensional motion 1 2 mv2 = 1 2 mv 2 x mv 2 y mv 2 z = 3k BT 2m Vibra@on and rota@on add degrees of freedom

21 MB sta/s/cs works when the average distance d between par/cles is greater than the quantum uncertainty. Δx << d ΔxΔp x 2 Equipartition theorem 2 p x 2m = k BT 2 Δp x = p x 2 (p 2 ) 2 = mk B / T Δx = 2 Gives : mk B / T : d= N V 1/3 N 3 << 1 3/2 V 8(mk B T ) Low density N/V High temperature T Heavy mass m

22 Fermions ψ 1, 2 Bosons ψ 1,2 ( ) = ψ ( 2,1) ( ) = +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 Monday we will talk about heat capacity in solids. The electrons are and the lazce vibra@ons are Bosons.

23 On Monday we are going to go back to this set of Microstates and ask how do we count, how do we occupy if the are Fermions or Bosons.

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MIDTERM #1 PHYS 33 (MODERN PHYSICS II) DATE/TIME: February 16, 17 (8:3 a.m. - 9:45 a.m.) PLACE: RB 11 Only non-programmable calculators are allowed. Name: ID: Please read the following instructions: This