The answer (except for Z = 19) can be seen in Figure 1. All credit to McGraw- Hill for the image.

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Arnold Hensley
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Problem 9.3 Which configuration has a greater number of unpaired spins? Which one has a lower energy? [Kr]4d 9 5s or [Kr]4d. What is the element and how does Hund s rule apply?

1 Problem 9.3 Which configuration has a greater number of unpaired spins? Which one has a lower energy? [Kr]4d 9 5s or [Kr]4d. What is the element and how does Hund s rule apply?. Solution The element is Pd. The configuration [Kr]4d has a lower energy, but a higher number of unpaired spins. This is due to filled subshells having a lower energy. This is one of the cases where Hund s rule is violated. Problem 9.4 The answer (except for Z = 9) can be seen in Figure. All credit to McGraw- Hill for the image. Figure : The spin configurations for Z = through Z = 8. All credit to McGraw-Hill for the image For Z = 9, the additional electron will occupy the 4s subshell, giving a condensed configuration of [Ar]4s. The image will look just like before, but now there s a 4s block on the right with one arrow in it. 3 Problem 9.5 What is the average temperature of plasma, which is formed by ionized atoms?

2 3. Solution Here we actually use a result from stat mech, that for a monatomic gas K = 3 kt Where k is Boltzmann s constant. For the atoms to be ionized, the average temperature must be on the order such that they have enough energy for the electrons to constantly be getting kicked loose. We see that the typical ionization energy is 5 ev. Since Boltzmann s constant is 5 ev, we see that we must have a temperature of T 5 K. 4 Exercise. (a) Show that (b) Show that n(e)de = πn V (πkt ) 3/ E/ e E kt de 4. Solution 4.. Part (a) 3 NkT V Start with the Maxwell-Boltzmann distribution for velocity: n(v)dv = 4πN ( m ) 3/ v e mv kt dv V πkt Remember that for kinetic energy, we have E = mv Taking the derivative of both sides gives de dv = mv de = dv mv Plugging these in gives n(e, v)de = 4πN ( m ) 3/ v V πkt m e E kt de To get rid of that remaining v we rearrange the definition of kinetic energy to get

3 Plugging this in gives v = E m n(e)de = 4πN V ( m ) 3/ E πkt m 3 e E kt de The m s cancel out. If we cancel out some of the s as well, we will get as needed. n(e)de = πn V (πkt ) 3/ E/ e E kt de 4.. Part (b) In this section, we use Which becomes En(E)dE Solving the integral gives This just simplifies to πn V (πkt ) 3/ E 3/ e E kt de πn V (πkt ) (kt 3 3/ )5/ πde 4 as needed. 3 NkT V 5 Problem. Show that the most probable speed for a particle in a Maxwell-Boltzmann distribution is kt v mp = m 3

4 5. Solution In the Maxwell-Boltzmann distribution, the likelihood of finding a particle with a particular velocity v (per unit volume) is given by n(v)dv = 4πN ( m ) 3/ v e mv kt dv V πkt The way to maximize any function is to find where its derivative is. d dv n(v) = Throwing out the constants out front, this turns into = d dv v e mv kt Make sure to remember the chain rule ( = ve mv kt + v m ) kt v Canceling some terms out gives e mv kt = v m kt Solving for v (and only keeping the positive root, a negative speed makes no sense) gives kt v mp = m 6 Problem.7 A molecule with a dipole moment p is placed in an external electric field, ɛ. The molecule can only be parallel or antiparallel to the field. (a) Show that there are two allowed energy states separated by pɛ. (b) Let the ground state have energy. The fraction of degeneracies between the excited and ground states is. If there are N particles at temperature T, what is the ratio of particle sin the excited state to the ground state? (c) Find the temperature for which this ratio is.9. Estimate p 3 C m and ɛ 6 V/m. (d) Calculate Ē for some temperature T. Ē 4pɛ 3 as T (e) What is E total? Show that the heat capacity is C = ( Nk ) ( ) ( pɛ kt Show that Ē as T and exp[pɛ/kt ] ( +.5 exp[pɛ/kt ]) (f) Sketch C. Where is it maximized? Explain the relationship in physical terms. ) 4

5 6. Solution 6.. Part (a) We know that there are only two states because there are only two allowed orientations of the molecule: parallel and anti parallel. The energy of these states is given by We can see that E = p ɛ So their difference will be 6.. Part (b) E parallel = pɛ and E antiparallel = pɛ E = pɛ The ratio of molecules in the excited state to the ground state is defined as R = g(pɛ)p MB(pɛ) g()p MB () The probability that a state is occupied is given by P MB (E) = Ae E kt The constant out front will cancel out when we take the ratio. Also e = giving 6..3 Part (c) Solving for T gives R = e pɛ/kt T = pɛ ( ) R k ln Plugging in the given values yields T = ( 3 )( 6 ( ) ) ln.8k 5

6 6..4 Part (d) Since we are working with a two level system, we cannot use the integral definition of average. We must instead use the discrete definition. Since we are dealing with a Maxwell-Boltzmann distribution, the probability that a particle will be found in a state with energy E i is P (E i ) = Ae E i kt We still have the normalization condition that = i g(e i )P (E i ) to Notice that we are careful to include degeneracy. In our case, this reduces = Ae + Ae pɛ/kt We can use this to solve for A, giving A = + e pɛ/kt Now we can use the discrete definition of average to get our answer i g(e i )P (E i )E i In our case we have AE () + A(4pɛ)e pɛ/kt This becomes (4pɛ)e pɛ/kt + e pɛ/kt Dividing the top and bottom by e pɛ/kt gives pɛ + epɛ/kt If we let T, then the denominator erupts, and we get. On the other hand, if we let T, then the exponential goes to. From here it s just a little algebra to get 4 3 pɛ. 6

7 6..5 Part (e) First we have Then we have E total = NĒ C = d dt E total Now the fun begins. Make sure to be careful with the chain rule! C = d dt N pɛ + ( epɛ/kt epɛ/kt ) ( pɛ k ( T ) ( ) C = Npɛ( ) + epɛ/kt Rearranging this around a bit, we can get the answer from the book 6..6 Part (f) Let s let C = Nk ( ) pɛ e pɛ/kt ( kt ) + epɛ/kt ) This makes z = pɛ kt C(z) = Nk z e z ( + ez) From here, there are two ways to proceed. You can find the maximum the hard way, by using the chain rule, or by graphing the function. Let s choose the later. Wolfram Alpha plotted the function and tells us that the maximum value is at Problem.8 Use the result from Exercise. n(e)de = Find: (a) the most probable energy (b) the average energy (c) the rms energy. πn V (πkt ) 3/ E/ e E kt de 7

8 Figure : Plot of C(z) done by Wolfram Alpha 7. Solution 7.. Part (a) The way to maximize any function is to find where its derivative is. d de n(e) = We can throw away the constants out front to get d de E/ e E kt = Using the chain rule, we get E / e E kt + E / ( ) e E kt = kt Canceling out the exponential, consolidating the E terms gives Solving for E mp gives = E mp kt E mp = kt 7. Part (b) The average energy is given by V N En(E)dE 8

9 Or V N Solving the integral gives πn V (πkt ) 3/ E 3/ e E kt de Simplifying π (πkt ) (kt 3 3/ )5/ πde 4 3 kt This is essentially the same problem as Exercise. Part (b). I apologize for the redundancy. At least you could have just copied your own work. 7.3 Part (c) To find the rms energy, we must first find the average of the square of the energy Or E = V N E n(e)de Solving the integral gives E = V πn N V (πkt ) 3/ E 5/ e E kt de Simplifying gives E π 5 = (kt )7/ (πkt ) 3/ π 8 This means that our E rms will be 8 Problem. E = (kt ) 5 4 E rms = 5 4 kt Calculate the average photon energy in a Bose-Einstein gas at (a) Temperature T (b) T = 6K 9

10 8. Solution For a Bose-Einstein gas, we have n(e)de = 8πE (hc) 3 e E/kT de The average energy per photon is given by En(E)dE n(e)de If we plug in n(e)de we will see that all of the messy constants cancel out E 3 e E/kT de E e E/kT de If we let z = E, then we also have E = kt z and de = kt dz. Plugging kt these in gives (kt )4 z 3 e z dz (kt ) 3 z e z dz Plugging in the formulae in the book gives 8. Part (b) π4 kt 5.4 Here we use T = 6, K and k = ev/k π4 ( eV This actually fits in well with our calculation of ionization energies of plasma in Problem 9.5. These photons aren t hot enough to ionize gasses, but they re not too far off (E ion 5eV) and we predicted an ionization temperature of 5 K. 9 Problem.6 Show that the average kinetic energy in a Fermi gas is Ē 3 5 E F.

11 9. Solution For a Fermi gas we have n(e)de = 8 πm 3/ e E / h 3 e (E E F )/kt + de The average energy per electron is given by En(E)dE n(e)de If we plug in n(e)de we will see that all of the messy constants cancel out E 3/ e de (E E F )/kt + E / de e (E E F )/kt + Now we must remember that at K, the Fermi Dirac distribution goes to zero for energies above the Fermi energy. It goes to for energies below the Fermi energy. EF E 3/ de EF E / de Neither of these integrals is too tricky. They work out so that Or simplifying 5 E5/ F 3 E3/ F 3 5 E F Problem.9 Calculate the probability that an electron in Cu at 3 K has an energy equal to 99% of the Fermi energy.. Solution For a Fermi gas we have P F D (E)dE = We are curious about E c =.99E F, so P F D (E c )de = e (E E F )/kt + de e.e F /kt + de

12 Plugging in E F = 7.5 ev for Cu, T = 3K, and k = ev/k, we will get P F D (E c )de = e.(7.5)/( ) de.94 + So it is very likely that this energy will be filled. Additional Problem. Solution See Figure 3 for the solution. There are 4 = 6 possible outcomes.there are 4 + = 5 macrostates. The macrostates are divided up using lines. The probabilities of each macrostate are in the rightmost column. The multiplicity of the macrostate can be gotten by multiplying the probability by 6. The equation for checking each macrostate is Ω(n) = For each of the cases these are Ω() = Ω() = Ω() = Ω(3) = Ω(4) = Yes, these numbers agree. N! n!(n n)! 4!!(4 )! = 4!!4! = 4!!(4 )! = 4!!3! = 4 4!!(4 )! = 4!!! = 6 4! 3!(4 3)! = 4! 3!! = 4 4! 4!(4 4)! = 4! 4!! = Additional Problem. Solution.. Part (a) There are possible outcomes.

13 Figure 3: The possible outcomes from tossing 4 coins. Credit to B. Wagner for making the table.. Part (b) We using the multiplicity formula again In our case Ω(n) = N! n!(n n)! Ω(5) = 5!.6 4 5!5!..3 Part (c) The probability of getting n heads is given by 3

14 So for us it will work out to be..4 Part (d) P (5) = The multiplicity of getting 3 heads is P (n) = Ω(n) Ω(all) 5!. 5!5! 5 Ω(3) = 5! 3!()! This means the probability of getting 3 heads is..5 Part (e) P (3) = The multiplicity of getting 4 heads is 5!.49 3!()! 5 Ω(4) = 5! 4!()! This means the probability of getting 4 heads is..6 Part (f) P (4) = The multiplicity of getting 5 heads is 5! !()! 5 Ω(5) = 5! 5!! = This means the probability of getting 4 heads is P (5) = This is only from flipping 5 coins, and there is already a difference in 5 orders of magnitude between the most and least likely state. Imagine how sharp this gets when you have 3 coins (i.e. particles)! 4

15 3 Addition Problem 3 3. Solution This one you can just plug into your favorite calculator. The point is to just give you a sense of scale. It works out so that x.4,,,. That s insanely huge! 5

Physics 2203, Fall 2011 Modern Physics

Physics 2203, Fall 2011 Modern Physics. Friday, Nov. 2 nd, 2012. 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