Advance P. Chem. Problems (I) Populations, Partition Functions, Particle in Box, Harmonic Oscillators, Angular Momentum and Rigid Rotor

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1 University of Connecticut Cheistry Education Materials Departent of Cheistry March 8 Advance P. Che. Probles I) Populations, Partition Functions, Particle in Box, Haronic Oscillators, Angular Moentu and Rigid Rotor Carl W. David University of Connecticut, Carl.David@uconn.edu Follow this and additional works at: Recoended Citation David, Carl W., "Advance P. Che. Probles I) Populations, Partition Functions, Particle in Box, Haronic Oscillators, Angular Moentu and Rigid Rotor" 8). Cheistry Education Materials

2 Probles for the Advanced Physical Cheistry Student Part, Populations, Partition Functions, Particle in a Box, Haronic Oscillators, Angular Moentu and the Rigid Rotor C. W. David Departent of Cheistry University of Connecticut Storrs, Connecticut Dated: March, 8) I. SYNOPSIS This is a set of probles that were used near the turn of the century and which will be lost when the web site they were on disappears with y deise. Because these probles are being taken fro the web and are being edited, their stateents and the hints/answers offered are subject to the typical editorial errors that ensue when such work is undertaken in the vacuu of a non-teaching situation. Therefore, I clai any errors for yself, and hate to note that there ost likely is no point in contacting e about the for obvious reasons.. The rotational constant B) for a certain olecule is.4 reciprocal centieter c ). In a high resolution infra red absorption experient, the intensity of the {n=,j=} to{n=,j=} transition had an intensity ratio of.774 relative to that of the {n=,j=} to {n=,j=3} transition. What was the teperature of the saple in K)? II. POPULATIONS. Obtain a forula for the ratio of the nuber of haronic oscillators expected to populate the n=,j= state relative to the n=,j=3 state. We know that E n = for vibration, and n + ) hω E J = B J J + ) so the nuber of oscillators in the higher energy state, n=,j=) will be proportional to # n=;j= + )e β+ ) hω+b +)) while the nuber of oscillators in the lower energy state, n=,j=3) will be proportional to # n=;j=3 3 + )e β+ ) hω+b 3 3+)) where β = kt, and reeber the degeneracy, J + ) fro which the answer follows. Electronic address: Carl.David@uconn.edu Fro the above proble, we have: + )e β + ) hω+b +)) + )e β + ) hω+b +)) = ergs where k =.38 K, and the conversion factor fro c 6 ergs to ergs is.986 c Notice that ω is not given for this olecule [] since it cancels top and botto). Solving for T is left as an exercise, but here is soe Maple code which ight help: > #tep > restart; > eqn := 3*exp-.4*.986e-6)* > /.38e-6)*T)); > eqn := 5*exp-.4*.986e-6)*6 > /.38e-6)*T)); > ratio := eqn/eqn; > solveratio=.774,t); eqn := 3 e T ) eqn := 5 e T ) ratio := e T ) e T ) 3xy.z Sile; surely you didn t expect e to do all the work! 3. Obtain a forula for the ratio of the nuber of H atos free radicals) expected to populate the n= Typeset by REVTEX

3 8 6 4 electronic state relative to the n= electronic state. E n = Ry n Z = e + e + 3e 3 which has to be nuerically evaluated. 5. Considering the sae energy level structure as in the above proble, Copute the therodynaics average) energy of the syste at 35K. Report your answer in c. We know that the population of a state will be proportional to a degeneracy ties an exponential of the energy/kt), i.e., # n g n e En kt and we have to figure out the degeneracy of the states we are dealing with. For the n = state, g =, while for the n = state, g = 8. Thus, we have for the desired ratio: 8e Ry kt e Ry kt 4. Consider a syste which has several energy levels as shown see Figure ). Use a unit syste 6 ergs k =.38 K c ergs and express energies in c. The diagra shows a doubly degenerate ground state, and 4 excited states. What is the nuerical value of the partition function at the teperature corresponding to kt = c? ε=3c We know that 6 ergs c k = K ergs =.695c K so 35K is c which is 43.4c. Then E = e e e e It is interesting, in this case, to plot the average energy as a function of teperature, to see that it rises fro zero to a saturated, in a sense, value: It teperature energy ε=c ε=c ε= c FIG. : A ake believe energy level schee FIG. : The Therodynaic Energy as a function of teperature for an artificial energy level schee. is also interesting to plot the heat capacity ) E T as a function of teperature, but this is left as an exercise for the interested student. One should verify that the Energy asyptotically approaches a fixed and obtainable) value, li E =? T 6. In the accopanying diagra, two electronic energy levels of a diatoic olecule are shown, and in balloons at the side, the appropriate vibrational

4 3 and rotational energy levels. If the rotational constant B) is 6 c, calculate the ratio of the nuber of olecules expected in the J =, n = state of the excited electronic state relative to the nuber of olecules expected in the J =, n = state of the ground electronic state assuing a teperature of K. The rotational energy levels for a diatoic olecule whose internuclear distance is fixed is given by E J = h JJ + ) I c J= J= J= J= J= J= = n= = and the selection rules say that J = ±. For the transition, for instance, the change in energy is E E = h I h I J= J= J= n= while for the transition, we would have FIG. 3: Two electronic energy levels for a diatoic olecule undergoing vibration and rotation E E = h I 3 h I = h I 6 ) = 4 h I The frequency of the first line cited above would be The upper state s population is proportional to 5e )+6 3 kt while the lower state s population is proportional to so the ratio will be III. 3e )+6 kt 5e )+6 3 kt 3e )+6 kt SPECTROSCOPY. The spacing between adjacent icrowave absorptions for the AlH olecule is.64 c. Calculate the internuclear distance in AlH. Assue the 7-isotope of Aluinu 7 3Al) has ass 6.98 gras/ole, and the -isotope of H H) has ass.9 gras/ole. Give your answer in Angstro Å). ν = h I h while that of the second line cited above would be ν = 4 h I h so the difference in frequencies the spacing between adjacent lines) of the two lines, which is given as.64 c in the stateent of the proble, would be ν = ν ν = h I h =.64c Since the oent of inertia I is µr eq where µ is the reduced ass of the AlH olecule, we can, once all the units have been worked out, copute r eq. We would have which would be.64c h = πµreq.64c = erg sec π πµr eq

5 4.64c c c sec = dyne c sec π) µr eq.64 c 3 c sec ) gra c sec c sec = π) µreq.64 3 sec sec = gra c π) µreq We now have µr eq =.64 3 π gra c where µ = ) gras Note that this value is per olecule! The rest is arithetic. But don t forget to convert your answer to Angstro Å).. If the internuclear distance in AlH is.66 Å, calculate the rate of rotation in radians per second) of an AlH olecule when it is in the J=7 state, B = 6.3 c, which is.5 5 in ergs). The oent of inertia for this olecule is in gra c ). Assue the 7- isotope of Aluinu, with ass 6.98 gras/ole, and the -isotope of H, ass.9 gras/ole as above). which needs to be converted to ergs. The conversion factor is hc). We have E = 6.37)8)c 6 ergs.986 c This energy is Iω, and we can solve for ω. 3. In the far infrared, HCl absorbs at 83 c. It is known that the B e value for this particular isotopic for of the olecule is about c. What J value increased by one to give this transition? We will have J)J + ) J + )J + ) 83c i.e., 83 = J + )J + ) J)J + ) Here we need to first copute the energy rotational) of this olecule, and then we need to reeber that E = I gra c ) ω rad sec ) fro eleentary physics. The units are ) rad gra c = gra c = dyne c = erg sec sec But the energy, quantu echanically, is which is an equation solvable for J, i.e., J + 3J)) + J J = 83 i.e., J + = 83 Notice that the equation will give a fractional value for J, which eans that we are going to have to figure out where the error is. 4. The force constant for H7 35 Cl is dynes/c. How any vibrations per second does D7Cl 35 ake? Check that your answer is not in radians/sec. E = BJ)J + ) = 6.37)8)c

6 5 Ε, ν J= > J = > J = >3 n=5 n=4 n=3 n= 9 c Birge Sponer 4 c FIG. 4: The high resolution infra-red spectru of a diatoic olecule. The hypothetical J = J = transition is the one used for the force constant coputation. 549 c Dissociation Energy ν = k π µ dyne c gra where the units work out to be so i.e., sec and µ gra c sec c gra gra gra sec We know the force constant k, we know the reduced ass, µ, and we can therefore obtain the frequency of vibration ν. n= FIG. 5: The Birge Sponer extrapolation. n νc ) νc ) The Birge Sponer extrapolation procedure is eployed in the P. Che. Lab, traditionally, for analyzing the visual spectru of iodine, which is usually obtained at high resolution in absorption. Since this is the only exaple of this procedure one is likely to see, it has reained ysterious to generations of students. This ade up exaple should clear up the situation. For a certain olecule, the following electronic transitions have been observed: The Birge Sponer extrapolation is a specialized coputation for diatoic olecule energy levels which, given all the inforation shown in Figure 5, allows one to copute the Dissociation Energy approxiately. Fro Figure 6 we can, assuing the trend continues, deterine the equation of the line ν = j + b

7 ν j 4 j 3 8, c 5,43 c 9 ν j Birge Sponer Area FIG. 7: Potential Energy Curves for Diatoic Molecule Undergoing Photo Dissociation j i.e., FIG. 6: The Birge Sponer extrapolation. ν = 5j + ν j j area This eans that the area denoted as the Birge Sponer Area is 9)j ) where j is the extrapolated value when ν. Obviously, j = 4. Fro Figure 5 we have )9) = 4 + E dissoc which gives us the value of the dissociation energy directly. 6. In the photo dissociation of a hoo-nuclear diatoic olecule, the exciting photon s energy is 8, c, and the dissociation energy of the olecule is 5,43 c. If the atoic ass of the nuclei is 6 gras/ole i.e. oxygen), and if the velocity of each of the oxygen atos are equal in agnitude, copute the value of the velocity of the resultant oxygen atos when the internuclear distance in infinite. Assue that the olecule was initially at rest we actually copute the velocity relative to the center of ass). Report your answer in c/sec. The excess energy, above that required for dissociation, is E excess = c which needs to be converted into ergs, divided by partitioning between the two now) independent separating nuclei), and, knowing the ass of these nuclei, used to obtain the velocity. 7. The dipole oent of an arbitrary olecule is µ = µ x î + µ y ĵ + µ zˆk which, in spherical polar coördinates is µ = µ sinϑ cos ϕî + µ sinϑsinϕĵ + µ cosϑˆk What is the value of the integral π dϑ π π sin ϑdϑ π dϕψ pz µ x ψ py = dϕ cos ϑµ x sin ϑ sin ϕ sin ϑ which is related to the x-dipole selection rule for a transition fro ψ pz to ψ py )? Notice that the radial part of the wave functions has been suppressed.

8 7 π sin ϑdϑ π dϕ cos ϑ { µ sin ϕ cos ϕ} sin ϑ sin ϕ µ π which is dϑ sin ϑ cos ϑ µ sin3 ϑ 3 π π sin 3 ϕ 3 What does this result ean? dϕ sin ϕ cos ϕ 8. Matsushia, Oka, and Takagi Phys. Rev. Letters, 78,664997)) reported the observation of spectral transitions of 4 HeH + in the THz region. One such line occurs at.839 GHz Giga = 9 ). It corresponds to a J= to transition. What is the bond length of the olecule in Angstro)? IV. π PARTICLES IN BOXES. For a one diensional particle in a box of length L) variable = x), obtain a forula for the probability density P x)) of the described particle if the total range doain) of x is a x b where a < b a and b > ), assuing the wave function over this doain is ψx) = A x / = A x We know that we need to integrate ψ ψ which is the probability density, over the doain of interest, and divide it by the integral of the probability density over the entire doain of the variable x. We have b A a x P a, b) dx L A x dx which leads to logarith integrals which are straightforward.. For a one diensional particle in a box of length L, copute the probability of finding the particle in the center third of the box when the particle is in the first excited state. Here, we have a ore traditional proble sin ) πx L/3 L dx ) sin πx L dx L/3 L L Why have we dropped the noralization constants? Again, the integrals are straightforward, and are left to the interested and/or otivated student. 3. For a two diensional particle in a box, calculate the nuber of states which lie between in energy) but not including) the states fro {3,5} to {6,6}, when the box is square. The notation is as follows, {n x, n y } labels each state, where n x is the quantu nuber associated with the x-direction, and n y is the quantu nuber associated with the y- direction. The wave function is a product of sines, one for the x-direction and one for the y-direction. Further, the energy is equal to E nx,n y = a constant tiesn x + n y) This question is addressing the question of degeneracy, and asking you to count the nuber of states by, in essence, explicitly listing the in ascending usually) order of energy, according to their {n x, n y } values. We start, but do not finish the job. The lower bound is {3,5} 34a while the upper bound is {6,6} 7a Thus {3,6} )a = 45a {3,7} )a = 58a.. {,8} + 84)a = 65a.. {,8} )a = 68a.... {7,} 49 + )a = 5a So there are lots and lots) of states, and a full list would take too uch space here.

9 8 4. For butadiene, assuing that it is straight and ade up of 3 C-C bonds of length.5 Å, calculate the expected absorption wavelength assuing that the electrons are Feri filled into the olecular orbitals generated using the particle in a box odel, i.e., that the electrons are ) independent of each other, and ) ignore the presence of nuclei, and just notice the ends of the olecule. We are asked to think of this olecule as a particle in a box proble, where the box is the total length of the olecule, i.e., 4.5 Å. With 4 electrons, we know that the ground n = ) and the first state n = ) will be occupied in the ground electronic state of the olecule, and that one of the n = electrons will transit to n = 3 in absorbing a photon, so we need to copute the change in energy in going fro the n = n = 3 states. We have E n= 3 = where L = c. h 3 e L ) Knowing E allows us to copute the energy of the photon which did the work, i.e., and E n= 3 = hν photon λ photon ν photon = cvelocity of light) Notice, this LCAO is idway between the n = state and the n = state so we expect the energy to be likewise positioned. We can get the explicit value in several ways, but the best fro a pedagogical point of view is to evaluate H op ψ + ψ ) and see if we can obtain ɛ ψ + ψ ) This eans we apply the Hailtonian operator, in the for h x h ψ + ψ ) x and see what we recover when we factor out the ter ψ + ψ ) fro the resultant. 6. For an electron in a 3-diensional box, copute the wavelength in Å) assuing that the box has diension a, a/, a) where a is.5å, for a transition fro the 4 th excited state to the nd excited state. allows us ultiately to copute λ photon. 5. The Schrödinger Equation for the particle in a box stretching fro x = to x = L) is: h ψ x = ɛψ and, given two eigenfunctions: ) πx ψ = L sin L and ψ = ) πx L sin L a state is prepared which is one half ψ and one half ψ, i.e., a linear cobination with equal coëfficients ψ cobination = c ψ + c ψ with c = c, what is the expected energy ɛ ) of the resultant state? E nx,n y,n z The energy levels are given by the expression: { nx ) ) } = h ny nz ) + + e a a/ a and we need to evaluate this expression for the choice of quantu nubers which create the two different states there s no particular algorith for searching for these two sets of quantu nubers that I know of). The ground state would be { E = h e a ) ) + + / ) } Therefore, aking a list of candidate states, sorting the in ascending order, and picking out the appropriately required ones is left as an exercise.

10 9 y b b/ a/ a FIG. 8: A particle in a D box. 7. For a particle in a two diensional box area = ab, perieter = a + b)), find the probability of finding it in the area bounded by a/ < x < a and b/ < y < b. Assue the particle is in the ground state. x Since the energy expression is E nx,n y = n x h π particle A + n y h π particle A) or ) h π n x E nx,n y = particle A + n y with n x =,, 3, and n y =,, 3, etc., we can evaluate a sequence of increasing values of the energy as the two quantu nubers increase until we ve crossed the threshold of the desired axiu. 9. For a particle in a 3-diensional x, y, z) box of lengths a, a, 3a): x a, y a, z 3a calculate the probability of finding the particle in the range: < x < a/, < y < a, < z < 3a/ Your answer better be 4! You can obtain this by inspection, introspection, or direct integration which is the preferred ethod). You would be evaluating two integrals as a fraction, the nuerator of which would be a b dx dyψ a/ b/ and the denoinator would be where ψ = a dx b dyψ πx sin a b a sin πy b Parenthetically, notice that there are no degeneracy probles here until b a, whereupon the situation becoes a little uddled. a a/ b dx dy b/ a πx sin b a ) sin πy b which is separable into the product of two integrals which can be done independent of each other. What happens if we are in an excited state? 8. A particle is trapped in a two-diensional rectangular box of sides A and A. How any states exist whose energy E) obeys the relation E h π particle A diensions, 3 diensions, what s the big deal? See above!!!!. Assue that a particular argon ato is constrained to exist in a two-diensional square box of length L on each side. A particular state of this syste is a cobination of the ground state in one coördinate) and the first excited state in the other coördinate): i.e., ψ, = ) πx L sin sin L and the alternative for ψ, = L sin πx L ) sin ) πy L ) πy L There are two possibilities for the {n x, n y } shown, and therefore a proper state of the particle is a linear cobination of the two possibilities, i.e., ψ question = aψ, + bψ, here, we take a = b for our LCAO ) i.e., ) ] πx πy ψ question = K noralization [sin sin L L ) )) πx πy + sin sin L L where L is Å. Obtain the energy of the argon ato in ergs).

11 , LUMO hν,,,,,,, HOMO Reeber that the Hailtonian is ) h x + y Argon which can be applied to the wave function shown.. Assuing that benzene s electrons can be approxiated as being in a two diensional square box which copletely encloses the olecule, and assuing that the C-C distance in benzene is.3, calculate the expected HOMO-LUMO energy gap in ergs/olecule) for benzene. FIG. : Energy level schee for this particle in a box odel, showing the HOMO LUMO transition. Vx) x The actual benzene olecule fits into a rectangle of length.5 by.6 Angstro, so let s assue the square has the sae area, i.e.,.5*.6 Å. Thus, let s take the square s edge length at.98 Å for this proble. Then, we need to find the energy.3 A o FIG. : A particle in a sei-haronic oscillator potential. The ground state of the SHO is the sae as the n= state of the HO, while the first excited state of the SHO is the sae as the 3rd state of the HO. π 6 HO Vx) second excited state first excited state ground state FIG. 9: A particle in a benzene like D box. x levels in ters of the {n x, n y } sets, deterine where the 6 electrons go, so that the highest filled orbital becoes the HOMO, and the next in sequence of energy becoes the LUMO. The energy difference between these two needs to be coputed, and the rest, as they say, is trivial. V. HARMONIC OSCILLATORS. For the Sei-Haronic Oscillator SHO) shown in Figure where V x) is the potential energy), what is the energy of the second excited state. FIG. : The energy levels of a haronic oscillator with the sei-haronic oscillator energy levels selected out and displayed. For the state of the SHO in question E nd excited state = 5 + ) hω see Figure ). This is because half of the HO eigenfunctions are finite at x = which violates the boundary conditions of the SHO, so they have to be eliinated. Only the odd states survive

12 the winnowing process. The state in question, the second excited state of the SHO corresponds to the n = 5 state of the HO.. The Schrödinger Equation in diensionless for, so called atoic units) for the Haronic Oscillator is: ψ y + y ψ = ɛψ and the eigenfunctions are: ψ n = Herite polynoial in y) e y For n=, the Herite Polynoial is 4y. Obtain the value of ɛ for the n = state. The first partial of the trial wave function is ψ y = y 8ye y4y )e y The second partial ) 8ye y y4y )e y y is also obtained in eleentary fashion, allowing substitution into the appropriate equation indicated above), leading to the resultant energy through siple cancellations. 3. Assuing the Na 35 Cl olecule, with k =.86 5 erg/c and µ = gras, obtain the value of the distance fro the r e in Å) so that the probability of finding the oscillator is.5. where k is dynes c, µ is gras olecule, and x is in centieters. It is inconvenient to work in units like this vide infra, so we wish to convert to a diensionless distance, say ρ. Let ρ = αx where α is to be deterined by soe future) easure of convenience. Then, fro the chain rule we have x = ρ x ρ = α ρ so we have, for the transfored equation is ) h α ψ µ ρ ) ρ α + k ψ = Eψ so, cross-ultiplying, we have ) ψ ρ kµ ρ µ α 4 h ψ = E α h ψ The choice that akes ost sense is so kµ α 4 h ) ψ ρ ρ ψ = ɛψ where ɛ s definition in ters of E and α is obvious. We have achieved a ore textbook appearance of our differential equation. Let s check the units: α 4 = kµ h gra dyne c erg sec = dyne c gra c erg sec = gra c erg sec = gra c dyne c sec The Schrödinger Equation for the isotropic - diensional Haronic Oscillator is h ) ψ µ x + k x ψ = Eψ which has as ground state solution ψ x)), ψ x) = N e kµ h x = gra c gra c sec c sec = c 4 In the new units, the ground state solution is so Ne ρ / ) ψ = ρψ ρ

13 and ) ψ ρ = ψ + ρ ψ Since P x) ψ dx so, in this unit syste ɛ =. The question asks, what is x answer for the following integral: P ρ answer ) = = x answer Ne α x / dx Ne α x / dx why the factor of in front?) The denoinator is a standard Gaussian integral, but the nuerator is related to the error function. P ρ answer ) = ρanswer = Ne ρ dρ Ne ρ dρ The Error function, defined as The question requires that we define where α 4 = kµ h xanswer = e αx dx 4. The Schrödinger Equation for the -diensional Haronic Oscillator is h ψ µ x + kx ψ = Eψ so we have: Erft) t e u du π P ρ answer ) = = Erfx answer) Erf ) t Erft) ψ := x 4 β x ) e α x ) which has a proposed solution: ψ proble = x 4 βx )e αx4 What follows is soe Maple code to carry out the echanics of this proble. psi := x^4- beta*x^)*exp-alpha*x^); ha := expand-hbar^/*))* diffpsi,x$) + k/)*x^)*psi-e*psi); ha_ := subsexpalpha*x^)=,ha); ha_ := collectha_,x); ha := 6 hbar x e + hbar β α x ) e + 9 hbar α x 4 5 hbar α x β hbar α x 6 α x ) e α x ) e α x ) e α x ) + hbar α x 4 β e α x ) + k x 6 e α x ) k x 4 β e E x4 α x ) e + E β x α x ) e α x ) ha := 6 hbar x + hbar β + 9 hbar α x 4 5 hbar α x β hbar α x 6 + hbar α x 4 β + k x6 k x4 β E x 4 + E β x

14 3 ha := k hbar α + 6 hbar 5 hbar α β ) x6 + 9 hbar α + E β) x + hbar β E k β + hbar α β ) x 4 We conclude, fro the absence of cancellations, that the trial wave function is NO GOOD! What is the l value associated with the function VI. ANGULAR MOMENTUM sin ϑ cos ϕ. The operator for angular oentu squared) L op) in spherical polar coördinates is { L op = h sin ϑ sin ϑ } + sin ϑ ϕ so, for an eigenfunction of this operator one would have L opψϑ, ϕ) = ll + ) h ψϑ, ϕ) L op = h { cos ϕ sin ϑ sin ϑ sin ϑ) + sin ϑ sin ϑ } cos ϕ) ϕ is a straightforward exercise in partial differentiation.. The operator for angular oentu squared) L op) in spherical polar coördinates is { } L op = h sin ϑ + sin ϑ sin ϑ ϕ and the operator for L z op is L z op = h ϕ so, for an eigenfunction of this operator one would have and L opψϑ, ϕ) = ll + ) h ψϑ, ϕ) Is the function L z opψϑ, ϕ) = l hψϑ, ϕ) sin ϑ cos ϕ an eigenfunction of L z op? We need to for L z opψϑ, ϕ) = L z op sin ϑ cos ϕ )? = l hψϑ, ϕ) and ask if l is an integer? L z op sin ϑ cos ϕ ) = h sin ϑ cos ϕ ) ϕ which equals so the answer is NO! h sin ϑ sin ϕ )

15 4 3. If two eigenfunctions of angular oentu are orthogonal to each other then the integral b a sin ϑdϑ d c dϕ {ψ ψ } ust have a certain value. What are a, b, c, and d and what is the value of the integral assuing the functions are center of ass z θ A φ y and ψ = cos ϑ x B ψ = sin ϑ cos ϕ FIG. 3: A diatoic olecule, in the center of ass coördinate syste, undergoing rotation as a rigid rotor. π sin ϑdϑ π which can be written π sin ϑ cos ϑdϑ dϕ {cos ϑ sin ϑ cos ϕ} π which is trivially integrable. 4. The proposed function sin ϑ cos ϑ cos ϕ sin ϕ cos ϕdϕ is an eigenfunction of angular oent L op). What is the quantu nuber l) associated with this function? Again, all we need do is operate on the function with the appropriate operator see above) and look for regeneration in the eigenfunction sense. { sin ϑ sin ϑ sin3 ϑ cos3ϕ) + sin ϑ } sin 3 ϑ cos3ϕ) ϕ = ɛψ After appropriate partial differentiation one need only collect ters.. The energy operator in spherical polar coördinates when r is not a variable r = R, a constant) is { } h sin ϑ ψ µr + ψ sin ϑ sin ϑ ϕ which, for planar rotation, with ϕ = π, would be h µr { sin ϑ } ψ ϕ Obtain an expression for the eigenvalues associated with this operator. VII. RIGID ROTORS. The Schrödinger Equation in diensionless for) for the Rigid Rotor is: { } sin ϑ ψ + ψ sin ϑ sin ϑ ϕ = ɛψ and a purported eigenfunction is sin 3 ϑ cos3ϕ). Obtain the value of ɛ for this function, verifying at the sae tie that it is truly an eigenfunction. h ψ µr sin π ϕ is the Hailtonian for this proble, which becoes { h } ψ I ϕ

16 5 which is a particle in a box Hailtonian, i.e., is one whose eigenfunctions and eigenvalues are already known. Notice how the oent of inertia evolves naturally. 3. Consider a olecule of H 35 Cl in the n th vibrational state and the J th rotational state. Calculate the value of J such that E n,j > E n+, i.e., a highly rotating olecule s energy is greater than the next highest vibrational energy level of the non-rotating olecule. The vibrational force constant for this olecule is dynes/c said another way, the energy difference between the n =, J = and the n =, J = state is about 886 c ). The rotational constant, B, is about c. Reeber, J is an integer! E n,j = n + ) hω + JJ + )B where B is the rotational constant. We are asked to solve the equation n + ) hω + JJ + )B n + + ) hω + + )B which is a straightforward task. VIII. EPILOGUE After editing this aterial for weeks, and continuously finding errors, soe sall, soe huge, I have to wrap it up and send this off. If, in the years 8- or so, you coe across an error, and you e-ail e, I will try to have it corrected. But since this aterial is written in there is soe doubt whether or not I ll have access to a Linux achine, and access to the digitialcoons site. You can try; we ll see what happens, if anything. Thanks to all the students over the last 45 years who ve taught e Physical Cheistry. [] OK, it s HCl and ω is 886 c