CHM 3411 – Problem Set 9
Due date: Wednesday, April 12th. (NOTE: The third hour exam is Friday, april 15 th. It will cover the following
material: Chapter 10, sections 4e (photoelectron spectroscopy), 6-8; Chapter 11, section 1; Chapter 12, sections 1-6,
8-14; Chapter 13, all; handouts)
Do all of the following problems. Show your work.
1) Carbon monoxide (CO) has an electronic transition that occurs at around 140. nm. The spectrum has been
assigned based on the intensity pattern that is observed. The following peaks in the spetrum are found (v’ is the
vibrational quantum number in the upper electronic state and v” is the vibrational quantum number in the lower
electronic state).
v’
v”
peak location (cm-1)
0
1
2
3
4
0
0
0
0
0
64703.
66231.
67675.
69088.
70470.
a) Based on the above data find e’ and exe’ for the excited electronic state in the above transition in CO.
b) The vibational constants for the ground electronic state in CO are as follows: e“ = 2170. cm-1 and exe“
= 14. cm-1. Based on this information and your answer in a, find T e’ the value for energy for the minimum in the
excited electronic state in CO.
2) A laser operating at 532.0 nm is used to take the Raman spectrum of N 2, a homonuclear diatomic molecule. In
addition to the Rayleigh line at the laser wavelength, a second peak is observed which is centered at 626.3 nm and
which exhibits rotational structure. Explain the origin of this peak, and estimate the value for , the vibrational
constant for N2.
3) Consider an argon ion laser operating at  = 488. nm in pulsed mode. One laser pulse has an energy of 0.475 J
and a duration of 12 ns.
a) Assuming the laser pulse is of constant power over the 12 ns duration of the pulse, what is the laser
power (in watt, 1 watt = 1 J/s) during the pulse?
b) What is the number of laser photons per pulse?
Also do the following from Atkins:
Exercise 10.12b What is the energy (in eV) of an electron that has been ejected from an orbital of
ionization energy 4.69 eV by a photon of wavelength 584 pm?
Problem 13.6
J. D. Dojahn et al characterized the potential energy curves of the ground and excited
electronic states of homonuclear diatomic halogen anions (F2-, Cl2-, etc). These anions have a 2u+ ground state and
2
g, 2u, and 2g+ excited states. To which excited states are transitions from the ground electronic state allowed?
For the forbidden transitions list all of the selection rules that make the transition forbidden.
Solutions.
1)
a) It is best to proceed by finding a general formula for analyzing the data. If we call E(v',v") the energy at
which a transition occurs from v" in the lower state to v' in the upper state, then.
E(v',v") = Te' + Evib(v') - Evib(v")
where Evib(v') is the vibrational energy in the upper state and Evib(v") is the vibrational energy in the lower state. if
we use
Evib(v) = e (v + 1/2) - exe (v + 1/2)2
and note that v" = 0 for all of the lower states, then
E(v',v") = Te' + [e' (v' + 1/2) - exe' (v + 1/2)2 ] - [ (1/2) e" - (1/4) exe" ]
= c + e' (v' + 1/2) - exe' (v + 1/2)2 , where c = Te' - [ (1/2) e" - (1/4) exe" ] is a constant.
Therefore, by fitting the data to a second order polynomial of the form
E(v',v") = a0 + a1 (v' + 1/2) + a2 (v' + 1/2)2
then
a0 = c
a1 = e'
a2 = - exe'
The program POLY was used to fit the data. The results were as follows
a0 = 63938. cm-1
a1 = 1554. cm-1
a2 = - 23.1 cm-1
and so
e' = 1554. cm-1, exe' = 23.1 cm-1.
b) Based on the analysis in a, we may say
a0 = Te' - [ (1/2) e" - (1/4) exe" ]
or
Te' = a0 + [ (1/2) e" - (1/4) exe" ]
Using the values for e" and exe" given in the homework, we get
Te' = 63938. cm-1 + 1081.5 cm-1 = 65020. cm-1.
2) The peak at 532.0 nm is Rayleigh scatter (starting and ending in v" = 0) while the peak at 626.3 nm is Stokes
scatter (v" = 0 going to v' = 1), as shown in the figure on the next page. So in the harmonic oscillator approximation
the difference in energy between the Rayleigh and Stokes photons is equal to e.
E( = 532.0 nm) = 1/(532.0 x 10 -7 cm) = 18797. cm-1.
E( = 626.3 nm) = 1/(626.3. x 10 -7 cm) = 15967. cm-1.
The difference in energy is 2830. cm-1, and so is approximately equal to e for N2.
3)
a) power = energy/time. Assuming a squar pulse (constant power during the duration of the pulse)
power = 0.475 J/(12. x 10-9 s) = 4.0 x 107 J/s = 40. MWatt
b) The energy of one photon is
E = h = hc/ = (6.626 x 10-34 J.s)(2.998 x 108 m/s)/(488. x 10-9 m) = 4.07 x 10-19 J/photon
So the number of photons per pulse is
# photons = (0.475 J/pulse)/(4.07 x 10 -19 J/photon) = 1.17 x 1018 photon/pulse
Exercise 10.12b
E(photon) = E(ionization) + E(kinetic)
So
E(kinetic) = E(photon) - E(ionization)
The energy of the photon is
E = hc/ = (6.626 x 10-34 J.s)(2.998 x 108 m/s)/584. x 10-12 m)
= 3.40 x 10-16 J (1 eV/1.602 x 10-19 J) = 2123.3 eV
So E(kinetic) = 2123.3 eV - 4.69 eV = 2118.6 eV.
Problem 13.6
u+ 
g,
allowed
2
u
forbidden (because u  u is forbidden)
g+
allowed
2
2
2
u+ 
2
u+ 
2