CHM 3400 – Problem Set 10
Due date: Wednesday, April 17th. Late homework will not be accepted. NOTE: The final exam is Monday, April
22nd, from 5:00pm to 7:00pm, in PG5-155. It will be comprehensive.
A few useful conversion factors and constants:
1 eV = 1.602 x 10-19 J
h = 6.626 x 10-34 Js
1 amu = 1.661 x 10-27 kg
c = 2.998 x 108 m/s
me = 9.109 x 10-31 kg
1) When light of wavelength  = 254. nm is used to illuminate a clean metal surface, the maximum kinetic energy of
the electrons ejected from the metal is EK(max) = 3.18 eV.
a) What is the maximum speed of the ejected electrons? Give your answer in units of m/s.
b) What is 0, the work function, for the metal? Give your answer in units of eV.
c) What is max, the longest wavelength of light capable of ejecting electrons from the above metal? Give
your answer in units of nm.
2) The conversion efficiency of a lightbulb represents the fraction of the electrical power that the bulb converts into
visible light. It is thus a measure of the effectiveness of the bulb as a light source. Due to their greater efficiency in
converting electrical power into visible light, compact fluorescent lightbulbs (CFLs) have replaced incandescent
lightbulbs (ILs) as light sources.
Consider a 100. watt ( = 100. J/s) lightbulb. Calculate the number of photons per second emitted by
a) A CFL (conversion efficiency = 8%)
b) An IL (conversion efficiency = 2%)
For ease in calculation, assume that all of the emitted photons have a wavelength  = 500. nm, roughly the center of
the visible region of the spectrum.
3) The energy levels of the hydrogen atom are given by the equation
EH = - RH/n2
n = 1, 2, 3,...
(3.1)
where RH is the Rydberg constant, with RH = 109677 cm-1.
a) Starting with eq 3.1, derive the following equation for the energy of a photon emitted by a hydrogen atom
with initial quantum number ni, to move to a state with final quantum number nf (note ni > nf).
Ephoton = RH [ (1/nf2) – (1/ni2) ]
(3.2)
b) The Balmer series is a series of light emissions in the visible region of the spectrum, corresponding to
hydrogen atoms making the transition from an initial state with quantum number ni to a final state with quantum
number nf = 2. Find the wavelength of light corresponding to the Balmer series emissions with ni = 3, 4, 5, 6, and .
Give your wavelengths in nm.
4) The rotational constants for several hydrogen halides are given below. Based on this information do the following
a) Find re, the equilibrium bond length, for each H – X bond. Give your answer in units of nm.
b) Are the bond lengths consistent with your expectations? Explain.
5) The vibrational constants for several hydrogen halides are given below. Based on this information do the
following:
a) Find k, the force constant, for each H – X bond. Give your answer in units of N/m.
b) Are the force constants consistent with your expectations? Explain.
molecule
 (cm-1)
B (cm-1)
mH (amu)
mX (amu)
1
4138.3
2991.0
2649.0
20.956
10.593
8.465
1.008
18.998
34.969
80.916
H19F
H35Cl
1 81
H Br
1
Solutions.
1)
a) Since the only energy possessed by the ejected electrons is kinetic energy, we may say
EK = mv2/2
or
v = (2EK/m)1/2
For the maximum kinetic energy, (EK)max = 3.18 eV 1.602 x 10-19 J = 5.094 x 10-19 J
And so v = [ 2 (5.094 x 10-19 J)/(9.109 x 10-31 kg) ]1/2 = 7.48 x 105 m/s
b) The expression in the photoelectric effect involving the work function is
Ephoton = (hc/) = 0 + (EK)max
And so 0 = (hc/) – (EK)max
For a 254. nm photon
Ephoton = (hc/) = (6.626 x 10-34 Js)(2.998 x 108 m/s) = 7.821 x 10-19 J
1 ev
= 4.88 eV
(254. x 10-9 m)
1.602 x 10 -19 J
And so 0 = 4.88 – 3.18 = 1.70 eV
c) The longest wavelength of light capable of ejecting an electron can be found from the same formula used
in part b of this problem, and would correspond to ejection of an electron with zero kinetic energy.
Ephoton = (hc/) = 0
max = hc
0
From part b
0 = 1.70 ev 1.602 x 10-19 J = 2.72 x 10-19 J
1 eV
max = (6.626 x 10-34 Js)(2.998 x 108 m/s) = 1.24 x 10-6 m = 1240. nm
(2.72 x 10-19 J)
2)
For both parts of the problem it is useful to know the energy of a single photon with  = 500. nm
Ephoton = (hc/) = (6.626 x 10-34 Js)(2.998 x 108 m/s) = 3.97 x 10-19 J/photon
(500. x 10-9 m)
a) For a 100 watt CFL (8% efficiency)
photons/s = (0.08) 100. J
1 photon
1s
3.97 x 10-19 J
= 2.01 x 1019 photon/s
a) For a 100 watt IL (2% efficiency)
photons/s = (0.02) 100. J
1 photon
1s
3.97 x 10-19 J
= 5.03 x 1018 photon/s
3)
a) Based on conservation of energy, the energy of the emitted photon must be equal to the difference
between the energy of the initial state and the energy of the final state. Therefore
Ephoton = Ei – Ef = [ - RH (1/ni2) ] – [ - RH (1/nf2) ]
= - RH [ (1/ni2) – (1/nf2) ] = RH [ (1/nf2) – (1/ni2) ]
b) For the Balmer series, nf = 2. Substitution into our expression above, we get
Ephoton = RH [ (1/4) – (1/ni2) ]
If the Rydberg constant is given in units of wavenumbers (cm-1), then the wavelength of the emitted photon
will just be equal to 1/Ephoton (in units of cm).
We may summarize our results in the form of a table. Note RH = 109677 cm-1. Also note that 1 nm = 10-7
cm.
4)
ni
[ (1/4) – (1/ni2) ]
Ephoton (cm-1)
1/Ephoton (cm)
 (nm)
3
4
0.1389
0.1875
15233.
20564.
6.565 x 10-5
4.863 x 10-5
656.5
486.3
5
6
0.2100
0.2222
23032.
24373.
4.342 x 10-5
4.103 x 10-5
434.2
410.3

0.2500
27419.
3.647 x 10-5
364.7
For both problem 4 and problem 5 we will need the reduced mass for each molecule.
HF =
mHmF
= (1.008)(18.998) = 0.9572 amn
(mH + mF)
(1.008 + 18.998)
1.661 x 10-27 kg = 1.590 x 10-27 kg
1 amu
HCl =
mHmCl
=
(mH + mCl)
(1.008)(34.969) = 0.9798 amn
(1.008 + 34.969)
1.661 x 10-27 kg = 1.627 x 10-27 kg
1 amu
HBr =
mHmBr
=
(mH + mBr)
(1.008)(80.916) = 0.9956 amn
(1.008 + 80.916)
1.661 x 10-27 kg = 1.654 x 10-27 kg
1 amu
a) The rotational constant for a diatomic molecule is given by the expression
B=

4cr2
 = h = (6.626 x 10-34 Js) = 1.055 x 10-34 Js
2
2
where B is in units of wavenumbers (cm-1)
If we solve the above equation for r, we get
r = (/4cB)1/2
If we give c = 2.998 x 1010 cm/s, then cm in c and B will cancel. Therefore, if we give all of our other units as MKS
units, the value for r we will get will have units of m. We can then convert to nm using 1 nm = 10-9 m.
For 1H19F
r = [ (1.055 x 10-34 Js)/4(2.998 x 1010 cm/s)(1.590 x 10-27 kg)(20.956 cm-1) ]1/2
= 9.17 x 10-11 m = 0.0917 nm
If we repeat the calculation for the other two molecules, we get
1
H35Cl
r = 0.127 nm
1
H81Br
r = 0.141 nm
b) The bond lengths are consistent with expectations. F, Cl, and Br are isoelectronic. An F atom has n = 2
valence electrons, a Cl atom has n = 3 valence electrons, and a Br atom has n = 4 valence electrons. For an
isoelectronic series, the larger the value of n the larger the atom. Since all of the bonds are with a H atom, we would
expect the larger halogen should have the longer bond distance, as we in fact observe.
5)
a) The vibrational constant for a diatomic molecule is given by the expression
 = (1/2c) (k/)1/2
where  is in units of wavenumbers (cm-1).
If we solve for k, we get
k = 42c22
As before, if we give c in units of cm/s, then cm in c and B will cancel out. If everything else is given in MKS units,
then our answer will also have MKS units (kg/s2 = (kgm/s2) (1/m) = N/m)
For 1H19F
k = 42(2.998 x 1010 cm/s)2(4138.3 cm-1)2(1.590 x 10-27 kg) = 966 N/m
If we repeat the calculation for the other two molecules, we get
1
H35Cl
k = 516. N/m
1
H81Br
k = 412. N/m
b) The force constants are consistent with expectations. All of the bonds are single bonds between a
hydrogen atom and a halogen atom, and are isoelectronic. We would expect the shorter bond length to correspond to
the stronger bond, and therefore the larger value of k. This is in fact what we observe.