CHM 3400 – Problem Set 9
Due date: Monday, April 9th. The third hour exam is Wednesday, April 11 th. It will cover Chapter 9, sections 9.39.9, and 9.13; Chapter 10, sections 10.3-10.11; Chapter 11, sections 11.1, and 11.3-11.7; Chapter 12, all sections
except section 12.7.
Do all of the following problems. Show your work.
"Luxery...means one of two things: either pretty stuff that, without marketing to infuse value, would be virtually
worthless (yellow diamonds, modern art, Steuben glass) or else something functional but no better that the same
product at a fraction of the price. Just because you call it a timepiece and charge $1000 for it doesn’t mean you
arrive at work any more promptly than with a watch selling for $10.” - Randall Lane, The Zeros.
1) Argon ion lasers are often used as monochromatic light sources. By changing the operating characteristics of the
laser a number of different wavelengths of laser light in the region 454 – 514 nm can be produced.
Consider an argon ion laser operating at  = 488. nm.
a) What is the energy of one photon of light from this laser? Give your answer in both units of J/photon and
in cm-1.
b) What is the energy of 1.00 moles of photons from the above laser? Give your answer in units of kJ/mol.
c) A light pulse from the above laser has a total energy of 3.5 mJ. How many photons does this correspond
to?
2) A typical home microwave oven uses radiation at a frequency  = 2.45 GHz (1 GHz = 109 Hz = 109 s-1) and
operates at a power of 700 watts (1 watt = 1 J/s).
a) What is the photon flux (photons/s) for a typical microwave oven?
b) Consider the use of a microwave oven to heat a 1.000 L sample of water originally at T = 20. C. How
long will it take until the water begins to boil? At this point, how many photons will the water have absorbed (give
your answer both in terms of the total number of photons absorbed, and the photons absorbed per water molecule)?
Briefly indicate any assumptions made in obtaining your answers. Note that the density of water is  = 1.00 g/mL,
and the specific heat of water is c = 4.18 J/g.K.
3) A variety of particles can be used in diffraction experiments, including electrons and neutrons. Find the speed of
an electron and the speed of a neutron required for these particles to have a de Broglie wavelength  = 0.100 nm,
roughly corresponding to a typical bond length in a molecule. Give your answers in units of m/s.
4) The uncertainty in the position of an electron is x = 0.0500 nm.
a) What is the minimum value for p, the uncertainty in the momentum of the electron? Give your answer
in units of kg.m/s.
b) What is the corresponding minimum value for v, the uncertainty in the velocity of the electron? Give
your answer in units of m/s.
5) The rotational and vibrational energies for a diatomic molecule are, in the RRHO (rigid rotor – harmonic
oscillator) approximation, given by the expressions
Ẽrot = J(J+1)B
J = 0, 1, 2, ...
B = h/(82cr2)
(5.1)
Ẽvib = (v + ½)
v = 0, 1, 2, ...
 = (1/2c) (k/)1/2
(5.2)
where AB, the reduced mass (units of kg), can be found in terms of the masses of the individual atoms making up the
diatomic molecule
AB =
mAmB
(mA + mB)
(5.3)
J and v are quantum numbers for rotation and vibration, respectively, r is the bond distance for the diatomic
molecule, k is the force constant for the molecular bond, and and Ẽ, B, and  are in units of cm-1.
The experimental values for B and  for the 12C16O isotopic form of carbon monoxide are B = 1.9313 cm-1
and  = 2170.21 cm-1. Based on this, find the following:
a) r, the equilibrium bond length in 12C16O. Give your answer in units of nm.
b) k, the force constant for the bond in 12C16O. Give your answer in units of N/m.
c) The values for B and  for the 12C18O isotopic form of carbon monoxide. You may assume that r and k
are the same for 12C18O and 12C16O.
Note that m(12C) = 12.0000 amu, m(16O) = 15.9949 amu, and m(18O) = 17.9992 amu.
Also do the following problems from Atkins:
12.9 Use the following data on the maximum kinetic energy of photoelectrons emitted by radiation of different
wavelengths from a metal to determine the value of Planck’s constant and the work function of the metal.
 (nm)
E (eV)
300.0
1.613
350.0
1.022
400.0
0.579
450.0
0.235
HINT: Plot E vs 1/ to find the work function and the value for the Planck constant. Give the value for  in units of
eV, and the value for h in units of J.s. Also use the data to find 0 (in nm), the longest wavelength capable of
producing electrons from the metal.
Solutions.
1)
a) E = hc = (6.626 x 10-34 J.s) (2.998 x 108 m/s) = 4.071 x 10-19 J/photon

( 488. x 10-9 m)
The energy in cm-1 can be found in two ways
E(cm-1) = (4.071 x 10-19 J/photon) (1 cm-1/1.986 x 10-23 J) = 20500. cm-1
Alternatively, E(cm-1) = (1/) = (1/488 x 10-7 cm) = 20500. cm-1
b) For one mole of photons
E = (4.071 x 10-19 J/photon) (6.022 x 1023 photon/mol) = 2.451 x 105 J/mol = 245.1 kJ/mol
c) # photons = (3.5 x 10-3 J)
2)
1 photon
4.071 x 10-19 J
= 8.60 x 1015 photons.
a) The energy of one photon is E = h = (6.626 x 10-34 J.s) (2.45 x 109 s-1) = 1.62 x 10-24 J/photon
So for 700. watts of power, the photon flux is
flux =
700. J/s
1.62 x 10-24 J/photon
= 4.31 x 1026 photon/s
b) We will assume that all of the microwave radiation is absorbed by the water, and that no thermal energy
is lost by the water as it is heated. Since these are both approximations, the actual time to heat the water to boiling is
likely longer than what we will calculate here.
Cwater = (1000. mL) (1.00 g/1. mL) (4.18 J/g.K) = 4180. J/g.K
The amount of heat needed to change the temperture of the water from 20. C to 100. C is
E = (4180. J/g.K) (100.0 – 20.0) K = 3.34 x 105 J
So the time required is
t = 3.34 x 105 J
1s
700. J
= 478. s (about 8 minutes)
The total number of photons absorbed is
# photons = 3.34 x 105 J
1 photon
1.62 x 10-24 J
= 2.06 x 1029 photons
The number of water molecules in 1.000 L is
# molecules = (1000. mL) (1.00 g/mL) (1 mol/18.02 g) (6.022 x 10 23 molecule/mol)
= 3.34 x 1025 molecules
So the number of photons per molecules is n(per molecule) =
2.06 x 1029 photon = 6170. photons/molecule
3.34 x 1025 molecule
3)
The de Broglie wavelength is
dB = h/mv
v=
h
mdB
For an electron
v =
(6.626 x 10-34 J.s)
= 7.27 x 10 6 m/s (about 2% of the speed of light!)
-31
-9
(9.111 x 10 kg) (0.100 x 10 m)
For a neutron
v =
(6.626 x 10-34 J.s)
= 3960. m/s
(1.674 x 10-27 kg) (0.100 x 10-9 m)
4)
a) According to the uncertainty principle, (x)(p)  /2
So
p 

= (1.055 x 10-34 J.s)
= 1.055 x 10-24 kg.m/s
-9
2(x)
2 (0.0500 x 10 m)
This value represents the minimum uncertainty in momentum.
b) p = m(v), and so
v = (p)/m = (1.055 x 10-24 kg/m.s)/(9.11 x 10-31 kg) = 1.16 x 106 m/s
5) Before we start, let us find the values for reduced mass that we will be using.
(12C16O) = (12.0000 amu) (15.9949 amu) = 6.8582 amu
1 kg
= 1.1385 x 10-26 kg
26
(12.0000 + 15.9949) amu
6.022 x 10 amu
(12C18O) = (12.0000 amu) (17.9992 amu) = 7.1999 amu
1 kg
= 1.1956 x 10-26 kg
26
(12.0000 + 17.9992) amu
6.022 x 10 amu
a) B = h/(82cr2)
So
r = [h/(82cB)]1/2 = {(6.626 x 10-34 J.s)/[82(2.998 x 1010cm/s)(1.1385 x 10-26 kg)(1.9313 cm-1)]1/2
= 1.1283 x 10-10m = 0.11283 nm
b)  = (1/2c) (k/)1/2
So
k = 42c22 = 42(2.998 x 1010 cm/s)2(2170.21 cm-1)2(1.1385 x 10-26 kg) = 1903. N/m
c) If we look at the formula for B, we can see that
B(12C18O) = (12C16O)
B(12C16O)
(12C18O)
or
B(12C18O) = B(12C16O) (12C16O) = (1.9313 cm-1) (1.1385 x 10-26 kg) = 1.8391 cm-1
(12C18O)
(1.1956 x 10 -26 kg)
and
(12C18O) = [(12C16O)]1/2
(12C16O)
[(12C18O)]1/2
or
(12C18O) = (12C16O) [(12C16O)/ (12C18O)]1/2
= (2170.21 cm-1) [(1.1385 x 10-26 kg)/(1.1956 x 10-26 kg) ]1/2 = 2117.75 cm-1
NOTE: To convince yourself that the above expressions are correct simply substitute for B and  for each isotopic
combination in each of the two equations and cancel out common terms. Alternatively, you could use the values for
r and k found in parts a and b and the equations for B and , using the value for (12C18O).
12.9
The data are presented and plotted below
(nm)
1/ (nm-1)
E (eV)
300.0
350.0
0.003333
0.002857
1.613
1.022
400.0
450.0
0.002500
0.002222
0.579
0.235
Plot of E vs 1/wavelength
2
E (eV)
1.5
1
0.5
0
0.002
-0.5
0.0025
0.003
0.0035
1/wavelength (nm-1)
The formula for the best fitting line is E = m(1/) + b
Based on the best fit to the data I get slope = m = 1240.5 eV.nm and intercept = b = - 2.5218 eV
The work function is  = - b = 2.5218 eV
The slope is m = hc, and so h = m/c = (1240.5 eV .nm)(1.6022 x 10-19 J/eV)(1 m/109 nm)/(2.998 x 108 m/s)
= 6.63 x 10-34 J.s (very close to the presently accepted value)
Finally, to get the value for 0, we recall that for this value of wavelength E = 0, and so
0 = m(1/0) + b
0 = - (m/b) = - [ (1240.5 eV.nm)/(- 2.5218 eV) ] = 491.9 nm