Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2010
Laura Gagliardi
Lecture 3, September 13, 2010
Solved Homework
The total power emitted by a blackbody is obtained by integrating the frequency
dependent energy density over all frequencies. Thus, we must solve
ρ
T
(
)
=
8
π
h
ν
3
c
3
e
h
ν
/
kT
−
1
(
)
d
ν
0
∞
∫
To solve this integral, it is most convenient to introduce a new variable
x
=
h
ν
/
kT
. In
that case, we also have
dx
= (
h
/
kT
)
d
ν
. We may rearrange these to see what
substitutions we will make in the integral for
ν
and
d
ν
, namely
ν
=
xkT
/
h
and
d
ν
=
(
kT
/
h
)
dx
. If we introduce these substitutions we have
ρ
T
(
)
=
8
π
h
kTx
h
3
c
3
e
x
−
1
(
)
kT
h
dx
0
∞
∫
=
8
π
k
4
T
4
c
3
h
3
x
3
e
x
−
1
(
)
0
∞
∫
dx
The value of the final integral can be looked up:
it is
π
4
/15. If one plugs in all the
appropriate values for the various constants, one obtains
ρ
=
aT
4
where
a
= 7.5657
x
10
−
16
J m
3
K
4
. This derivation from Planck's formula agrees essentially perfectly
with experiment.
Page 125 of the Book (Chapter 1)
The Orbiting Electron Model of the Hydrogen Atom
In the first decade of the 1900s, it began to become clear that atomic structure
consisted of massive nuclei, composed of protons and neutrons, surrounded by
electrons that had comparatively enormous volumes of empty space available to them.
Protons and electrons attract one another through the Coulomb force (the only one of
the four physical forces to really matter in Chemistry). Since atoms do not
spontaneously annihilate their charged particles in fractions of a second, there must be
an opposing force that repels the electrons from the nucleus. The situation just
described has an obvious analogy (particularly to physicists), namely the centrifugal
force of an orbiting body that opposes the gravitational force attracting it to a central
mass, e.g., a planet.
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32
The total energy of a "planetary" hydrogen atom, assuming the nucleus to be
stationary, would derive from two terms:
the kinetic energy of the orbiting electron
and the potential energy from the Coulomb attraction. We will address each term
individually.
Kinetic Energy for an Orbiting Body
A convenient expression for kinetic energy, usually denoted
T
(not to be
confused with temperature!), is
T
=
p
2
2
m
(31)
where
p
is momentum (mass times velocity, units of massdistance per time—it is a
trivial matter to verify that this agrees with the other common expression
T
= (1/2)
mv
2
)
(32)
For an orbiting body, however, it is usually more convenient not to discuss
mass, velocity, and momentum, but rather the analogous quantities, moment of
inertia, angular velocity, and angular momentum. To see the relationship between
them easily, it is helpful to visualize the system.
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 Fall '08
 Staff
 Physical chemistry, Atom, pH, Angular Momentum, Photon, Electric charge, Fundamental physics concepts

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