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Chem 3502/5502
Physical Chemistry II (Quantum Mechanics)
3 Credits
Fall Semester 2010
Laura Gagliardi
Lecture 15, October 13, 2010
Solved Homework
We are asked to give the minimum energy required for the ionization reaction
Be
3+
→
Be
4+
+
e

Since Be is atomic number
Z
= 4, this corresponds to ionization of the electron from a
oneelectron atom. Oneelectron atoms are the systems that we know how to deal with
exactly based on our work of last lecture.
Note that the reason the poorly worded problem presumably says the “minimum
energy” is that it has in mind the energy required to just separate the electron from the
nucleus and leave both particles at rest (this defines the system’s zero of energy). You
could, however, provide
more
energy, so that not only would the particles be infinitely
separated, but they’d have some kinetic energy too, instead of being at rest. However, the
problem is poorly worded because the correct answer is that the minimum energy is
ε
,
where
ε
is a positive number arbitrarily close to zero. How can this be? Well, what if the
electronic wave function corresponds to principle quantum number
n
= 10
9
? The binding
energy in that case is incredibly small (
ε
), and it’s easy to remove the electron.
To clean up the wording, one need only specify that Be
3+
is in its
ground state
. In
that case, we know that the binding energy of the electron to the nucleus is
E
n
=
1
Be
3
+
(
29
= 
Z
Be
2
e
4
μ
Be
32
π
2
ε
0
2
1
( 29
2
h
2
Now, we could go look up all the necessary constants and plug them in, and that
would give a very nice answer. A quicker approach, however, is to use the datum
provided that
E
n
=1
(H) =

13.6 eV. Then we can consider
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E
n
=
1
Be
3
+
(
29
E
n
=
1
H
( 29
=

Z
Be
2
e
4
μ
Be
32
π
2
ε
0
2
1
( 29
2
h
2

Z
H
2
e
4
μ
H
32
π
2
ε
0
2
1
( 29
2
h
2
=
16
μ
Be
μ
H
=
16
m
Be
m
e
m
Be
+
m
e
m
H
m
e
m
H
+
m
e
≈
16
m
Be
m
e
m
Be
m
H
m
e
m
H
=
16
Thus, the final ionization potential for Be should be 16 times the ionization potential for
H, or

217.6 eV.
Hydrogenic Orbitals
Last lecture, we went through a fairly complicated process to derive the quantum
restrictions on the energy levels of oneelectron atoms, but we did not take the final step
of putting our final quantum numbers back into our various differential equations to pull
out the eigenfunctions themselves.
If one attempts to do this in a completely general way (i.e., for arbitrary quantum
numbers
n
and
l
) one can write a general wave function in terms of what are called the
associated Laguerre polynomials. This general form, however, is not terribly informative,
and it is more useful to simply consider some of the specific functions associated with
smaller quantum numbers. For example, what if we pick the principle quantum number
n
= 1? From our previous work, and in particular eq. 1430, this requires that
l
and
m
both
be zero, and
β
=
n
= 1.
The requirement that
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This note was uploaded on 12/18/2010 for the course CHEM 3502 taught by Professor Staff during the Fall '08 term at Minnesota.
 Fall '08
 Staff
 Physical chemistry, Atom, pH, Reaction

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