Introduction to Quantum Ideas
61
The energy levels are now different for a given k + n but different k or n
values. Thus, the degeneracy due to different angular momentum states is
removed by the addition of the potential g/r
2
. Indeed, the Coulomb degeneracy
is removed by any additional interaction.
Example 9
A quick estimation of the binding energies of the hydrogen atom is obtained by
the following simple argument.
A stationary state may be thought of as one for which an integral number of
de Broglie wavelengths can be fitted over the orbit, e.g. for circular orbits
2πr = n (h/p), n = 1, 2, ...
(2.95)
Using this relation, the total energy is
2
2
0
1
2
4
r
Ze
E
p
m
r
=
− πε
2 2
2
2
0
4
2 r
n
Ze
r
m r
=
− πε
(2.96)
If the state is stable, it corresponds to a minimum of this energy:
2 2
2
3
2
0
4
r
dE
n
Ze
dr
m r
r
= −
+ πε
= 0
(2.97)
so that
2 2
0
min
2
4
r
n
r
m
Ze
πε
=
(2.98)
This leads to the energy levels
2
2
2 2
0
4
2
n
mr
Ze
E
n
= −
πε
(2.99)
PROBLEMS
1. If the continuum spectrum of the sun approximates that of a black body,
peaking at λ m ≈ 5000 Å, what is the surface temperature of the sun? What
can you infer about the temperature of the material surrounding the sun
from the observation of Balmer absorption lines in the spectrum?
61
The energy levels are now different for a given k + n but different k or n
values. Thus, the degeneracy due to different angular momentum states is
removed by the addition of the potential g/r
2
. Indeed, the Coulomb degeneracy
is removed by any additional interaction.
Example 9
A quick estimation of the binding energies of the hydrogen atom is obtained by
the following simple argument.
A stationary state may be thought of as one for which an integral number of
de Broglie wavelengths can be fitted over the orbit, e.g. for circular orbits
2πr = n (h/p), n = 1, 2, ...
(2.95)
Using this relation, the total energy is
2
2
0
1
2
4
r
Ze
E
p
m
r
=
− πε
2 2
2
2
0
4
2 r
n
Ze
r
m r
=
− πε
(2.96)
If the state is stable, it corresponds to a minimum of this energy:
2 2
2
3
2
0
4
r
dE
n
Ze
dr
m r
r
= −
+ πε
= 0
(2.97)
so that
2 2
0
min
2
4
r
n
r
m
Ze
πε
=
(2.98)
This leads to the energy levels
2
2
2 2
0
4
2
n
mr
Ze
E
n
= −
πε
(2.99)
PROBLEMS
1. If the continuum spectrum of the sun approximates that of a black body,
peaking at λ m ≈ 5000 Å, what is the surface temperature of the sun? What
can you infer about the temperature of the material surrounding the sun
from the observation of Balmer absorption lines in the spectrum?