Elements of Modern Physics
166
This relation can be used to determine the nuclear spin.
For the hydrogen atom, I= 1/2. The ortho-modification has I = 1 (with
3 states) and the para-modification has I = 0 (with 1 state). At room temperature
the two modifications will occur in the ratio of 3 : 1. However, at low temperatures
most of the hydrogen molecules will go into the para state which has a slightly
lower energy, so that its spectral band will have alternate lines missing. When
para-hydrogen is heated, it retains its modification for a long time (several
weeks). As the modification changes, the spectral bands, as well as some physical
characteristics such as the heat capacity, show interesting variations.
5.8 EXAMPLES
Here, a few examples will be discussed to illustrate and extend some of the
ideas introduced in this chapter.
Example 1
Hund’s rule can be used to deduce the ground state of the elements. In particular,
consider the period from Na to Ar.
Sodium has one 3s electron in the valence shell, and hence its ground state
is
2
S 1/2 . Magnesium has (3s)
2
in the valence shell. Since this corresponds to a
closed subshell, its ground state is
1
S 0 . For Al, there is one electron in the
3p subshell so that its ground state is
2
p 1/2 (the smallest J value allowed is 1/2).
For Si, the valence shell has (3p)
2
so that the ground state has S = 1. The largest
allowed orbital angular momentum has L = 1 (since the space part is
antisymmetric, the largest M L corresponds to the two electrons having m l = 1
and m l = 0, so that the largest value of L is 1). Thus, the ground state is
3
P 0
(smallest value of J is zero). For P, the valence shell is (3p)
3
so that the ground
state has S = 3/2. The only allowed value of L is L = 0 (the antisymmetric spatial
wave function corresponds to electrons having m l = 1, m l = 0, and m l = – 1).
Therefore, the ground state is
4
S 3/2 .
For sulphur the shell is more than half-filled. Since a closed shell has J = L
= S = 0, it is easier to consider the unfilled shell as hole states (two holes for
sulphur). As in the case of holes in the Dirac sea, these holes may be regarded
as having positive charge. The spin of the two-hole state for sulphur is S = 1, the
orbital angular momentum is L = 1 (as for the two electron state), and J = 2 (the
holes have positive charge so that the constant C is Eq. (5.26) is negative and the
ground state has the largest allowed J value). Therefore, the ground state is denoted
by
3
P 2 . for Cl, there is only one hole which gives for its ground state,
2
P 3/2 (largest
J value). Finally or Ar, the subshell is closed, giving its ground state as
1
S 0 .
As an example of two unfilled subshells, consider molybdenum whose unfilled
shells are (4d)
2
(5s). The largest spin has S = 3, and the only allowed value of L
166
This relation can be used to determine the nuclear spin.
For the hydrogen atom, I= 1/2. The ortho-modification has I = 1 (with
3 states) and the para-modification has I = 0 (with 1 state). At room temperature
the two modifications will occur in the ratio of 3 : 1. However, at low temperatures
most of the hydrogen molecules will go into the para state which has a slightly
lower energy, so that its spectral band will have alternate lines missing. When
para-hydrogen is heated, it retains its modification for a long time (several
weeks). As the modification changes, the spectral bands, as well as some physical
characteristics such as the heat capacity, show interesting variations.
5.8 EXAMPLES
Here, a few examples will be discussed to illustrate and extend some of the
ideas introduced in this chapter.
Example 1
Hund’s rule can be used to deduce the ground state of the elements. In particular,
consider the period from Na to Ar.
Sodium has one 3s electron in the valence shell, and hence its ground state
is
2
S 1/2 . Magnesium has (3s)
2
in the valence shell. Since this corresponds to a
closed subshell, its ground state is
1
S 0 . For Al, there is one electron in the
3p subshell so that its ground state is
2
p 1/2 (the smallest J value allowed is 1/2).
For Si, the valence shell has (3p)
2
so that the ground state has S = 1. The largest
allowed orbital angular momentum has L = 1 (since the space part is
antisymmetric, the largest M L corresponds to the two electrons having m l = 1
and m l = 0, so that the largest value of L is 1). Thus, the ground state is
3
P 0
(smallest value of J is zero). For P, the valence shell is (3p)
3
so that the ground
state has S = 3/2. The only allowed value of L is L = 0 (the antisymmetric spatial
wave function corresponds to electrons having m l = 1, m l = 0, and m l = – 1).
Therefore, the ground state is
4
S 3/2 .
For sulphur the shell is more than half-filled. Since a closed shell has J = L
= S = 0, it is easier to consider the unfilled shell as hole states (two holes for
sulphur). As in the case of holes in the Dirac sea, these holes may be regarded
as having positive charge. The spin of the two-hole state for sulphur is S = 1, the
orbital angular momentum is L = 1 (as for the two electron state), and J = 2 (the
holes have positive charge so that the constant C is Eq. (5.26) is negative and the
ground state has the largest allowed J value). Therefore, the ground state is denoted
by
3
P 2 . for Cl, there is only one hole which gives for its ground state,
2
P 3/2 (largest
J value). Finally or Ar, the subshell is closed, giving its ground state as
1
S 0 .
As an example of two unfilled subshells, consider molybdenum whose unfilled
shells are (4d)
2
(5s). The largest spin has S = 3, and the only allowed value of L