Elements of Modern Physics
388
The Ω
–
is made of three strange quarks (sss). The three quarks are in the
ground state with spatial angular momentum equal to zero and in the totally
symmetric spin 3/2 state. This would violate Fermi statistics. In order to get
out of this difficulty, an additional property called colour was assigned to the
quarks. The three quarks in the baryons are supposed to be of different colours
so that the exchange statistics is not applicable.
Example 2
The β-decay in Eq. (10.21) is to be regarded as a spontaneous emission of e and
e
v and not an escape of an electron bound in the nucleus. This is indicated by
the fact that a bound electron would have a momentum p ~ 0
/r
(follows from
the uncertainty principle) and a kinetic energy
KE ~ (m
2
c
4
+ c
2
2
0
/r
2
)
1/2
–mc
2
(10.47)
which for r 0 ~ 10
–15
m comes out to be approximately 200 MeV. Such a highlyenergetic electron would quickly escape in about 10
–23
s, and the slow rate of
the β-decay cannot be explained.
Example 3
The parity of π
–
is determined from the reaction
π
–
+ d → N + N
(10.48)
The low-energy π
–
is captured is a Bohr orbit (with a radius about 2 ×
10
–13
m) in the deuterium. The reaction in Eq. (10.48) proceeds with l = 0
(nuclear forces become effective only at a short range of about 10
–15
m) so that
the initial total parity for the reaction is that of the intrinsic parity of the pion.
Now if the orbital angular momentum of the NN state had l = 0, the spin of
the NN system would be 1 (equal to the spin of the deuteron, by the conservation
of the total angular momentum). This implies that the spins of the two neutrons
would have to be parallel, and hence, this state is forbidden by Fermi statistics.
Therefore, the orbital angular momentum of the NN state has l = 1. This means
that the parity of the final state is –1 and therefore the intrinsic parity of
the π-meson is –1 (parity being conserved in strong interaction).
Example 4
The behaviour of neutral K-mesons provides a very interesting application of
the superposition principle. Neutral K-mesons have the decay modes
K
0
→ π
+
+ π
–
, π
0
+ π
0
0
K → π
+
+ π
–
, π
0
+ π
0
(10.49)
388
The Ω
–
is made of three strange quarks (sss). The three quarks are in the
ground state with spatial angular momentum equal to zero and in the totally
symmetric spin 3/2 state. This would violate Fermi statistics. In order to get
out of this difficulty, an additional property called colour was assigned to the
quarks. The three quarks in the baryons are supposed to be of different colours
so that the exchange statistics is not applicable.
Example 2
The β-decay in Eq. (10.21) is to be regarded as a spontaneous emission of e and
e
v and not an escape of an electron bound in the nucleus. This is indicated by
the fact that a bound electron would have a momentum p ~ 0
/r
(follows from
the uncertainty principle) and a kinetic energy
KE ~ (m
2
c
4
+ c
2
2
0
/r
2
)
1/2
–mc
2
(10.47)
which for r 0 ~ 10
–15
m comes out to be approximately 200 MeV. Such a highlyenergetic electron would quickly escape in about 10
–23
s, and the slow rate of
the β-decay cannot be explained.
Example 3
The parity of π
–
is determined from the reaction
π
–
+ d → N + N
(10.48)
The low-energy π
–
is captured is a Bohr orbit (with a radius about 2 ×
10
–13
m) in the deuterium. The reaction in Eq. (10.48) proceeds with l = 0
(nuclear forces become effective only at a short range of about 10
–15
m) so that
the initial total parity for the reaction is that of the intrinsic parity of the pion.
Now if the orbital angular momentum of the NN state had l = 0, the spin of
the NN system would be 1 (equal to the spin of the deuteron, by the conservation
of the total angular momentum). This implies that the spins of the two neutrons
would have to be parallel, and hence, this state is forbidden by Fermi statistics.
Therefore, the orbital angular momentum of the NN state has l = 1. This means
that the parity of the final state is –1 and therefore the intrinsic parity of
the π-meson is –1 (parity being conserved in strong interaction).
Example 4
The behaviour of neutral K-mesons provides a very interesting application of
the superposition principle. Neutral K-mesons have the decay modes
K
0
→ π
+
+ π
–
, π
0
+ π
0
0
K → π
+
+ π
–
, π
0
+ π
0
(10.49)