The Nucleus
359
The radial Schrödinger equation for the ground state with l = 0, is
2
2
2
1
2
r
m
r
r
r
∂
∂ψ


−


∂
∂


= (
)
E V
− ψ
(9.122)
where m is the reduced mass
1
,
2
p
m
m


≈




V = 0 for r > a and V = – V 0 for r ≤ a.
It is easy to show that
ψ =
1
2
1
for
sin
for
2
kr
A e
r a
A
r
r a
−

>

α
<

(9.123)
where
k =
1/2
1/2
0
2
2
2 (
)
2 | | ,
m V E
m E
+



 α =








Continuity of the wave function and the derivative of the wave function, at
r = a, give the condition
α cot αa = – k
(9.124)
Since | E | is known to be small compared to V 0 , one has an approximate
relation
αa ≈ π/2
(9.125)
which implies V 0 ≈ 23 MeV for a ≈ 2 fm. A better approximation would be
αa ≈ 2
k
π + α
(9.126)
which yields a value of V 0 » 33 MeV for | E | ~ 2 MeV. In any case, it is seen that
the interaction is quite strong but the deuteron is a shallow bound state, i.e.
(| E |/V 0 ) << 1.
Example 3
As an application of the shell model, the spin and magnetic moments of
17
O and
127
I are considered.
The odd nucleon in
17
O is the ninth neutron which is in the d 5/2 state.
Therefore,
17
O has j = 5/2 and its magnetic moment [see Eq. (9.38)] is expected
to be about 2 p
e
m
−
(1.91). Experimentally j is found to be 5/2 and the magnetic
moment to be 2 p
e
m
−
(1.89).