3.3 Solutions
173
3.21 u = C e
−kr
∞
0 |u|
2 dr = c
2
∞
0 e
−2kr
=
c
2
2k
= 1
C =
√
2k
The probability that the neutron – proton separation in the deuteron exceeds
R is
P =
∞
R
|u|
2 dr = 2k
∞
R
e
−2kr dr
= e
−2k R
= e
−(2×0.232×2)
≈ 0.4
Average distance of interaction
< r >=
∞
0
r |u
2
|dr = 2k
∞
0
re
−2kr dr
=
1
2k
=
1
2 × 0.232
= 2.16 fm
3.22 The inside wave function u 1 = A sin kr is maximum at r ≈ R. Therefore
kR = π/2
or k
2 R
2
= M(V 0 − W )R
2
/
2
= π
2
/4
V 0 =
π
2
2 c
2
4Mc 2 R 2 + W
Substituting c = 197.3 MeV − fm, Mc
2
= 940 MeV, R = 1.5 fm and
W = 2.2 MeV, we find V 0 ≈ 47 MeV
3.23 < r
2
>=
ψ
∗ r
2
ψdτ
=
∞
0
r
2
r 2
α
2π
e
−2αr 4πr
2 dr
=
1
2α 2
√ < r 2 > =
1
√
2α
=
4.3 × 10
−15 m
√
2
= 3.0 × 10
−15 m = 3.0 fm
3.24 Referring to Problem 3.18, the energy of the nth level is
E n =
n
2 h
2
8m L 2
(1)
and
E n+1 =
(n + 1)
2 h
2
8M L 2
(2)
Therefore E n+1 − E n =
(2n + 1)h
2
8m L 2
(3)