178
3 Quantum Mechanics – II
R
0
|ψ 1 |
2 dτ +
∞
R
|ψ 2 |
2 dτ = 1
R
0
u
2
1 .4πr
2 dr/r
2
+
∞
R
u
2
2 4πr
2 dr/r
2
= 1
A
2
R
0
sin
2 krdr + C
2
∞
R
e
−2γ r dr = 1/4π
Integrating and using (2), we find
A
2
=
γ
2π (γR + 1)
(4)
Using (4) in (3)
P =
1
γ R + 1
(5)
Now γ R =
M W
2
1/2
R =
Mc
2 W
2 c 2
1/2
R
=
940 × 2.2
(197.3) 2
1/2
× 2.1 = 0.48
where we have inserted Mc
2
= 940 MeV /c
2 ,
W = 2.2MeV and R = 2.1 f m
Therefore p =
1
0.48+1
= 0.67
Thus neutron and proton stay outside the range of nuclear forces approximately 70% of time.
3.27 By Problem 3.25, for the finite well, for class I
α tan αa = β
with α =
(2m E)
1
2
; β =
[2m(V 0 − E)]
1
2
As V 0 → ∞, β → ∞ and αa = nπ/2 (n odd)
Therefore, α
2 a
2
=
2mEa
2
2
=
n
2 π
2
4
Or E = n
2
π
2
2
/8ma
2 (n odd)
For class II
α cot αa = −β
As V 0 → ∞, β → ∞ and αa = nπ/2 (n even)
3 Quantum Mechanics – II
R
0
|ψ 1 |
2 dτ +
∞
R
|ψ 2 |
2 dτ = 1
R
0
u
2
1 .4πr
2 dr/r
2
+
∞
R
u
2
2 4πr
2 dr/r
2
= 1
A
2
R
0
sin
2 krdr + C
2
∞
R
e
−2γ r dr = 1/4π
Integrating and using (2), we find
A
2
=
γ
2π (γR + 1)
(4)
Using (4) in (3)
P =
1
γ R + 1
(5)
Now γ R =
M W
2
1/2
R =
Mc
2 W
2 c 2
1/2
R
=
940 × 2.2
(197.3) 2
1/2
× 2.1 = 0.48
where we have inserted Mc
2
= 940 MeV /c
2 ,
W = 2.2MeV and R = 2.1 f m
Therefore p =
1
0.48+1
= 0.67
Thus neutron and proton stay outside the range of nuclear forces approximately 70% of time.
3.27 By Problem 3.25, for the finite well, for class I
α tan αa = β
with α =
(2m E)
1
2
; β =
[2m(V 0 − E)]
1
2
As V 0 → ∞, β → ∞ and αa = nπ/2 (n odd)
Therefore, α
2 a
2
=
2mEa
2
2
=
n
2 π
2
4
Or E = n
2
π
2
2
/8ma
2 (n odd)
For class II
α cot αa = −β
As V 0 → ∞, β → ∞ and αa = nπ/2 (n even)