236
3 Quantum Mechanics – II
σ L (45
◦ ) = 4| f (90
◦ )| CM .
Thus quantum mechanics explains the experimental result
3.106 σ =
4π
k 2
l=0
(2l + 1) sin
2
δ l
(1)
By problem
sin δ l =
(iak )
l
√
(2l + 1)l!
(2)
Therefore,
sin
2
δ l =
(−a
2 k
2 )
l
(2l + 1)l!
(3)
Using (3) in (1)
σ =
4π
k 2
l=0
(−a
2 k
2 )l
l!
Summing over infinite number of terms for the summation and writing
k
2
=
2mE
2 ,
σ =
2π
2
mE
exp(−a
2 k
2 )
=
2π
2
mE
exp
−
2mEa
2
2
3.107 Let b be the impact parameter. In the c-system
bP cm = l =
where we have set l = 1 for the p-wave scattering
E CM =
P
2
CM
2μ
=
P
2
CM
M
=
2
/Mb
2 (Since the reduced mass μ = M/2, where
M is the mass of neutron or proton)
E Lab = 2E CM =
2
2
Mb
2
=
2
2 c
2
Mc
2 b 2
Inserting c = 197.3 MeV.fm, Mc
2
= 940 MeV and b = 2 fm, we find
E Lab = 20.6 MeV. Thus up to 20 MeV Lab energy, s-waves (l = 0) alone
are important
3.108 Only s-waves (l = 0) are expected to be involved as the scattering is
isotropic.
σ =
4π sin
2
δ 0
k 2
Now k
2
2
= p
2
= 2mE
sin
2
δ 0 =
2mE σ
4π 2 =
2mc
2 Eσ
4π 2 c 2
Inserting mc
2
= 940 MeV; E = 1.0 MeV,
σ = 0.1 b = 10
−25 cm
2
= 10 fm
2
, c = 197.3 MeV − fm
sin
2
δ 0 = 0.03845
δ 0 = ±11.3
◦