1000 Solved Problems in Modern Physics

236 3 Quantum Mechanics – II

σL(45◦)= 4 |f(90◦)|CM.

Thus quantum mechanics explains the experimental result

3.106σ=

(

4 π

k^2

)∑

l= 0

(2l+1) sin^2 δl (1)

By problem

sinδl=

(iak)l

√

(2l+1)l!

(2)

Therefore,

sin^2 δl=

(−a^2 k^2 )l

(2l+1)l!

(3)

Using (3) in (1)

σ=

(

4 π

k^2

)∑

l= 0

(−a^2 k^2 )l

l!

Summing over infinite number of terms for the summation and writing

k^2 =

2 mE

^2

,

σ=

(

2 π^2

mE

)

exp(−a^2 k^2 )

=

2 π^2

mE

exp

(

−

2 mEa^2

^2

)

3.107 Letbbe the impact parameter. In thec-system

bPcm=l=

where we have setl=1forthep-wave scattering

ECM=

PCM^2

2 μ =

PCM^2

M =

(^2) /Mb^2 (Since the reduced massμ=M/2, where
Mis the mass of neutron or proton)
ELab= 2 ECM=

2 ^2

Mb^2

=

2 ^2 c^2

Mc^2 b^2

Insertingc = 197 .3MeV.fm,Mc^2 =940 MeV andb=2 fm, we find

ELab= 20 .6 MeV. Thus up to 20 MeV Lab energy,s-waves (l=0) alone

are important

3.108 Onlys-waves (l = 0) are expected to be involved as the scattering is
isotropic.

σ=

4 πsin^2 δ 0

k^2

Nowk^2 ^2 =p^2 = 2 mE

sin^2 δ 0 =^24 mEπ 2 σ=^2 mc

(^2) Eσ
4 π^2 c^2
Insertingmc^2 =940 MeV;E= 1 .0MeV,
σ= 0. 1 b= 10 −^25 cm^2 =10 fm^2 ,c= 197 .3MeV−fm
sin^2 δ 0 = 0. 03845
δ 0 =± 11. 3 ◦