398 7 Nuclear Physics – I
dΩ∗/dEp= 4 π/E 0 (2)
And dσp/dΩ∗=σ/ 4 π (3)
as the scattering is isotropic in the CMS.
Using (2) and (3) in (1) we get dσp/dEp=σ/E 07.10 As the scattering is isotropic in the CMS the differential cross-section of the
recoiling nuclei is constant and is given byσ(φ∗)=σ/ 4 π=constant.
Now the differential cross-sections in the LS and CMS are related by
σ(φ)=(sinφ∗dφ∗/sinφdφ).σ(φ∗)
Butφ∗= 2 φand dφ∗=2dφ
σ(φ)=(sin 2φ.2dφ/sinφdφ)(σ/ 4 π)=σπcosφ
Thus,∫ σ(φ) has cosφdependence. It is of interest to note that
σ(φ)dΩ=
∫π/ 2
0 σcosφ.^2 πsinφdφ/π=σ
as it should. The upper limit for the integration is confined to 90◦as the target
nucleus can not recoil in the backward sphere in the LS.7.11 In Fig. 7.13,bdenotes the impact parameter. Consider particlesI 0 going
through a ring perpendicular to the central axis, its area being 2πbdb.On
hitting the sphere, the same number of particles are scattered through a solid
angle dΩ= 2 πsinθdθ
Fig. 7.13Scattering of
particles of negligible size
from an infinitely heavy hard
sphere of radius R
Therefore,I 0 σ(θ). 2 πsinθdθ=−I 02 πbdb
Orσ(θ)=−bdb/sinθdθ (1)
The angles of incidence and reflection are measured with respect to the
normal at the point of scattering. From the geometry of the figure,
θ=π−(i+r)=π− 2 i (2)
sinθ=sin (π−2i)=sin 2i=2sinicosi (3)
Sincer=iand dθ=−2di
b=Rsiniand db=Rcosidi
We findσ(θ)=R^2 /4(4)
The right hand side of (4) is independent of the scattering angleθ; that is
the scattering is isotropic or equally in all directions.
The total cross-section