1000 Solved Problems in Modern Physics

(Romina) #1

7.1 Basic Concepts and Formulae 371


tanθ=sinθ∗/(cosθ∗+m 1 /m 2 ) (7.2)
Ifm 2 >m 1 ;0<θ<π
m 2 =m 1 ;0<θ<π/ 2
m 2 <m 1 ;0<θ<θmax
whereθmax=sin−^1 (m 2 /m 1 ) (7.3)

CM velocity combined withv 1 ∗orv 2 ∗givesv 1 orv 2 , respectively


φ=φ∗/2 (regardless of the ratiom 1 /m 2 ) (7.4)
φmax=π/ 2 (7.5)

Total kinetic energy available in the CMS


T∗=(1/2)μv 12 (7.6)

whereμis the reduced mass given by


μ=m 1 m 2 /(m 1 +m 2 ) (7.7)

Energy associated with the CMS is


(1/2)(m 1 +m 2 )vc^2 (7.8)

Scattering cross-section


LetI 0 be the beam intensity of the projectiles, that is the number of incident particles
crossing unit area per second, andIbe the intensity of the scattered particles going
into a solid angle dΩper second, andnthe number of target particles intercepting
the beam, then


I=I 0 nσ(θ, φ)dΩ (7.9)

If we assume here an azimuthal symmetry, then we can omit the azimuth angle
and simply writeσ(θ).
The constant of proportionalityσ(θ), also written as dσ(θ)/dΩ, is known as the
differential cross-section. It is a measure of the probability of scattering in a given
direction (θ, β) per unit solid angle from the given target nucleus. The integral over
the solid angle is known as total scattering cross-section (Fig. 7.3).


σ=


σ(θ, φ)dΩ (7.10)
Free download pdf