7.3 Solutions 397
7.7 Let the alpha particle of massmmoving with velocityv 0 and momentum
p 0 collide with proton of massm. After the collision the maximum velocity
v 2 and momentump 2 will be acquired by proton when it is emitted in the
incident direction. Since the alpha particle is heavier than proton, it must also
proceed in the same direction as the proton with velocity v 1 and momentum
p 1. Assuming that the collision is elastic,P 02 / 2 m 1 =p 12 / 2 m 1 +p 22 / 2 m 2 (energy conservation) (1)
P 0 =p 1 +p 2 (momentum conservation) (2)Notingm 1 = 4 m 2 , p 1 can be eliminated between (1) and (2) to yield
p 2 = 0. 4 p 0
orm 2 v 2 = 0. 4 m 1 v 0 = 0. 4 × 4 m 2 v 0
v 2 = 1. 6 v 0
7.8 Use the transformation equation for the differential cross-sections in the CMS
and LS.
σ(θ)=(1+γ^2 + 2 γcosθ∗)^3 /^2 σ(θ∗)/| 1 +γcosθ∗|
whereγ=m 1 /m 2 =m/m=1. Then the equation simplifies to
σ(θ)=4 cos(θ∗/2)σ(θ∗)
7.9 We can write by chain rule
dσp/dEp=(dσp/dΩ∗).(dΩ∗/dEp)(1)
Let the proton be scattered through an angleθ∗in the CMS with the direc-
tion of incidence of neutron (left to right). The CMS velocity will bev 0 /2.
vc =v 0 m 1 /(m 1 +m 2 )= v 0 /2 as the masses of neutron and proton are
approximately equal. The proton has velocityvcin the CMS both before and
after scattering. The velocity of the scattered proton is combined with the CMS
velocity to give the velocityv 0 in the LS as shown in Fig. 7.12. From the
velocity triangle we have
v^2 =vc^2 +vc^2 + 2 vc^2 cosθ∗= 2 vc^2 (1+cosθ∗)
=v 02 (1+cosθ∗)/ 2
Therefore,Ep=E 0 (1+cosθ∗)/ 2
Differentiating
dEp=−E 0 sinθ∗dθ∗/ 2 =−E 0 dΩ∗/ 4 π
The negative sign means that asθ∗increasesEpdecreases
Thus,Fig. 7.12n−p scattering
in CMS