7.3 Solutions 395
(becausemandMare oppositely directed in the CMS, and the recoil angle of
the proton in the CMS is always twice the angle in the LS)
Therefore sinθ∗=sin(π− 2 φ)=sin 2φ
and cosθ∗=cos(π− 2 φ)=−cos 2φ
Equation (1) then becomes
tanθ=sinθ/cosθ=sin 2φ/(M/m−cos 2φ)
Cross multiplying the second equation and re-arranging
(M/m)sinθ=sinθ cos 2φ+cosθsin 2φ=sin(θ+ 2 φ)
M/m=sin(θ+ 2 φ)/sinθ
Usingθ= 5. 60 andφ= 22. 10 , we findM= 7 .8m≈ 7 .8amu.
7.2 (a) We can work out this problem in the LS, but we prefer the CMS for
convenience. Writing the equation for transformation of angles
tanθ=sinθ∗/(cosθ∗+M/m)(1)
The condition for the maximum angle of scattering,θmaxis d tanθ/dθ=0.
This gives us
cosθmax=−m/M (2)sinθmax∗ =(M^2 −m^2 )^1 /^2 /M (3)
When (2) and (3) are used in (1) we find cotθmax=(M^2 −m^2 )^1 /^2 /m,
whence
sinθmax=m/M (4)(b)sinφmax=sin(φ∗max/2)=sin[(π−θ∗max)/2]=cos(θ∗max/2)
=[(1+cosθ∗max)/2]^1 /^2 =[(1−m/M)/2]^1 /^2 =[(M−m)/ 2 M]^1 /^2
(5)(c) Usingm=1 andM=2 in (4) and (5)
we findθmax+φmax= 30 ◦+ 30 ◦= 60 ◦
7.3 We prefer to work in the CMS.
Letmbe the proton mass.
Energy available in the CMS in the d-d collision isE∗=^1 / 2 μu^2 =^1 / 2 mu^2
The centre of mass velocityvc=u/ 2
The energyE∗is partitioned between the product particles, proton and tri-
ton as follows.
Ep∗=3E∗/4 andEθ∗=E∗/4. The corresponding velocities in the CMS will
be,vp∗=√
3u/2 andvt∗=u/ 2√
3, respectively. Using the formula for the
transformation of angles from CM to LS (see formula 7.2)tanθ=sinθ∗
cosθ∗+vc/v∗(1)
And usingθ =θp=^45 ◦,vc=u/2 andv∗=vp∗=√
3u/2 and solving
forθ∗, we findθp∗=θ∗= 69 ◦.