1000 Solved Problems in Modern Physics

396 7 Nuclear Physics – I

As triton will be emitted in the opposite direction in the CMS

θt∗= 111 ◦

Using formula (1) again, with the substitution.

θt∗= 111 ◦,vc =u/2 andvt∗=u/ 2

√

3, we can solve forθ, and obtain

θt= 34 ◦in the LS.

Finally, we can use the inverse transformation

tanθ∗=cossinθ−θvc/v

And substituteθ∗= 111 ◦,θ= 34 ◦andvc=u/2 to findv= 0 .48 u.

7.4 Use the result of Problem 7.1,

M/m=sin (θ+ 2 φ)/sinθ

M= 4. 004 ,m= 1. 008 ,θ= 100.

Substituting these values in the above equation and solving forφ, we find

φ= 16. 8 ◦which is the recoil angle of proton

7.5 Alphas of 20 MeV energy means that we are dealing with non-relativistic par-

ticles. From the results of Problem 7.2, the maximum scattering angleθmaxfor

m 1 >m 2 , is given by

sinθm=m 2 /m 1

sin 30^0 = 0. 5 =m 2 /4orm 2 = 2

The gas is deuterium. The limiting angleθmaxis independent of the incident

energy.

7.6 The description of the scattering event in the CM system is shown in Fig. 7.11.

From the velocity triangle

vLsinθL=vMsinθM

vLcosθL=vMcosθM+vC

Dividing the two equations and simplifying

tanθL=

sinθM

cosθM+vC/vM

=

sinθM

cosθM+m 1 /m 2

(

∵vC=

vm 1

(m 1 +m 2 )

andvm=

vm 2

(m 1 +m 2 )

)

The maximum scattering angleθmis given by

sinθM=m 2 /m 1 =^1 /^4 =^0.^25

θM= 15 ◦

Fig. 7.11Velocity triangle
for elastic scattering