1000 Solved Problems in Modern Physics

240 3 Quantum Mechanics – II

binding energyWof the deuteron, which is much smaller than the well depth

(∼25 MeV). In the deuteron problem the outside functionCe−γr, where

γ =

√

MW/^2 , is matched with the inside functionAsinkr.Herewe

match the functions sin(kr+δ 0 ) andCe−γratr =R, both in magnitude

and first derivative.

This gives uskcot(kR+δ 0 )=−γ. Further,R≈0. This is also reasonable

since for the square well the main features of the deuteron problem remain

unaltered by narrowing the well width and deepening the well. It follows that

sin^2 δ 0 =k^2 /k^2 +γ^2 )

But thes-wave cross-section is given by

σ= 4 πsin^2 δ 0 /k^2 = 4 π/(k^2 +γ^2 )

Substitutingk^2 =ME/^2 andγ^2 =MW/^2

σ=

4 π^2

M

1

W+E

(1)

where M is proton or neutron mass,W is the deuteron binding energy

(2.225 MeV), andEis the lab kinetic energy.

Formula (1) agrees well with experiment at relatively higher energies (say

5–10 MeV) but fails badly at very low energies. ForEW, for example,

(1) predictsσ =2 barns which is far from the experimental value of 20

barns. Wigner pointed out that inn−pscattering the spins of the colliding

nucleons could be either parallel or antiparallel. Formula (1) holds for the

parallel case because the analogy is made with the deuteron problem which

has parallel spins. Now for random orientations of spins:

σ=

3

4

σt+

1

4

σs (2)

whereσtandσsare the cross-sections for the triplet and singlet scattering,

the factors^34 and^14 being the statistical weights. In (1),W is the binding

energy of then−psystem for the triplet state. Corresponding to the singlet

state the qualityWsis introduced, although it is a virtual state.

Combining (1) and (2)

σ=

3 π^2

M(E+W)

+

π^2

M(E+Ws)

(3)

Wstakes a value of 70 keV if agreement is to reach with the experiments.

Agreement at higher energies is preserved because forEWorWs,(3)

reduces to (1).

3.3.9 Scattering(BornApproximation).....................

3.113 (a) F(q)≈

∫∞

0

ρ(r) sin(qr/)4πr^2

qr/ dr

ρ(r)=Aexp(−r/a)

F(q)≈ 4 πA

∫∞

0

rexp(−r/a)[sin(qr/)/(q/)]dr