8.3 Solutions 471
For the forward reaction, energy available in the CMS isE∗=Q+Td∗+TN∗=Q+TdmN
mN+md= 13. 57 +20 × 14
14 + 2
= 31. 07
The energy of 31.07 MeV is shared betweenαand^12 C. Using the energy
and momentum conservation, we findPα∗= 416 .8MeV/c.
The inverse reaction will be endoergic, and so an energy of 31. 07 + 13. 57 =
44 .64 MeV must be provided. This corresponds to the Lab kinetic energy for
αgiven byTα= 44. 64 ×12 + 4
12
= 59 .52 MeVIn the CMS the energy of 31.07 MeV is shared between^2 H and^14 N. The
momentum of deuteron would be 441.9 MeV/c.
σdN
σαC=
(2Sα+1)(2SC+1)
(2Sd+1)(2SN+1)Pα∗^2
Pd∗^2=1 × 1
3 ×(2SN+1)
×
(
441. 9
319. 4
) 2
=
1. 91
2 SN+ 1
sinceSα=Sc=0 andSd=1. The spin of^14 N can be determined from
the experimental value of the cross-sections. ForSN=1, the ratio of cross-
sections is expected to be 0.64.
The intrinsic parities of all the four particles is positive. If theα′sare cap-
tured in the s-state for which the parity will be positive as it is given by (−1)l,
theα′scan not be produced in thel=1 state for which the parity would be
negative, resulting in the violation of parity.8.74 According to Butler’s theory, the neutron energy
En=1
2
Ed= 0. 5 × 460 =230 MeV
The spread in energy
ΔEn= 1. 5√
BdEd= 1. 5√
2. 2 × 460 = 47 .7MeV
And the angular spread isΔθ= 1. 6√
Bd
Ed= 1. 6
√
2. 2
460
= 0 .11 radians8.3.14 Fission and Nuclear Reactors........................
8.75 IfQis the number of atoms of^23 Na per gram at any timet, the net rate of
production of^24 Na is
dQ
dt
=φΣa−λQ