8.3 Solutions 467
Fig. 8.11Energy levels
8.61 For the nuclear reaction
(^3) H+p→ (^3) He+n− 0 .7637 MeV (1)
(^3) H− (^3) He=n−p− 0. 7637 (2)
∴Mass difference
mn−mH=mH 3 −mHe 3 −me+ 0. 7637 (3)
where the masses are atomic.
Add and subtractmein the right hand side so that
mn−mH=mH 3 −mHe 3 −me+ 0 .7637 MeV (4)
The masses are now atomic.
Now consider the decay
(^3) H→ (^3) He+β−+ν+ 18 .5keV (5)
On the atomic scale
mH 3 −mHe 3 = 18 .5keV= 0 .0185 MeV (6)
use (6) in (4) to find
mn−mH=(0. 0185 + 0 .7637) MeV
= 0 .7822 MeV.
8.62 For the reactionX(a,b)Y,
Q=Eb
(
1 +
mb
mY)
−Ea(
1 −
ma
mY)
−
2
mY√
mambEaEbcosθFor the reaction^3 H(d,n)^4 He, we identifya=d,x=^3 H,b=n,Y=^4 He.
SubstituteEa=^0 .3 MeV. Maximum neutron energy is obtained by putting
θ= 0 oand minimum energy forθ= 180 ◦in the above equation.
En(max)= 15 .41 MeV andEn(min)= 13 .08 MeV.
Thus, the range of neutron energy will be 13.08 – 15.41 MeV, the energy at
other angle of emission will be in between.8.63 Energy available in the CMS
−W∗=EpmTa
mTa+mp(1)
whereEpis the Lab proton kinetic energy. Using the values of the masses of
EaandmpandEp, we find the excited levelW∗= 4 .972 MeV. The energy of
the excited state will bemw+W∗= 169 ,490 MeV.