8.3 Solutions 483
8.3.15 Fusion ...........................................
8.93 The minimum energy of neutrino is zero whendande+are emitted in oppo-
site direction and the neutrino carries zero energy. The maximum energy of
neutrino corresponds to a situation in whiche+is at rest anddmoves in a
direction opposite to neutrino.
Q= 2 × 938. 3 − 1875. 7 − 0. 51 = 0 .39 MeV
Td+Tν=Q= 0 .39 (energy conservation) (1)
Ped^2 =Pν^2 (momentum conservation) (2)
2 mdTd=Tν^2 (3)
Equations (1) and (3) can be solved to giveTν(max)= 0 .38996 MeV. Thus
neutrino energy will range from zero to 0.39 MeV.
8.94 En=mQmHe+Hemn=^33.^2 +× 13 = 2 .4MeV
8.95 (a)E=^1.^44 rz^1 z^2 =^1.^4480 ×^1 ×^1 = 0 .018MeV=18 keV
(b) Equating the kinetic energy to heat energy
E=^32 kT
18 × 1. 6 × 10 −^16 J=^32 × 1. 38 × 10 −^23 T
T= 1. 39 × 108 K
(c) The temperature can be lowered because with smaller energy Coulomb bar-
rier penetration becomes possible.
8.96 The Lawson criterion is just satisfied if
L=energy output
energy input=
n^2 d<σdtv>tcQ
6 ndkT=
nd<σdtv>tcQ
6 kT= 1
Substitutend= 7 × 1018 m−^3 ,kT=10 keV= 104 eV
<σdT.v≥ 10 −^22 m^3 s
− 1
,Q= 17. 62 × 106 eV to findtc= 4 .86 s