8.3 Solutions
483
8.3.15 Fusion
8.93 The minimum energy of neutrino is zero when d and e
+ are emitted in opposite direction and the neutrino carries zero energy. The maximum energy of
neutrino corresponds to a situation in which e
+ is at rest and d moves in a
direction opposite to neutrino.
Q = 2 × 938.3 − 1875.7 − 0.51 = 0.39 MeV
T d + T ν = Q = 0.39 (energy conservation)
(1)
P ed
2
= P ν
2
(momentum conservation)
(2)
2 m d T d = T ν
2
(3)
Equations (1) and (3) can be solved to give T ν (max) = 0.38996 MeV. Thus
neutrino energy will range from zero to 0.39 MeV.
8.94 E n =
Qm He
m He +m n
=
3.2×3
3+1
= 2.4 MeV
8.95 (a) E =
1.44z 1 z 2
r
=
1.44×1×1
80
= 0.018MeV = 18 keV
(b) Equating the kinetic energy to heat energy
E =
3
2
kT
18 × 1.6 × 10
−16 J =
3
2
× 1.38 × 10
−23 T
T = 1.39 × 10
8 K
(c) The temperature can be lowered because with smaller energy Coulomb barrier penetration becomes possible.
8.96 The Lawson criterion is just satisfied if
L =
energy output
energy input
=
n
2
d < σ dt v > t c Q
6n d kT
=
n d < σ dt v > t c Q
6kT
= 1
Substitute n d = 7 × 10
18 m
−3
, kT = 10 keV = 10
4 eV
< σ dT .v ≥ 10
−22 m
3 s
−1 , Q = 17.62 × 10
6 eV to find t c = 4.86 s