7.3 Solutions 425
7.111 i. Momentum=300 Br=(300)(0.1)(0.061)= 1 .83 MeV/c
Use the relativistic equation to find total energy,E=(p^2 +m^2 )^1 /^2 =
(1. 832 + 0. 5112 )^1 /^2 = 1 .90 MeV
Kinetic energy of electron
T=E−m= 1. 90 − 0. 51 = 1 .39 MeV= 1. 39 × 106 eV
Neglecting the energy of the recoiling nucleus the neutrino energy
Eν= 2. 4 − 1. 39 = 1 .01 MeV
ii. Maximum kinetic energy will be carried by the nucleus when the nucleus
recoils opposite toβ-particle and the neutrino is at rest. Momentum
conservation gives
PN^2 = 2 MNTN=Pe^2 =Te^2 + 2 Tem (1)
Energy conservation gives
TN+Te= 2 .4(2)
EliminatingTe, we findTN=5keV.
7.112n(E)=C
√
E(Emax−E)^2
whereC=constant.
Differentiaten(E) with respect toEand set dn/dE=0. We find that the
maximum occurs atE=Emax/5.