Question

In the decay ${}^{64}Cu{\to }^{64}Ni+e^{+}+v$, the maximum kinetic energy carried by the positron is found to be 0.650 MeV. (a) What is the energy of the neutrino which was emitted together with a positron of kinetic energy 0.150 MeV? (b) What is the momentum of this neutrino in kg m $s^{-1}$ ? Use the formula applicable to a photon.

Answer 1

${}^{64}Cu{\to }^{64}Ni+-e+v$

Emission of neutrino is along with a positron emission.

(a) Energy of positron = 0.650 MeV

Energy of Neutrino

= 0.650 - KE of given positron

= 0.650 - 0.150

= 0.5 MeV

= 500 KeV

(b) Momentum of Nutrino

= $\frac{500\times 1.6\times {10}^{-19}}{3\times {10}^8}\times {10}^{8}J$

$=2.67\times {10}^{-27}\text{Kg/ms}$