Question

In decay ${}^{64}Cu{\to }^{64}Ni+e^{+}+v$ the maximum kinetic energy carried by the positron os 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 momentum of this neutrino in $kgm{s}^{-1}$?
Use the formula applicable to a photon

Answer 1

Given,

Maximum kinetic energy of the positron, $K=0.650MeV$

Now,

(a) Neutrino and positron are emitted simultaneously.

$\therefore$ Energy of neutrino =0.650-kinetic energy of the given positron

$=0.650-0.150$

$=0.5MeV=500keV$

(b) Momentum of the neutrino, $P=\frac{E}{c}$

Here, $E=$Energy of neutrino

$c=$Speed of light

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

$=2.67\times {10}^{-22}Kgm{s}^{-1}$