In the decay ^ 64 Cu → ^ 64 Ni + e^ + + , the maximum kinetic...

In beta decay, the total energy released is shared between the emitted positron and the neutrino, neglecting the small recoil energy of the daughter nucleus.

The maximum kinetic energy of the positron corresponds to the case where the neutrino carries zero energy. Thus, the total energy available for the positron and neutrino is equal to the maximum kinetic energy of the positron.

E_{\text{total}} = K_{\text\max}(e^{+}) = 0.650 \text{ MeV}

By conservation of energy, the sum of the kinetic energy of the positron and the energy of the neutrino is constant and equal to $E_{\text{total}}$.

$K(e^{+}) + E_{\nu} = E_{\text{total}}$

Substituting the given values:

$0.150 \text{ MeV} + E_{\nu} = 0.650 \text{ MeV}$

$E_{\nu} = 0.650 \text{ MeV} - 0.150 \text{ MeV} = 0.500 \text{ MeV}$

Converting to keV:

$E_{\nu} = 500 \text{ keV}$