^ 32 P undergoes beta-decay to ^ 32 S . Calculate the sum of the...

The beta-decay equation is:

^{32}_{15}\text{P} \rightarrow ^{32}_{16}\text{S} + e^{-} + \bar{\nu}_{e}$<br><br>The total energy released ($Q$-value) in the$\beta^{-}$-decay is shared between the$\beta$-particle, the antineutrino, and the recoiling daughter nucleus. Disregarding the recoil energy of the daughter nucleus, the sum of the kinetic energy of the$\beta$-particle and the energy of the antineutrino is equal to the$Q$-value.<br><br>The$Q$-value for$\beta^{-}$-decay in terms of atomic masses is:<br><br>$Q = [m(^{32}\text{P}) - m(^{32}\text{S})]c^2$<br><br>Substituting the given atomic masses:<br><br>$\Delta m = 31.974 - 31.972 = 0.002 \text{ u}$<br><br>Converting the mass defect into energy using$1 \text{ u} = 931.5 \text{ MeV}/c^2$:<br><br>$Q = 0.002 \times 931.5 \text{ MeV} = 1.863 \text{ MeV}$<br><br>The sum of the energy of the antineutrino and the kinetic energy of the$\beta$-particle is approximately$1.86 \text{ MeV}.