The nuclide Sc^50, mass 49.9516 amu, is neutron rich. It decays to form Ti^50, mass 49.94479 amu. If — Nuclear Chemistry and Radioactivity Chemistry Question
Question
The nuclide Sc^50, mass 49.9516 amu, is neutron rich. It decays to form Ti^50, mass 49.94479 amu. If the emitted β-particle has a kinetic energy of 0.80 MeV, what is the kinetic energy of the antineutrino emitted simultaneously?
💡 Solution & Explanation
**Step 1: Write the beta-minus decay equation.** $$\ce{^{50}_{21}Sc -> ^{50}_{22}Ti + e^- + \bar{\nu}_e}$$ **Step 2: Calculate the Q-value.** $$Q = \bigl[M(\ce{^{50}Sc}) - M(\ce{^{50}Ti})\bigr] \times 931.5 \text{ MeV/u}$$ $$Q = (49.9516 - 49.94479) \times 931.5 = 0.00681 \times 931.5 = 6.34 \text{ MeV}$$ **Step 3: Apply energy conservation.** The total decay energy is shared between the beta particle and the antineutrino: $$Q = KE(\beta^-) + KE(\bar{\nu}_e)$$ $$6.34 = 0.80 + KE(\bar{\nu}_e)$$ $$\boxed{KE(\bar{\nu}_e) = 6.34 - 0.80 = 5.54 \text{ MeV}}$$ **Note:** In beta decay, the energy is not fixed but distributed as a spectrum. This calculation gives the energy when the antineutrino carries 5.54 MeV and the electron carries 0.80 MeV (one particular sharing of the total 6.34 MeV). **Answer: B — The antineutrino carries 5.54 MeV kinetic energy.**