Consider the β decay, Au^198 -> Hg^198*, where Hg^198* represents a mercury nucleus in an excited st — Nuclear Chemistry and Radioactivity Chemistry Question

💡 Solution & Explanation

Step 1 - Calculate Q-value for the Transition to Ground State of Hg-198 $$Q = [m(\text{Au}^{198}) - m(\text{Hg}^{198})] \times 931.5\ \text{MeV/u}$$ $$= (197.968 - 197.966) \times 931.5 = 0.002 \times 931.5 = 1.863\ \text{MeV}$$ Step 2 - Account for the Excited State The decay goes to $\ce{Hg^{198*}}$, an excited state 1.063 MeV above the ground state. The available energy for the kinetic energy of the beta particle and antineutrino: $$E_{\text{available}} = Q - E^* = 1.863 - 1.063 = 0.800\ \text{MeV}$$ Step 3 - Maximum Kinetic Energy of the Electron The maximum KE of the $e^-$ occurs when the antineutrino carries zero energy: $$KE_{\max}(e^-) = E_{\text{available}} = \boxed{0.8\ \text{MeV}}$$ Step 4 - Evaluate Options - **(A) 0.8 MeV**: Matches our calculation. **Correct.** - **(B) 1.863 MeV**: This is the total Q-value to the ground state, not accounting for the excited state energy. Incorrect. - **(C) 1.063 MeV**: This is the excitation energy of the Hg-198* state. Incorrect. - **(D) 1.0 MeV**: Incorrect. $$\boxed{\text{Answer: A — Maximum KE of electron} = 0.8\ \text{MeV}}$$