Consider the beta decay, , where represents a mercury nucleus in an excited state at energy 1.063 Me — Nuclear Chemistry and Radioactivity Chemistry Question
Question
Consider the beta decay, $\text{Au}^{198} \to \text{Hg}^{198*}$, where $\text{Hg}^{198*}$ represents a mercury nucleus in an excited state at energy 1.063 MeV above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic masses of $\text{Au}^{198}$ and $\text{Hg}^{198}$ are 197.968 u and 197.966 u, respectively. (1 u = 931.5 MeV)
Answer: A
💡 Solution & Explanation
The total energy released (Q-value) is $\Delta m \cdot c^2 = (197.968 - 197.966) \text{ u} \cdot 931.5 \text{ MeV/u} = 1.863 \text{ MeV}$. Since the nucleus is left in an excited state of 1.063 MeV, the maximum kinetic energy available for the electron is $1.863 - 1.063 = 0.8 \text{ MeV}$. Therefore, correct answer is A.
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