Gold `._(79)^(198)Au` undergoes `beta^(-)` decay to an excited state of `._(80)^(198)Hg`. If the excited state decays by emission of a `gamma`-photon with energy `0.412 MeV`, the maximum kinetic energy of the electron emitted in the decay is (This maximum occurs when the antineutrino has negligble energy. The recoil enregy of the `._(80)^(198)Hg` nucleus can be ignored. The masses of the neutral atoms in their ground states are `197.968255 u` for `._(79)^(198)Hg`).

Options

A. `0.412 MeV`
B. `1.371 MeV`
C. `0.959 MeV`
D. `1.473 MeV`

Explanation

To solve the problem step by step, we will follow these steps:

### Step 1: Identify the Masses
We need to identify the masses of the involved isotopes. The mass of Gold-198 (Au) is given as:
- Mass of \( _{79}^{198}Au \) = 197.968255 u

The mass of the neutral atom of Mercury-198 (Hg) in its ground state is not provided, but we can assume it is:
- Mass of \( _{80}^{198}Hg \) = 197.966752 u (as mentioned in the transcript).

### Step 2: Calculate the Mass Defect
The mass defect \( \Delta m \) can be calculated using the formula:
\[ \Delta m = \text{mass of reactants} - \text{mass of products} \]
In this case, the reactant is Gold-198 and the product is the excited state of Mercury-198. The mass defect is:
\[ \Delta m = \text{mass of } _{79}^{198}Au - \text{mass of } _{80}^{198}Hg \]
Substituting the values:
\[ \Delta m = 197.968255 \, \text{u} - 197.966752 \, \text{u} = 0.001473 \, \text{u} \]

### Step 3: Convert Mass Defect to Energy
To find the energy equivalent of the mass defect, we use the equation:
\[ E = \Delta m \cdot c^2 \]
Using the conversion factor \( 1 \, \text{u} = 931.5 \, \text{MeV} \):
\[ E = 0.001473 \, \text{u} \times 931.5 \, \text{MeV/u} = 1.372 \, \text{MeV} \]

### Step 4: Account for Gamma Emission
The excited state of Mercury-198 emits a gamma photon with energy:
\[ E_{\gamma} = 0.412 \, \text{MeV} \]
To find the maximum kinetic energy of the emitted electron during the beta decay, we subtract the energy of the gamma photon from the total energy calculated from the mass defect:
\[ K_{\text{max}} = E - E_{\gamma} \]
Substituting the values:
\[ K_{\text{max}} = 1.372 \, \text{MeV} - 0.412 \, \text{MeV} = 0.960 \, \text{MeV} \]

### Final Answer
Thus, the maximum kinetic energy of the emitted electron is:
\[ K_{\text{max}} \approx 0.959 \, \text{MeV} \]