Consider the beta decay
`^198 Au rarr ^198 Hg ** + Beta^(-1) + vec v`.
where `^198 Hg^**` represents a mercury nucleus in an excited state at energy `1.088 MeV` above the ground state. What can be the maximum kinetic energy of the electron emitted? The atomic mass of `^198 Au` is `197.968233 u` and that of `^198 Hg` is `197.966760 u`.

Explanation

To find the maximum kinetic energy of the emitted electron during the beta decay of gold to mercury, we can follow these steps:

### Step 1: Understand the Beta Decay Process
The beta decay process can be represented as:
\[ ^{198}Au \rightarrow ^{198}Hg^* + \beta^{-} + \nu \]
where \( ^{198}Hg^* \) is the excited state of mercury, \( \beta^{-} \) is the emitted electron, and \( \nu \) is the anti-neutrino.

### Step 2: Identify the Masses
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