Consider the following nuclear fission reaction
`._(88)Ra^(226) to ._(86)Rn^(222)+._(2)He^(4)+Q`
In the fission reaction. Kinetic energy of `alpha`-particle is 4.44 MeV. Find the energy emitted as `gamma`-radiation in keV in this reaction.
`m(._(88)Ra^(226))=226.005` amu
`m(._(86)Rn^(222))=222.000` amu

Explanation

To solve the problem, we need to find the energy emitted as gamma radiation in the given nuclear fission reaction. We will follow these steps:

### Step 1: Identify the masses involved
We have the following masses:
- Mass of Radium-226 (Ra): \( m(\text{Ra}) = 226.005 \, \text{amu} \)
- Mass of Radon-222 (Rn): \( m(\text{Rn}) = 222.000 \, \text{amu} \)
- Mass of Alpha particle (He): \( m(\alpha) = 4.000 \, \text{amu} \) (This is a standard value for the alpha particle)

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