Find the kinetic energy of the `alpha` - particle emitted in the decay `^238 Pu rarr ^234 U + alpha`. The atomic masses needed are as following:
`^238 Pu 238.04955 u`
`^234 U 234.04095 u`
`^4 He 4.002603 u`.
Neglect any recoil of the residual nucleus.

Explanation

To find the kinetic energy of the alpha particle emitted in the decay of \( ^{238}Pu \) to \( ^{234}U + \alpha \), we will follow these steps:

### Step 1: Identify the masses involved in the decay.
We need the atomic masses of the isotopes involved:
- Mass of \( ^{238}Pu = 238.04955 \, u \)
- Mass of \( ^{234}U = 234.04095 \, u \)
- Mass of \( ^{4}He \) (alpha particle) = \( 4.002603 \, u \)

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