# Problem: The following nuclei are observed to decay by emitting an α particle: 239/94Pu.Part AWrite out the decay process for 239/94Pu.Part BDetermine the energy released in this reaction.

###### FREE Expert Solution

For an α-decay, the mass number of the nucleus is always reduced by 4 and the charge number reduced by 2 for the given nucleus.

The energy released in the reaction:

$\overline{){\mathbf{∆}}{\mathbf{E}}{\mathbf{=}}{\mathbf{\left(}}{\mathbf{∆}}{\mathbf{m}}{\mathbf{\right)}}{\mathbf{\left(}}{\mathbf{931}}{\mathbf{.}}{\mathbf{5}}{\mathbf{M}}{\mathbf{e}}{\mathbf{V}}{\mathbf{\right)}}}$

Part A

The decay process for 239/94Pu is:

94Pu2392He4 + 92U235 + ΔE

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###### Problem Details

The following nuclei are observed to decay by emitting an α particle: 239/94Pu.

Part A

Write out the decay process for 239/94Pu.

Part B

Determine the energy released in this reaction.