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:
The decay process for 239/94Pu is:
94Pu239 → 2He4 + 92U235 + ΔE
The following nuclei are observed to decay by emitting an α particle: 239/94Pu.
Write out the decay process for 239/94Pu.
Determine the energy released in this reaction.
Frequently Asked Questions
What scientific concept do you need to know in order to solve this problem?
Our tutors have indicated that to solve this problem you will need to apply the Nuclear Physics concept. You can view video lessons to learn Nuclear Physics. Or if you need more Nuclear Physics practice, you can also practice Nuclear Physics practice problems.