# Decay of Cr

## Main Question or Discussion Point

One more question (I'm wary I'm totally spamming the forum!).. its in reference to the decay of Cr to V, see attached files.

The decay is clearly beta plus decay.

p->n+positron+antineutrino

The question asks to determine the neutrino spectrum. Now the energy of the neutrino is simply the mass difference between the {original state} and {the GS of the final state PLUS the rest mass energy of a positron}.

Why on earth is the question talking about a K shell electron? I also attach the solution to the problem. It doesn't take into account mass of the positron when calculating the energy of the neutrino: Why?

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## Answers and Replies

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Staff Emeritus
2019 Award
Since it's a homework problem, I'll only give you a hint.

It doesn't take into account mass of the positron when calculating the energy of the neutrino: Why?
Hint: What positron?

Second hint: reread some of the other threads you've started.

Since it's a homework problem, I'll only give you a hint.

Hint: What positron?

Second hint: reread some of the other threads you've started.
well you said in the other thread that the positron is very short lived (I take it undergoes electron positron anhiliation?).

Therefore I would think this energy radiates away to the surroundings - ie. it is not captured by the neutrino...

I don't really get it to be honest:(

EDIT: ALSO according to the solutions (the second file), the K shell energy of the original particle has been subtracted from the difference in masses. Why is a K shell electron liberated in this decay... a positron should be produced, not an electron liberated.

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Staff Emeritus
2019 Award
Let me give you the hint again. What positron? Point to the positron in the diagram.

If you need another hint, look up the decay modes of Cr-51.

vanesch
Staff Emeritus
Gold Member
The decay is clearly beta plus decay.
No, it isn't. Look at the NNDC website if needed (the chart of nuclides).
http://www.nndc.bnl.gov/chart/

aaah I see, its K shell capture.

So V captures a K shell electron. Another stupid question, why is it K shell capture? V already has a (two) K shell electron(s); do you not need an x ray photon to knock of one of these in the first place? It would seem a lot easier for the electron to tag onto on of the valence shells, no?

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Astronuc
Staff Emeritus