Calculating the Neutrino Spectrum for Cr Decay to V

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Discussion Overview

The discussion centers around the decay of chromium (Cr) to vanadium (V), specifically focusing on the calculation of the neutrino spectrum associated with this decay process. Participants explore the nature of the decay, the role of the positron, and the implications of K shell electron capture.

Discussion Character

  • Homework-related
  • Debate/contested
  • Exploratory

Main Points Raised

  • One participant asserts that the decay is a beta plus decay, involving a proton transforming into a neutron, positron, and antineutrino.
  • Another participant questions the relevance of the positron in the context of the neutrino spectrum calculation, suggesting that the positron's short-lived nature means its energy does not contribute to the neutrino.
  • There is confusion regarding the mention of a K shell electron in the problem, with participants debating why a K shell electron is liberated during the decay process.
  • Some participants clarify that the decay involves K shell capture, leading to a transformation where a K shell electron is captured by V, and a proton is converted to a neutron.
  • Questions arise about the necessity of an X-ray photon to remove a K shell electron, given that V already possesses K shell electrons.
  • Participants discuss the binding energy of K shell electrons and their probability of interacting with the nucleus compared to valence shell electrons.

Areas of Agreement / Disagreement

Participants express differing views on the nature of the decay process, with some asserting it is beta plus decay while others suggest it is K shell capture. The discussion remains unresolved regarding the implications of the positron's role and the specifics of K shell electron behavior.

Contextual Notes

There are unresolved questions about the assumptions regarding the positron's energy contribution and the mechanics of K shell capture versus beta decay. The discussion reflects varying interpretations of the decay modes and the associated energy calculations.

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

vertices said:
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.
 
Vanadium 50 said:
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.
 
Last edited:
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.
 
vertices said:
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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vertices said:
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?
Cr captures K-electron and a proton is transformed to a neutron (A remains unchanged, but Z decreases by 1).

All elements from He on have 2 K-electrons, and they are most tightly bound to the atom/nucleus. QM-wise, K electrons have a greater probability of interacting with the nucleus. An ionizing X-ray would be one that 'lifts' the electron up out of the potential well.
 

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