Decay of Cr

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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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  • #2
Vanadium 50
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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.
 
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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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Vanadium 50
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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.
 
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vanesch
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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
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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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