Calc Neutrino Energy Threshold for νe + ^71Ga → ^72Ge + e⁻

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Discussion Overview

The discussion revolves around the calculation of the energy threshold for the nuclear reaction involving an electron neutrino and gallium, specifically the reaction νe + 71Ga → 72Ge + e-. Participants explore how to determine the minimum energy required for the neutrino for this reaction to occur, focusing on the Q-value and the distinction between atomic and nuclear masses.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Homework-related

Main Points Raised

  • One participant inquires about the energy threshold of approximately 0.23 MeV and how to calculate the minimum energy of the neutrino for the reaction.
  • Another participant suggests calculating the Q-value of the reaction and mentions that the threshold energy is approximately 0.25 MeV, depending on the values used.
  • A participant calculates the energy threshold to be approximately 0.236 MeV using the Q-value formula and specific mass values.
  • One participant challenges the previous calculation, arguing that atomic masses should not be used since the reaction involves nuclear masses, and provides a revised calculation resulting in a threshold of about 0.235667 MeV.
  • Another participant expresses gratitude for the clarification regarding the inclusion of electron masses in the calculations.
  • A participant suggests looking for introductory nuclear physics books to aid understanding.
  • One participant seeks clarification on why nuclear masses are used instead of atomic masses in the calculations.
  • A later reply explains that only the nucleus participates in the reaction, which is why nuclear masses are relevant for the Q-value.

Areas of Agreement / Disagreement

Participants express differing views on the appropriate masses to use in the calculations, with some advocating for nuclear masses while others initially used atomic masses. The discussion remains unresolved regarding the implications of these choices on the energy threshold calculation.

Contextual Notes

Participants reference specific mass values and calculations, but there is no consensus on the best approach to determining the energy threshold, highlighting potential dependencies on definitions and assumptions regarding mass types.

Nakalay
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Hi. I am curious about the following nuclear reaction:

[tex]\nu_{e} + ^{71}_{31}Ga \rightarrow ^{71}_{32}Ge + e^{-}[/tex].

Can anyone explain to me why this reaction has an energy threshold of approximatly 0.23 MeV? How does one calculate the minimum energy of the neutrino for this reaction to take place?

Thanks in advance.
 
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Just calculate the Q-value of the reaction.

Is this homework?

If you do the calculation, you will see that the threshould energy is approx 0.25MeV (depending on how many/good numbers you have from your table, I just did)

If you have no idea what Q-value is, then I suggest you look it up and present your attempt to a solution before we proceed. This is very straightforward and these kinds of reactions are covered in all introductory nuclear physics books.
 
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All right, that was surprisingly easy. I found the energy threshold to be approximatly 0.236 MeV:

[tex]Q = -\Delta m\times c^{2} = (70.924954 u - 70.924701 u)\times931.494 \frac{MeV}{u} \approx 0.236 MeV[/tex]

Thanks for the help.
 
I think this is wrong.

Tabulated are the ATOMIC masses, i.e all electrons are included.'

Before the reaction you have 31 electrons, and after you have 32 + 1 = 33; so you must remove 2electron masses.

Now i calculate with the nuclear masses.

[tex]M( ^{71}_{31}Ga ) = 70.924701 - 31*0.0005486 = 70.9076944 u[/tex]


[tex]M( ^{71}_{32}Ge ) = 70.924954 - 32*0.0005486 = 70.9073988 u[/tex]

[tex]Q = M( ^{71}_{32}Ge ) + m_e - M( ^{71}_{32}Ge ) = 70.9073988 + 0.0005486 - 70.9076944 = 2.53*10^{-4} u = 0.235667 MeV[/tex]

Might not differ so much in this case, but the principle is imortant. In another reaction than this, it might make big difference.
 
Great, thanks again. I was actually wondering why I hadn't accounted for the electrons.
I've lost my physics book and I haven't been able to find much information elsewhere on the web so your help is much appreciated.
 
Too bad you have lost your book! =(
But have you looked at your library? There should be plenty of books about introductory nuclear physics. Remeber that even quite old books from the 1950-1960 is good for this.
 
I've found my book again so no worries. There is one thing I have not yet fully understood, though. Why is it that you have to use the nuclear masses instead of the atomic masses in the calculation?
 
Only the nucleus takes part in the reaction. The atomic electrons don't parcipitate.

This is reflected in the Q-value, which is defined for the nucleus mass.
 
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