Conservation of charge in a nuclear reaction

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    Charge Nuclear Reaction
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Discussion Overview

The discussion revolves around the conservation of charge in nuclear reactions, specifically examining a reaction involving neutrons and helium nuclei. Participants explore the implications of including or excluding electrons in charge calculations and how this affects the understanding of nuclear processes.

Discussion Character

  • Debate/contested
  • Conceptual clarification
  • Technical explanation

Main Points Raised

  • One participant presents a specific nuclear reaction and questions the charge balance, noting a potential misunderstanding regarding whether a proton or an atom of hydrogen is produced.
  • Several participants assert that electrons are typically excluded from nuclear reaction considerations, emphasizing that nuclear reactions focus on nuclear charges.
  • Another participant references beta decay as an example where electrons are included in charge conservation, suggesting a distinction in treatment between different types of nuclear processes.
  • There is a discussion about whether charge conservation should always consider electrons, with some participants arguing that it depends on whether new electrons are produced or destroyed in the reaction.
  • One participant expresses confusion about the use of superscripts in notation, indicating a need for clarity in communication.

Areas of Agreement / Disagreement

Participants generally disagree on whether electrons should be included in charge conservation for all nuclear reactions, with some asserting that they are irrelevant unless produced or destroyed, while others argue that they should be considered in certain contexts, such as beta decay.

Contextual Notes

Participants note the importance of notation clarity, particularly regarding superscripts in chemical symbols, which may lead to misunderstandings if not used correctly.

eneacasucci
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Conservation of charge: The sum of the charges on all the particles before and after a reaction are the same

I was considering this reaction:
n + 3 He → 3 H + p + 0,764 MeV

Making the charge balance I have:
0 + 0 (2protons + 2 electrons) → 0 (1proton + 1electron) + 1 (proton)

There is something really fundamental that I'm missing...
I've read in other references that the reaction is also written as
n + 3 He → 3 H + 1 H + 0,764 MeV
So may it be that the product is not a proton but an atom of hydrogen 1H? (but I found this reaction in a neutron detector, that is a proportional counter that needs the ionising charged particle, so it should be a proton...)
 
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Electrons are not included. These are nuclear reactions.

Also, please use superscripts where appropriate, ie, 3He rather than 3He. The latter may be misunderstood as three He.
 
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We typically ignore electrons when talking about nuclear reactions, as they play virtually no part in them.
 
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I thought we considered electrons such as in nuclear decays where for example a beta- decay:
neutron ##\rightarrow## proton + electron + antineutrino
where in this case for charge conservation we count also the electron.

Can we conclude then that about charge conservation law in nuclear decays we consider the charge of the electrons while in other nuclear reactions we just consider the protons? I didn't read it anywhere so I wasn't aware about it, thank you both.
 
Orodruin said:
Also, please use superscripts where appropriate, ie, 3He rather than 3He. The latter may be misunderstood as three He.
I agree with you but I read the LaTex Guide and I didn't find how to put the superscript, but I just noticed that it is included in the tool-bar. I'll edit it
 
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eneacasucci said:
neutron → proton + electron + antineutrino
where in this case for charge conservation we count also the electron.
This is different. The electron is explicitly included.
 
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eneacasucci said:
neutron → proton + electron + antineutrino
It's a new electron that is produced in the nuclear reaction here. It's relevant for the process.
In the example in post 1 the reaction does not produce or destroy an electron. The helium nucleus may or may not have electrons, we can't tell and it doesn't matter. Same for the hydrogen in the final state.
 
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mfb said:
It's a new electron that is produced in the nuclear reaction here. It's relevant for the process.
In the example in post 1 the reaction does not produce or destroy an electron. The helium nucleus may or may not have electrons, we can't tell and it doesn't matter. Same for the hydrogen in the final state.
OK thank you, so unless a new electron is produced, I only carry out the charge balance on nuclear charges.
 
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eneacasucci said:
unless a new electron is produced
… or destroyed!
Consider electron capture.
 
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