- #1
werekdells
- 4
- 0
Alpha Decay "extra" electrons
I am sure this question has been asked before, and I have searched many sources for a good answer, but have yet to be satisfied with what i have read.
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Question 1 deals with this version of the decay formula
238/92 U --> 234/90 Th + (alpha)
Its seems incomplete. There are 92 electrons on the left but only 90 on the right since alpha is just a nucleus.
Why is this reaction never written as:
238/92 U --> 234/90 Th + (alpha) +2 (1/0 e)
As I have read that there are 2 atomic electrons stripped that then get attached to another atom soon after, shouldn't they be accounted for in the original reaction and not just missing ?
When determining energy released, rather than using the alpha particle (He nucleus) mass, we use the full Helium atom mass to account for the two "extra" electrons, but why doesn't the reaction simply have 2 extra electrons as a product to tell the story more accurately
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My second question is basically the same one from a different view point
Often the reaction is written as:
238/92 U --> 234/90 Th + 4/2 He
why isn't it written
238/92 U --> 234/90 Th + 4/2 He (2+) + 2 (1/0 e)
Is the revised reaction I wrote not what actually happens? Traditionally the extra 2 electrons are lumped into the helius nucleus to simplify and everything balances, but it has always seemed to be skipping a step in the process?
=======
Maybe I am missing something entirely, thanks in advance for the clarification if you have it.
====
p.s. A similar question can be asked about beta decay
normally shown 14/6 C --> 14/7 N + Beta
why not ?? 14/6 C --> 14/7 N (1+) + Beta
I am sure this question has been asked before, and I have searched many sources for a good answer, but have yet to be satisfied with what i have read.
=====================================
Question 1 deals with this version of the decay formula
238/92 U --> 234/90 Th + (alpha)
Its seems incomplete. There are 92 electrons on the left but only 90 on the right since alpha is just a nucleus.
Why is this reaction never written as:
238/92 U --> 234/90 Th + (alpha) +2 (1/0 e)
As I have read that there are 2 atomic electrons stripped that then get attached to another atom soon after, shouldn't they be accounted for in the original reaction and not just missing ?
When determining energy released, rather than using the alpha particle (He nucleus) mass, we use the full Helium atom mass to account for the two "extra" electrons, but why doesn't the reaction simply have 2 extra electrons as a product to tell the story more accurately
========================
My second question is basically the same one from a different view point
Often the reaction is written as:
238/92 U --> 234/90 Th + 4/2 He
why isn't it written
238/92 U --> 234/90 Th + 4/2 He (2+) + 2 (1/0 e)
Is the revised reaction I wrote not what actually happens? Traditionally the extra 2 electrons are lumped into the helius nucleus to simplify and everything balances, but it has always seemed to be skipping a step in the process?
=======
Maybe I am missing something entirely, thanks in advance for the clarification if you have it.
====
p.s. A similar question can be asked about beta decay
normally shown 14/6 C --> 14/7 N + Beta
why not ?? 14/6 C --> 14/7 N (1+) + Beta