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Variation in mass and binding energy in nuclear reactions

  • #1
Krushnaraj Pandya
Gold Member
697
71

Homework Statement


I'm perpetually confused keeping track of the energetics of a nuclear reaction and I broke down my conceptual questions into the following parts
statement a-In a fission reaction, the two medium sized daughter nuclei each have more binding energy per nucleon than the original nucleus
statement b-In a fusion reaction the heavier nucleus has more binding energy per nucleon than the lighter ones

Which of these statements are correct and why, I know that mass of the reactants is always higher (the loss in mass is released as energy) and total binding energy of reactants is always lower (I can't visualize this very well) for a spontaneous reaction but I can't figure out the validity of the above statements and I'd really appreciate some help with that-thank you.

Homework Equations


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The Attempt at a Solution


Mentioned above
 

Answers and Replies

  • #2
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Both statements are correct for reactions that release energy. There are also reactions that do not. You can e.g. fuse heavy nuclei with an accelerator delivering the necessary energy.
and total binding energy of reactants is always lower (I can't visualize this very well)
The sign convention for binding energy is a bit unfortunate.

If in doubt you can always consider the total energy.
 
  • #3
Krushnaraj Pandya
Gold Member
697
71
Both statements are correct for reactions that release energy. There are also reactions that do not. You can e.g. fuse heavy nuclei with an accelerator delivering the necessary energy.The sign convention for binding energy is a bit unfortunate.

If in doubt you can always consider the total energy.
How can we use total energy without using the sign convention for binding energy?
 
  • #4
Krushnaraj Pandya
Gold Member
697
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Do you mean to say we'd use rest mass*c^2 of reactants+ KE due to any velocity =rest mass of products*c^2 +KE
 
  • #5
34,059
9,932
That always works, sure.
 
  • #6
Krushnaraj Pandya
Gold Member
697
71
That always works, sure.
Alright, thank you very much :D
 

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