Binding energy and favorable reactions

AI Thread Summary
The discussion centers on determining the energetics of the reaction 2H + 4He -> 6Li by analyzing binding energies. Participants emphasize the importance of using the binding energy per nucleon curve and calculating the combined binding energies for accurate assessment. It is concluded that the reaction is energetically favorable since the total binding energy of the reactants is less than that of the product. Additionally, the need for an additional photon to conserve energy and momentum in the reaction is mentioned. Overall, the conversation highlights the significance of binding energy calculations in nuclear reactions.
letec
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Hi everyone!
I'm currently learning nuclear physics and I have a question:
Watching the binding energy per nucleon curve, is the following reaction energetically favorable?
2H + 4He -> 6Li
And why?
500px-Binding_energy_curve_-_common_isotopes.svg.png

Thanks!
 
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Why don't you take the values from the diagram and check for yourself? Better (more accurate): Look up the binding energies (or the masses) and calculate it.
 
mfb said:
Why don't you take the values from the diagram and check for yourself? Better (more accurate): Look up the binding energies (or the masses) and calculate it.
The question asks to answer according to the curve :\
 
I moved you to the homework section.

Well, then use the diagram and check it.
 
mfb said:
I moved you to the homework section.

Well, then use the diagram and check it.
That's exactly my question... The binding energy of the 4He is bigger than the B.E of the product. Is that it? or is there something more?
 
What about the binding energy of deuterium?
 
mfb said:
What about the binding energy of deuterium?
well it's smaller! and that is my question!
 
No, that is the homework question.

What is the combined binding energy of D+4He?
 
mfb said:
No, that is the homework question.

What is the combined binding energy of D+4He?
do you mean for 2H + 4He -> 6Li:
2*1 MeV + 4*7MeV = 30MeV < 6*5.3 = 31.8 therefore the reaction is energetically favorable?
 
  • #10
Sure.

To conserve energy and momentum, you would need an additional photon in the final state.
 
  • #11
mfb said:
Sure.

To conserve energy and momentum, you would need an additional photon in the final state.
Thanks for the help! :)
 
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