How Does Carbon-12 Have a Lower Rest Energy Than Its Separate Nucleons?

Join the discussion
Ask a follow-up here, or get your own question answered by working scientists, mathematicians and engineers — people, not an autocomplete.
Real named experts · corrections over time · the nuance an AI answer skips
3 replies · 3K views
fridakahlo
Messages
10
Reaction score
1

Homework Statement


"You might wonder how six protons and six neutrons, each having a mass larger than 1 u, can be combined with six electrons to form a carbon-12 atom having a mass of exactly 12 u. The bound system of carbon-12 has a lower rest energy than that of six separate protons and six separate neutrons. According to E=mc^2, this lower rest energy corresponds to a smaller mass for the bound system. The difference in mass accounts for the binding energy when the particles are combined to form the nucleus."

Homework Equations


E=mc^2
1amu= 1.660539e-27 kg

The Attempt at a Solution


I came across this passage in my textbook, and I'm having difficulty in seeing how the rest energy for carbon-12 can be lower than that of six separate protons or six separate neutrons. My understanding is that the more mass the greater the rest energy.
 
on Phys.org
As a little hint...
To solve your confusion on how the total mass can be equal to 12u, when the protons are bonded, some of their mass is expended as energy for the bonding, resulting in the 12u.
:smile:
 
  • Like
Likes   Reactions: fridakahlo
Thank you! (: I just finished reading up on nuclear binding energy, and it has become a lot clearer now.
 
  • Like
Likes   Reactions: Uranic_Wabbit
fridakahlo said:
Thank you! (: I just finished reading up on nuclear binding energy, and it has become a lot clearer now.
Great Job! :smile:
Always to be sure to check your textbook for topics that are somewhat confusing or which you are not particularly sure on, and if that doesn't work it is always fine to look up the term itself on the browser. Heads up, make sure not to simply look up the solution until you have an answer you are somewhat confident about. Not doing this results in all your previous work on the problem turning obsolete.
:wink: