A-level physics energy-mass equivalence question

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localmemelord
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"the mass of a helium atom is slightly lower than the mass of the constituent parts. The binding energy has increased and therefore the mass must decrease to obey mass-energy conservation."
Why does mass decrease when BE increases when BE is directly proportional to mass defect?
Shouldnt it increase?
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Kian Stevens
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The binding energy becomes E = \Delta mc^2, where \Delta m is the difference in mass between the total mass of each individual nucleon and the mass of the atom

It's important to note that this energy is lost to the surroundings during the 'binding' process, so I guess 'binding energy' is quite a non-intuitive term
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localmemelord
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(Original post by Kian Stevens)
The binding energy becomes E = \Delta mc^2, where \Delta m is the difference in mass between the total mass of each individual nucleon and the mass of the atom

It's important to note that this energy is lost to the surroundings during the 'binding' process, so I guess 'binding energy' is quite a non-intuitive term
i understand this but in the quote for which is the mass decreasing, the nucleus or the nucleons.
i think its the nucleus since its separated nucleons have more BE energy so they have more mass due to mass-energy equivalence and so the mass of the nucleus decreases.
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Kian Stevens
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(Original post by localmemelord)
i understand this but in the quote for which is the mass decreasing, the nucleus or the nucleons.
i think its the nucleus since its separated nucleons have more BE energy so they have more mass due to mass-energy equivalence and so the mass of the nucleus decreases.
It's the mass of the nucleons that decreases, as it's these which form the nucleus (and the knock-on effect is that the resulting nucleus is of a lower mass)

I mean, logically speaking, the nucleus wasn't initially there... So how can the mass of it be decreased?
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localmemelord
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(Original post by Kian Stevens)
It's the mass of the nucleons that decreases, as it's these which form the nucleus (and the knock-on effect is that the resulting nucleus is of a lower mass)

I mean, logically speaking, the nucleus wasn't initially there... So how can the mass of it be decreased?
in your previous comment, you said "this energy is lost to the surroundings during the 'binding' process", is this energy equal to the BE?
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Kian Stevens
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(Original post by localmemelord)
in your previous comment, you said "this energy is lost to the surroundings during the 'binding' process", is this energy equal to the BE?
Correct, this energy is the binding energy

Since the binding energy is the energy required to separate a nucleus into its individual nucleons, then just from conservation of energy alone this must mean that the energy released to the surroundings during the 'binding' process is equivalent to the energy required to 'unbind' the nucleus (the binding energy)
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localmemelord
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(Original post by Kian Stevens)
Correct, this energy is the binding energy

Since the binding energy is the energy required to separate a nucleus into its individual nucleons, then just from conservation of energy alone this must mean that the energy released to the surroundings during the 'binding' process is equivalent to the energy required to 'unbind' the nucleus (the binding energy)
youre a god, thanks i understand it now.
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Kian Stevens
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(Original post by localmemelord)
youre a god, thanks i understand it now.
No worries!
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Eimmanuel
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(Original post by Kian Stevens)
The binding energy becomes E = \Delta mc^2, where \Delta m is the difference in mass between the total mass of each individual nucleon and the mass of the atom

It's important to note that this energy is lost to the surroundings during the 'binding' process, so I guess 'binding energy' is quite a non-intuitive term
.....difference in mass between the total mass of each individual nucleon and the mass of the atom.
If you want to use the word nucleon, then the mass of "atom" is incorrect. A common mistake.
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