# Question about nuclear fission/fusion and binding energy

So, i know that during nuclear fission, a heavy particle is split into two smaller ones that have a higher specific binding energy. And i know that a certain amount of mass is transformed to energy for the binding. The same thing happens during nuclear fusion.

However, from my intuitive (and probably wrong) point of view, one has to put energy INTO the atom in order to make one with a higher binding energy.

My reasoning is this: binding energy = the amount of energy to build the atom from scratch (i.e. moving the nucleons piece by piece from infinity to the fm-sized shell). So, if the binding energy is higher, more energy is required to put the atom together or rip it apart. So, wouldn't it make sense to go from a high binding energy state to a lower one, so as to RELEASE this binding energy?

It just doesn't make sense to me that energy is RELEASED to go to a high bound state. For instance, in atomic physics, an electron releases energy (an x-ray) when going to a lower binding energy-state.

I'm a bit confused despite having understood this for years in the past.... can anyone help?

## Answers and Replies

f95toli
Science Advisor
Gold Member
All your statements are correct except the first one. If the final energy for the created nucleus is higher than the total energy of the initial nuclei you do indeed need to add energy; and this is precisely what happens if you try to create an element with more protons than iron.
However, for some elements lighter than iron the total energy (mass+binding energy) of the created element (e.g. helium) can be SMALLER than that of the starting elements (e.g. deuterium); the energy coming out of a fusion process is just the "spare" energy.

I suspect you are implicitly assuming that heavier elements "contain" more binding energy than lighter elements but this is only true for heavy elements (beyond iron); the total energy of helium is smaller than that of two deuterium.

I suspect you are implicitly assuming that heavier elements "contain" more binding energy than lighter elements but this is only true for heavy elements (beyond iron); the total energy of helium is smaller than that of two deuterium.

Well, from my memory, the element Helium is actually lighter than the sum of two deuterons, and the difference in mass is the binding energy. However, my question remains: intuition tells me that going from two deuterons to one helium is an endothermic reaction because it requires some form of energy to convert that amount of mass into (binding) energy.

I know this is not the way things work, but what mechanism causes some mass to be converted into released energy? I know it's not gravity and not the weak interaction. Electromagnetic force seems to be to weak for these large amounts of energy on that scale, so left is the strong force?

f95toli
Science Advisor
Gold Member
intuition tells me that going from two deuterons to one helium is an endothermic reaction because it requires some form of energy to convert that amount of mass into (binding) energy.
No, you DO need some initial energy to overcome the potential barrier that is due to the Coulomb force between the nuclei, but the net reaction is exothermic (otherwise fusion wouldn't work).

But what mechanism causes some mass to be converted into released energy?

As far as I understand there is no need for a mechanisms ; the difference in mass is DUE to the difference in binding energy; it is quite literally the energy itself that has a "weight".
Remember that SR not only tells that we can convert mass to energy; it also tells us that they are equivalent. Now, I can't claim to know much about on general relativity (I work in condensed matter/device physics); but as far as I understand the equivalence is quite clear there as well; mass and energy both have the same effect on space-time with a "conversion factor" equal to c^2.

Hence, all nuclei have a total mass given by the mass of the protons&neutrons + the binding energy divided by c^2

So, i know that during nuclear fission, a heavy particle is split into two smaller ones that have a higher specific binding energy. And i know that a certain amount of mass is transformed to energy for the binding. The same thing happens during nuclear fusion.

However, from my intuitive (and probably wrong) point of view, one has to put energy INTO the atom in order to make one with a higher binding energy.

My reasoning is this: binding energy = the amount of energy to build the atom from scratch (i.e. moving the nucleons piece by piece from infinity to the fm-sized shell). So, if the binding energy is higher, more energy is required to put the atom together or rip it apart. So, wouldn't it make sense to go from a high binding energy state to a lower one, so as to RELEASE this binding energy?

It just doesn't make sense to me that energy is RELEASED to go to a high bound state. For instance, in atomic physics, an electron releases energy (an x-ray) when going to a lower binding energy-state.

I'm a bit confused despite having understood this for years in the past.... can anyone help?
Hey I think you need to be introduced with the concept of “binding energy per nucleon” (B.N).You know this (B.N)=(B.E)/A where B.E represents binding energy of whole nucleus and A represents total number of nucleons .
Here when the Uranium atom is bombarded with the high speed neutrons it break into Barium and Krypton .I have looked the Binding energy per curve which tells that these Barium and Krypton have more (B,N) that the of the Uranium nucleus .This is not that the total binding energy of nucleus of the Barium or Krypton is greater than the uranium .It happens due to the less number of the nucleons of the Barium and Krypton nucleus.
Please notice the formulae I have given to figure out I have told.
There is less number of nucleons such that the total binding energy of the nucleus of the Barium or Krypton when Divided to the total number of nucleons ,each nucleons receive more binding energy than the each nucleons in the Uranium.Thats the way , despite the energy released in the uranium fission, the Barium or krypton nucleus have high bound state
Beside you mentioned the X ray production phenomenon. Please note that when speedy electrons strike the atom the metal ,the atom rises to the excitation potential and then releases the energy when returning the ground state . So they are different phenomenon.So you cannot link this to nuclear fission.

Last edited:
Hey I think you need to be introduced with the concept of “binding energy per nucleon” (B.N).You know this (B.N)=(B.E)/A where B.E represents binding energy of whole nucleus and A represents total number of nucleons .
Here when the Uranium atom is bombarded with the high speed neutrons it break into Barium and Krypton .I have looked the Binding energy per curve which tells that these Barium and Krypton have more (B,N) that the of the Uranium nucleus .This is not that the total binding energy of nucleus of the Barium or Krypton is greater than the uranium .It happens due to the less number of the nucleons of the Barium and Krypton nucleus.
Please notice the formulae I have given to figure out I have told.
There is less number of nucleons such that the total binding energy of the nucleus of the Barium or Krypton when Divided to the total number of nucleons ,each nucleons receive more binding energy than the each nucleons in the Uranium.Thats the way , despite the energy released in the uranium fission, the Barium or krypton nucleus have high bound state
Beside you mentioned the X ray production phenomenon. Please note that when speedy electrons strike the atom the metal ,the atom rises to the excitation potential and then releases the energy when returning the ground state . So they are different phenomenon.So you cannot link this to nuclear fission.

Thanks for the reply. I still don't quite understand, however.

Let me put it simple, in terms of nuclear fusion.

The net equation is: deuteron + deuteron -> helium + garbageload of energy.

The deuteron has fewer B.E. PER NUCLEON, which i will call BE/A.
Helium has more BE/A.

So, before the reaction, you have:

deuteron (2 nucleons) low BE/A
deuteron (2 nucleons) low BE/A
So, in total:

4 nucleons with low BE/A.

Then after the reaction, you have:

Helium (4 nucleons), high BE/A.

In other words, you go from 4 nucleons with low total binding energy to 4 nucleons with high binding energy. (Since the total number of nucleons is same before as after, we can drop the binding energy "per nucleon").

So how can energy be RELEASED when the binding energy is released? My thought is that it requires energy to increase binding energy, in any system.

Thanks for the reply. I still don't quite understand, however.

Let me put it simple, in terms of nuclear fusion.

The net equation is: deuteron + deuteron -> helium + garbageload of energy.

The deuteron has fewer B.E. PER NUCLEON, which i will call BE/A.
Helium has more BE/A.

So, before the reaction, you have:

deuteron (2 nucleons) low BE/A
deuteron (2 nucleons) low BE/A
So, in total:

4 nucleons with low BE/A.

Then after the reaction, you have:

Helium (4 nucleons), high BE/A.

In other words, you go from 4 nucleons with low total binding energy to 4 nucleons with high binding energy. (Since the total number of nucleons is same before as after, we can drop the binding energy "per nucleon").

So how can energy be RELEASED when the binding energy is released? My thought is that it requires energy to increase binding energy, in any system.

you Know that when two protons is made to come closer then by our knowledge we can say that these protons face electrostatic repulsion.To overcome this electrostatic repulsion external energy that usually comes as force have to applied to overcome the electrostatic repulsion barrier,surely this can cancel the electrostatic repulsion(opposing force effect.But it is useful to know that the energy is conserved whether force is cancelled.And note that these energy appears as the nuclear force which binds the nucleons and surely which can sense that the total binding energy has increased and therefore the binding energy per nucleon supporting the stable nuclei (as u meant as high bound state)
you may also be confused because the energy is released but it is useful to know that these energy comes due to mass lost (E=mc^2)
Note that the Gravitational force is negligible in comparision to electostsat and nuclear force
I also hope for your response,please reply.

you Know that when two protons is made to come closer then by our knowledge we can say that these protons face electrostatic repulsion.To overcome this electrostatic repulsion external energy that usually comes as force have to applied to overcome the electrostatic repulsion barrier,surely this can cancel the electrostatic repulsion(opposing force effect.But it is useful to know that the energy is conserved whether force is cancelled.And note that these energy appears as the nuclear force which binds the nucleons and surely which can sense that the total binding energy has increased and therefore the binding energy per nucleon supporting the stable nuclei (as u meant as high bound state)
you may also be confused because the energy is released but it is useful to know that these energy comes due to mass lost (E=mc^2)
Note that the Gravitational force is negligible in comparision to electostsat and nuclear force
I also hope for your response,please reply.

Yes, i know there is a good amount of kinetic energy needed to overcome the Coulomb repulsion. 100 million degrees will do.

But my question remains: how can you go from a lower bound state to a higher bound state under the RELEASE of energy? In any other physical system, you need to insert energy to go to a higher bound state. If you climb a mountain, you need to put energy into it. If you want to put an electron in a higher bound state, you need to put a photon (energy) into it. If you want to go up the Harmonic Oscillator ladder, you require energy. You get the point... why is energy released upon going to a higher bound state when two deuterons form one Helium atom?

Hi there,

It is all a matter of terminology. In "binding energy", one means the energy that binds the nuclei together. Therefore, this energy is a well-type potential energy. This means that you need to give energy to the nuclei so they can escape the attractive potential. Thus, this binding potential energy is less than 0.

Cheers

vanesch
Staff Emeritus
Science Advisor
Gold Member
So how can energy be RELEASED when the binding energy is released? My thought is that it requires energy to increase binding energy, in any system.

It's the name that confuses you. "Binding energy" is a negative energy. It is "missing energy".

The binding energy of a bound system is the energy the system is *missing* to be a system with its components separated.

If a hydrogen atom has a binding energy of 13.6 eV, that means that the actual energy of the bound system, as compared to a free electron and proton, is actually -13.6 eV.

A system with "high binding energy" is strongly bound, because it REQUIRES a lot of energy to break it.

It's clear to me now. Thanks a lot, guys. 