# In the formation of bonds, is Binding Energy absorbed or released

1. Mar 14, 2014

### fandi.bataineh

i think i dont understand the nature of "Binding Energy" very well, and i have a very significant question
what iam really intersted in; is the nuclear binding energy, but i think (hopefuly iam right) that it has the same physical nature (NOT origin) as atomic and molecular binding energies.

first; i will try to put everything i know about this (nuclear binding energy) together in a simple example, if iam wrong in any way please correct me.

here's my example:
in the very simple case; when 2 -free or unbound- nucleons bind together to form a nuleus (like the case when 2 hydrogen nuclei -NOT deuterium or tritium, a hydrogen atom is a single proton- are fused to make a helium nucleus) they release some energy while they (the nucleons) switch to a higher binding energy state (higher binding energy per nucleon), and this energy can be calculated diretly from Einstein's famous equation E=m.c^2 if we know how much of the whole system's mass was lost -converted to energy- in the reaction.

what i dont understand is: why this amount of energy is released rather than being absorbed?
i know that my question might seem quite confusing, but as i think of it; switching from one energy (binding energy) state to another with HIGHER energy would require absorbing energy.
lets think of gravitational potential energy as an example; if you want to move an object to a higher-gravitational-potential-energy state; you have to do work on it, that is, the object must gain -or absorb- energy in the form of work done on it. we know that the strong nuclear force -which is responsible for keeping nucleons together in all nuclei- is an attractive force, just like the gravitational force, with huge differences in strength and range between the two fundamental forces. so why we dont work on nucleons in order to make them switch to higher energy states?
REMARK: all work done on protons -or hydrogen nuclei- to make them fuse to form larger nuclei; is acctually needed to overcome Coulomb's repulsive force between them before they enter the range of dominance of the strong nuclear force, this is why a huge amount of energy is needed to start a fusion reaction, like the hydrogen bomb.

or did i miss some fundamental conception here; maybe higher-BINDING-energy-state is different from higher-POTENTIAL-energy-state, or something like this

2. Mar 14, 2014

### Bill_K

Binding energy is the negative of the potential energy. For a bound state the potential energy is negative. By convention we talk about a positive quantity instead, the binding energy. Binding energy is defined to be the amount of energy that would need to be supplied to break the nucleus into its constituent protons and neutrons.

3. Mar 14, 2014

### fandi.bataineh

@Bill_K, may you please explain further; and consider the following arguments:

"Binding energy is the negative of the potential energy"
do you mean that the sum of the two energies is equal to zero?
and consequently they have the same magnitude?
so they would be 100% corelated to each other and it does not make any sense to have both of them physically defined because dealing with one of them would be sufficient for all purposes.

"For a bound state the potential energy is negative"
lets think of this example: in a simple sense, rockets -for example- on earth are in the range of its gravitational field, hence they are gravitationally bound to it (the earth), and they need to gain certain amount of energy -binding energy- to break this bond, which is the kinetic energy they gain when they are launched with certain velocity (the escape velocity), and this velocity is calculated to be around 11.3 km/s for rockets -or any other object- to be launched from earth's surface, that is a distance R (earths mean radius) from earths center of mass, in order for them to escape the gravitational field -or break the gravitational bond-, although other bound states do exist for any given altitude. anyways, what i want to say is that; these objects -rockets- possess POSITIVE gravitational potential energy, and are still bound to the earth.

Last edited: Mar 14, 2014
4. Mar 14, 2014

### MikeGomez

They don’t. Potential energy is relative, and you can assign any value you want. In your examples it might make more sense to say that the gravitational potential at the surface of the earth (your rocket) is zero. Then the gravitational binding energy is negative, the potential gravitational energy at the surface of the earth is zero, and in order to escape the gravity of the earth you would need to apply enough kinetic energy to overcome the binding energy. Then, after you have overcome the binding energy and the rocket is in space, your rocket has positive potential gravitational energy (negative of the binding).

5. Mar 14, 2014

### fandi.bataineh

@MikeGomez

"In your examples it might make more sense to say that the gravitational potential at the surface of the earth (your rocket) is zero. Then the gravitational binding energy is negative, ..."

"Then, after you have overcome the binding energy and the rocket is in space, your rocket has positive potential gravitational energy (negative of the binding)."

is there any relation between potential and binding energies?
and if the relation between these two conceptions does exist; what exactly is physical basis and nature of this relation? i mean; upon what assumption(s) this relation is based?

Last edited: Mar 14, 2014
6. Mar 14, 2014

### Staff: Mentor

Binding energy is the energy that is released when a previously unbound particle gives up some of its potential energy and becomes bound.

For example.... A ball rolling freely around on the floor falls into a well in the middle of floor and can't get out. The potential energy while it's rolling around on the level floor is $E$, the potential energy in the bottom of the well is $E-mgh$ where $h$ is the depth of the well, and when the ball falls to the bottom of the well, the energy released is $mgh$. (Note that as long as I'm consistent, I can pick any value I please for $E$; it may be natural to choose the potential energy at floor level to be zero but I don't have to).

The binding energy is interesting when you want to know how much energy you'll have to add to unbind a bound particle - in this case, how much energy we'd have to add to get the ball out of the well if it's already in there. So you can calculate the binding energy from the potential energy, but the potential energy is a much more powerful and general concept, used in many more situations.

7. Mar 15, 2014

### dauto

Binding energy is defined as potential energy of bound state times (minus one) with the extra assumption that the potential energy was defined to be zero for the unbound state.