# Binding energy, fission and fusion

My high school physics book doesn't elaborate the idea of binding energy and how it's related to fission and fusion adequately in a way that made me have wrong thoughts about these ideas.

What i understand after doing some research is that: - Binding energy is the energy that has to be given for nucleons to separate them from each other.

• Nucleons in stable nucleus have negative energy considering the energy of a free static nucleon to be the reference energy.

My questions and confusions:

• in fission, does the nucleus divide because of the collision between the neutron and the heavy nucleus, or because the nucleus would become unstable after the mass number has increased.

• this question is related to the above one ; neutrons are supposed to be the main factor of nucleus stability because it contributes in the strong force. why does adding a new neutron or more to any nucleus without changing the number of protons, make the nucleus unstable?

• My book mentions that when alpha decay happens ,a decrease in mass turns into kinetic energy gained by the products. does it mean the decrease of mass because of the lost neutrons and protons or what?

• the process of losing mass for energy and vice versa in fission and fusions, shouldn't the mass for the neutrons shot out of the mass increase back and that's it? where would energy come from?

• So, checked out any other books ie library? – user207455 May 25 at 9:52
• Too advanced for a high school student – passepartout May 25 at 11:57

## 3 Answers

• The heavy nucleus (e.g. $$^{238}$$U) is already unstable. It will decay on some timescale $$\sim$$ $$4$$ billion years.
The extra (slow) neutron is shot onto the heavy nuclear, creating $$^{239}$$U, for it to be even more unstable and with a more shorter lifetime, $$\sim 20$$ minutes. This allows the energy to be released and put to use within human lifetimes.

• Nuclear instability is not just due to the interplay between nuclear force$$^\dagger$$ (attractive) and electromagnetic force (repulsive). Neutrons and protons are fermions and they are forced to occupy higher energy states. The latter plays a fundamental role in causing the extra neutron to result in an increased instability.

• If you spontaneously decay from a system $$U_i \rightarrow U_f + \alpha$$, the final state $$U_f$$ must have lower energy than the initial state $$U_i$$. The difference in their energies is given to the $$\alpha$$ particle's kinetic energy. The origin of this energy is the binding energy increase between $$U_f$$ and $$U_i$$ because of the increased stability.

• When people say that you "lose mass" what they actually mean is the following. The initial, unstable nucleus $$U_i$$ has energy $$E_i = m_ic^2$$. The final, (more) stable nucleus $$U_f$$ has energy $$E_f=m_fc^2$$. The decay happens because the $$E_f < E_i$$ (and especially $$E_f < 0$$ to be a bound ssytem), hence the "mass" of the final states $$m_f < m_i$$.
Another way to see this, from fusion: $$p+p+p+p\rightarrow \,^4$$He.
$$m_p+m_p+m_p+m_p > m_{He}$$.
The Helium atom is more stable i.e. you have to put energy in to break it. When you break it, it goes back to a system of 4 separate protons each with energy $$m_pc^2$$. Hence, if He is more stable than the 4 protons, you need $$m_{He} < 4m_p$$.

$$\dagger$$: the strong force hold the quarks together inside the nucleons. The nucler force is a remnant of the strong force and is what holds the nucleons together. Aking to how the van der Waals interaction is a remnant of the EM force.

• So when alpha decay happens, the binding energy of the final nucleus decreases (as in becomes more negative) in a way that would make the final nucleus harder to break apart right? And this amount of energy became kinetic energy? and this all is because some part of the mass became energy? Also I'm sorry but my textbook doesn't mention anything about quarks or fermions so I don't really know them. By the way shouldn't in the last of your answer the mass of helium is less than the mass of the protons? – passepartout May 25 at 11:55
• The binding energy increases, as it is defined as the energy you have to put in to break it. The differnece in binding energy becomes kinetic energy. Sorry about the Helium mistake, I fixed it. – SuperCiocia May 25 at 12:06
• increases but in negative right? since it's a "binding energy" – passepartout May 25 at 12:49
• Binding energy is always defined as the energy needed to put in the system to break it. So The total energy becomes more negative, i.e. more stable, which means that the binding energy becomes larger (more positive. – SuperCiocia May 25 at 12:54
• Right. sorry for late reply. – passepartout May 27 at 10:21

If you have n free protons and m free neutrons , the sum of the masses of the free nucleons is always larger than the same n+m bound in a nucleus.

The difference is what generates this plot: It shows against n+m how bound is each nucleon(proton or neutron) within the nucleus. How much energy would be needed to set a nucleon free.

Fission happens to nuclei after Fe to the right in the plot, fusion to light nuclei on the left of F

In fission, does the nucleus divide because of the collision between the neutron and the heavy nucleus, or because the nucleus would become unstable after the mass number has increased.

How stable a nucleus is , is a solution of the quantum mechanical problem of having so many nucleons bound in a "box", there are stable and unstable levels, as the other answer states, that depend on the number of protons and neutrons and how the levels are filled. There might be spontaneous decay to lighter nuclei , the sum of which will release energy. In general adding a proton or a neutron may change a nucleus to a less bound one that has a high probability of decay, or may immediately break up as with uranium nuclei ( see the other answer).

why does adding a new neutron or more to any nucleus without changing the number of protons, make the nucleus unstable?

a)The neutron when free decays within 13 minutes. When bound in the nucleus in a stable energy level it will not decay, and it helps overcome the electrostatic repulsion of protons between them. But if there are too many neutrons, they are not in a stable energy, but there is a probability that the weak decay will generate a beta decay

b)The solution of the collective potential after adding a neutron may push the nucleus into an unstable energy level, as happens with Uranium.

My book mentions that when alpha decay happens ,a decrease in mass turns into kinetic energy gained by the products.

For any decay, since energy is a conserved quantity, adding the new masses and finding a deficit, means that kinetic energy must be taken by the decay products.

the process of losing mass for energy and vice versa in fission and fusions, shouldn't the mass for the neutrons shot out of the mass increase back and that's it? where would energy come from?

The neutron always has the same mass when free.

In fusion, there will always be at least three particles involved because momentum has to be conserved also. See this article for detailed examples.

The energy comes from the difference in the binding energy between the initial and final nucleons. Helium has high binding energy, hydrogen zero. To bind a proton and an neutron and get an deuterium, releases energy (though it is more complicated than just two particles, look at the link)

In fission one goes towards higher on the left binding energy, so energy is released in the decay products.

Maybe this classical example will help about binding energy: There is a lake up in the mountain. The water is at an energy level. Making a dam and putting turbines the water falls to a lower energy level giving up the potential energy to the flow and the turbines.

• is it necessary for the decrease in mass to always result in increase in binding energy? – passepartout May 27 at 10:39
• @passepartout In general, yes. Energy is a conserved variable in a given inertial frame, and in special relativity mass and energy are connected . see hyperphysics.phy-astr.gsu.edu/hbase/Relativ/releng.html . – anna v May 27 at 10:52

The process of nuclear fission much depends on the interplay of the electromagnetic and the nuclear forces among the nucleons.

Nucleus consists of charged protons and uncharged neutrons. So the electromagnetic forces acts only among the protons. But the nuclear force being charge-independent, both protons and neutrons are affected by it.

• Electromagnetic force: Only protons are affected; long range.
• Nuclear force: Both protons and neutrons are affected; very short range.

In a nucleus in stable equilibrium, both these forces balances each other. We can consider the nucleus to be a union of small domains such that both the forces are in equilibrium at each domains. Now we consider the following two cases:

1. Light elements: A small perturbation could not disturb the system. This is because the electromagnetic and nuclear forces manages to balance each other as the nucleus itself is small and both the forces could act at the domains. Due to this all the domains retain the equilibrium situation and the nucleus remains stable.
2. Heavy elements: A small perturbation disturbs the system. The reason is that in this case there are a large number of domains as the nucleus is large. However, since the nuclear forces are short ranged, each domain feels the nuclear force only from the nearby domains. On the other hand, electromagnetic forces are felt from the whole nucleus on the protons resulting to repulsion. Due to this reason, the domains couldn't sustain the perturbation as the electromagnetic force dominates. As a result the domains become unstable and the nucleus divides to attain equilibrium.
• Could you please elaborate the idea of domains more? like a real life example? – passepartout May 27 at 10:41
• By domains I mean small regions inside the nucleus whose size can be considered to be of the order of the range of nuclear forces. – Richard May 27 at 11:08
• Oh alright. like zones and each zone is occupied equivalently by these 2 forces if the nucleus is stable. right? And in big sized nucleus most domains would be mostly occupied by electric force resulting in a state of unstablity? – passepartout May 27 at 13:18
• Exactly. In a large nucleus (i.e. large atomic number) there would be a large number of such domains. And at each domain there would be nuclear force only from the neighboring nucleons and electromagnetic force from all over the nucleus. And these two forces would just balance and after a slight perturbation the nuclear force couldn't balance the electromagnetic force leading to instability. Whereas the size of light nuclei is roughly of the order of the range of nuclear forces and also the number of domains is very small. So both the forces could easily balance each other. – Richard May 28 at 6:17