What causes mass defect in the nucleus?

In summary, the nucleons in the nucleus of an atom have less mass than if they were separated due to the conversion of part of their mass into nuclear force to bind them together. This is known as the "mass defect" of nucleons. When free nucleons are assembled into a nucleus through nuclear fusion, energy is released, causing the mass deficit. However, if the nucleus of an atom is split in two through fission, the mass of the nucleons will increase. This is because nuclei lighter than iron are in a lower energy state when they are split, whereas nuclei heavier than iron are in a higher energy state when they are split. Exceptions to this trend include Helium 4, Carbon 12, and Oxygen 16,
  • #1
Karagoz
Hi.

I read that the mass of nucleons in nucleus of Oxygen-16 (or nucleus of another atom) is less than the sum of its nucleons if they were separated (also the mass of 8 protons and 8 neutrons).

But why does nucleons have less energy in nucleus of atoms than if they were separated? What causes that "mass defect" of nucleons?

The nucleons on atoms have less mass because part of their mass is converted to nuclear force to bind nucleons together in nucleus of an atom?

Or the nucleons release some energy ( and hence mass ) when binding together?

If nucleus of an atom is split in two, the mass of the nucleons will increase?
 
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  • #2
Karagoz said:
The nucleons on atoms have less mass because part of their mass is converted to nuclear force to bind nucleons together in nucleus of an atom?
Yes, if you have a bunch of free nucleons and assemble a nucleus (nuclear fusion) then it will release energy. This released energy is the mass deficit.

Karagoz said:
If nucleus of an atom is split in two, the mass of the nucleons will increase?
Yes, this is why fission requires net energy for nuclei lighter than iron
 
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  • #3
For the smaller nuclei, they are a lower energy state than the separate components. Fusion is possible where they go to a state of lower energy when they combine. In the process, this energy released in the fusion process accounts for the mass difference with ## E=\Delta m \, c^2 ##.
 
  • #4
Fission of nuclei lighter than iron won't release energy (ulike in the case of fission of Uranium)?
 
  • #5
Karagoz said:
Fission of nuclei lighter than iron won't release energy (ulike in the case of fission of Uranium)?
That is correct. I believe unstable isotopes would be an exception to this.
 
  • #6
Bu the mass of a proton is ca 1.0072 u and a neutron is ca 1.0086 u. The mass of nucleons is less when these nucleons are part of a nucleus, or part of an atom.

Why their mass is less in atoms? Is it because they lose energy to bind to each other (to form a nuclei)?
 
  • #7
Karagoz said:
Bu the mass of a proton is ca 1.008 u and a neutron is ca 1.007 u. The mass of nucleons is less when these nucleons are part of a nucleus, or part of an atom.

Why their mass is less in atoms? Is it because they lose energy to bind to each other (to form a nuclei)?
Yes. The attractive nuclear forces brings them to a state of lower energy. Basically, in a fusion reaction, they get pulled towards each and speed up towards each other and then are brought to a stop. This kinetic energy gets released in the form of photons, etc., x-rays, uv, etc.. ## \\ ## And I do believe the neutron is generally slightly more massive than a proton, (you have it reversed), partly because it also contains an electron. The electron and proton of the neutron are attracted to each other, so again, in combining there will be a small mass deficiency. ## \\ ## It may interest you in the larger atoms that release energy in a fission reaction that the nuclei are bound together in a state that is a lower energy than if they were separated by a small distance=it is a stable potential well, (like a small valley high up on a large mountain), but the lower and more stable energy state is when the atom gets split into parts, and when this occurs, there is the release of energy from Coulomb repulsion as the protons of the component nuclei repel each other and drive each other apart. (Once the atom configuration gets pushed out of the high valley=the potential well, it then goes much lower down the hill).
 
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  • #8
If I did get it right, atoms that weigh lower than Fe (iron) give up energy in fusion and absorb energy in fission. But atoms that weigh higher than Fe give up energy in fission and absorb energy in fission. This is not without exception (according to the graph below taken from Wikipedia, as the line goes up and down between He, Li, C etc.).

But what I didn't get is, why is it that? That fusion and fission sometimes absorbs energy but sometimes gives away energy?

671px-Binding_energy_curve_-_common_isotopes.svg.png
 
  • #9
Karagoz said:
If I did get it right, atoms that weigh lower than Fe (iron) give up energy in fusion and absorb energy in fission. But atoms that weigh higher than Fe give up energy in fission and absorb energy in fission. This is not without exception (according to the graph below taken from Wikipedia, as the line goes up and down between He, Li, C etc.).

But what I didn't get is, why is it that? That fusion and fission sometimes absorbs energy but sometimes gives away energy?

View attachment 210897
Helium 4 is extremely stable in its binding energy, but if you fuse a neutron or two and a proton onto helium to make lithium, I believe it still is a state of lower energy and will release energy in the process. (Not certain of this=it would require detailed calculations and might turn up to be one of the exceptions. Editing: See below, where I computed it for Lithium 6.) ## \\ ## Likewise Carbon 12 and Oxygen 16 are very stable, but making nitrogen from Carbon 12 should still be a lower energy state. ## \\ ## The calculation is more complicated than just taking the average binding energy per nucleon. You need to compute the total binding energy and compare to the total binding energy of the component parts. Any extra binding energy is energy that gets released in the fusion process. ## \\ ## In the case of He 4 vs. Li 6, the calculation is quite simple. Since an isolated proton and an isolated neutron each have zero binding energy, the question is, if 6x binding energy per nucleon for Li 6 is greater than 4 times the binding energy per nucleon for Helium 4. The answer is yes. ## (6 )(5.3) > (4)(7.2) ##.
 
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  • #10
Karagoz said:
But what I didn't get is, why is it that? That fusion and fission sometimes absorbs energy but sometimes gives away energy?
A nucleus is subject to the repulsive electrical (Coulomb) force between protons, and the attractive strong nuclear force among both protons and neutrons. The relative "balance" between these two forces varies with different numbers of protons and neutrons, and determines whether fission or fusion is more "natural".

The details are too long to write in a forum post, but they're in many textbooks and web pages. Google for "semi-empirical binding energy formula" or "semi-empirical mass formula" and you should be able to find something that describes the different terms in that formula and their origin. This is usually derived using the the "liquid drop model" of the nucleus, so you might also Google for that.
 
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  • #11
Karagoz said:
If I did get it right, atoms that weigh lower than Fe (iron) give up energy in fusion and absorb energy in fission. But atoms that weigh higher than Fe give up energy in fission and absorb energy in fission. This is not without exception (according to the graph below taken from Wikipedia, as the line goes up and down between He, Li, C etc.).

That graph shows that Fe56 has the highest binding energy per nucleon, without exception, evidenced by the fact that there's nothing with a higher binding energy per nucleon! The wiggles around He and Li do not demonstrate an exception. They demonstrate that the maximum at Fe56 is not approached smoothly.

But what I didn't get is, why is it that?

Are you asking why the element with the highest binding energy per nucleon is the most stable? Or are you asking what it is about that particular number of nucleons that makes it the one with the highest binding energy per nucleon?

The first question almost answers itself and thus answers your fusion versus fission question. I don't know the answer to the second question beyond what the others have already mentioned.
 
  • #12
In this graph: you see that Fe has least mass per nucleon.
http://dev.physicslab.org/img/7b784857-7639-45de-bd2e-32cdeb2dabdc.gif

But Fe has highest binding energy per nucleon, like you see in this graph:
https://cnx.org/resources/72ec9ed4c72df1425d0318f15a94e3c8247182bb/Figure_32_06_03a.jpg

Is there a relationship between "average mass of per nucleon in nuclei" (or mass / nucleon) and "binding energy per nucleon"?
Also higher binding energy per nucleon, then less "mass / nucleon" will there be be?

If so, why is there such relationship or correlation?
 
  • #13
Karagoz said:
Is there a relationship between "average mass of per nucleon in nuclei" (or mass / nucleon) and "binding energy per nucleon"?
The binding energy is just the difference between the mass of the nucleus and the sum of the masses of the particles in the nucleus, if those particles were all separated from each other. (And then converted to energy units, if necessary, via ##E=mc^2##.)

Edit: deleted some irrelevant and potentially misleading words
 
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  • #14
DrGreg said:
The binding energy is just the difference between the mass of the nucleus and the sum of the masses of the particles in the nucleus, if those particles were all separated from each other. (And then converted to energy units, if necessary, via ##E=mc^2##.)

So, the mass of nucleons decrease because that mass is converted to "binding energy"?
 
  • #15
Karagoz said:
So, the mass of nucleons decrease because that mass is converted to "binding energy"?
The system when it binds, goes to a state of lower energy. Energy is released in the process because the energy is conserved. The quantity of energy that is released (as photons and heat, etc.) is said to be the binding energy for that system. To get the system separated again, this amount of energy (the binding energy) would need to be added to the system. (The binding energy is a measure of how strongly the system is bound. If it is given a negative sign, it can then be considered to be the potential energy of the system. Total energy=E=Kinetic energy +potential energy=mc^2). ## \\ ## Meanwhile the energy of the system is related to its mass by ## E=mc^2 ##. If some of the energy contained in the mass ## m ## of the particle is subsequently converted to energy in the form of radiated photons (electromagnetic waves,etc.), it will then have a lower mass. In the process of binding, energy is given off in the form of photons.
 
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  • #16
Karagoz said:
So, the mass of nucleons decrease because that mass is converted to "binding energy"?
The individual masses of the nucleons do not change. A proton always has a mass of 938.3 MeV/c2, and a neutron always has a mass of 939.6 MeV/c2. The mass of a system of particles (e.g. a nucleus) does not equal the sum of the masses of the individual particles, in general. The potential energy of the system and the kinetic energies of the individual particles (*) also contribute to the mass of the system.

(*) More precisely, the part of the kinetic energy that is not associated with the motion of the system (nucleus) as a whole.
 
  • #17
jtbell said:
The individual masses of the nucleons do not change. A proton always has a mass of 938.3 MeV/c2, and a neutron always has a mass of 939.6 MeV/c2. The mass of a system of particles (e.g. a nucleus) does not equal the sum of the masses of the individual particles, in general. The potential energy of the system and the kinetic energies of the individual particles (*) also contribute to the mass of the system.

(*) More precisely, the part of the kinetic energy that is not associated with the motion of the system (nucleus) as a whole.

You say that the individual masses of the nucleons don't change, but why the mass of a nucleus is less than the sum of the masses of the individual particles?
 
  • #18
Karagoz said:
You say that the individual masses of the nucleons don't change, but why the mass of a nucleus is less than the sum of the masses of the individual particles?
Because that is the way that "mass" or, more specifically, "invariant mass" is defined. It is the norm of the energy-momentum four-vector.

In more common terms, it is the total energy of the system as measured in frame of reference where the system is at rest. Divided by ##c^2## according to the formula: ##E=mc^2## if using a system of units in which c is different from 1. If you reduce the energy of the system while keeping it motionless, you reduce its mass.

Edit: "Binding energy" is negative. It is a deduction from the total system energy. But it is not tied to any particular nucleon. So the masses of the individual nucleons are reckoned without the deduction.
 
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  • #19
Karagoz said:
So, the mass of nucleons decrease because that mass is converted to "binding energy"?

The mass of the nucleus is smaller, but the mass of the nucleons is the same. So think of pulling those nucleons apart; the energy you put into the process is called the binding energy, and it equals the increase in mass that you measure. Assembling the nucleus is the reverse of this process. There's no conversion of mass to energy, just a change in what you're calling something; the thing you once called mass is now a thing you call energy.

The same process happens when you pull a neutral hydrogen atom apart by removing the electron. It's just that the amount of energy is so small compared to the separate masses of the proton and the electron that you don't notice the change. That allows you to pretend that the mass of the atom is equal to the sum of the mass of a proton and the mass of an electron, and that the energy that binds them together is something separate from the mass. That pretense would be exposed, though, if you were able to measure that tiny change in mass.
 
  • #20
Karagoz said:
You say that the individual masses of the nucleons don't change, but why the mass of a nucleus is less than the sum of the masses of the individual particles?

The notion that the mass of a composite body equals the sum of the masses of its constituents is part of the Newtonian approximation. It's an illusion, a pretense. Einstein's discovery of rest energy, and its equivalence to mass, has taught us this lesson.
 
  • #21
Karagoz said:
You say that the individual masses of the nucleons don't change, but why the mass of a nucleus is less than the sum of the masses of the individual particles?
jtbell said:
The potential energy of the system and the kinetic energies of the individual particles (*) also contribute to the mass of the system.
The potential energy of a bound system like a nucleus is negative. This reduces the total energy of the nucleus, and therefore its mass also.
 
  • #22
One need to put energy (e.g. hit it hard) to break a table, but the sum of the masses of broken parts of the table is not more than the mass of the whole table. The mass of the broken parts don't increase.

jtbell said:
The potential energy of a bound system like a nucleus is negative. This reduces the total energy of the nucleus, and therefore its mass also.

And that negative energy is the "energy needed to split the nuclei"?

So in physically the mass of nucleons doesn't change (they are same both in a bound system and as separated), but it's just only in calculation that the "negative energy" is deducted from the total energy (and also from the total mass) of the nucleus?
 
  • #23
Karagoz said:
One need to put energy (e.g. hit it hard) to break a table, but the sum of the masses of broken parts of the table is not more than the mass of the whole table.
Are you sure? What would be the predicted increase in mass, and do you think you can measure it that precisely?

Karagoz said:
So in physically the mass of nucleons doesn't change (they are same both in a bound system and as separated)
How would you measure the mass of a single bound nucleon?
 
  • #24
Dale said:
How would you measure the mass of a single bound nucleon?

But jtbell said the individual masses of the nucleons do not change. That means the mass of nucleons are the same both in a bound system and as separated?
 
  • #25
Karagoz said:
But jtbell said the individual masses of the nucleons do not change. That means the mass of nucleons are the same both in a bound system and as separated?
@jtbell may know of a way to measure the mass of a bound nucleon, but I don't. I am not as thoroughly versed in this literature as he is.
 
  • #26
Dale said:
@jtbell may know of a way to measure the mass of a bound nucleon, but I don't. I am not as thoroughly versed in this literature as he is.
Isn't this some kind of random? depending on the nucleon and it is position? You may average the energy loss across them and calculate the mass but that isn't accurate.

You could do it though for simple atoms right?
 
  • #27
Biker said:
Isn't this some kind of random? depending on the nucleon and it is position? You may average the energy loss across them and calculate the mass but that isn't accurate.
Does a nucleon actually have a position in a nucleus?
 
  • #28
jbriggs444 said:
Does a nucleon actually have a position in a nucleus?
Uhh probably something with quantum mechanics..

All the animations shows them vibrating around which doesn't give a definite position and most likely that they are wrong on some level(Yes, I know it is wrong to think of them as little balls). My question should be doesn't each nucleon contribute differently than others? and does it even make sense to talk about mass loss of a single nucleon?
 
  • #29
Karagoz said:
You say that the individual masses of the nucleons don't change, but why the mass of a nucleus is less than the sum of the masses of the individual particles?
Well, if a nucleon does any work, then its energy and its mass decrease. Conversely, if nucleon's mass never changes, then we can guess that the nucleon never does any work.

Now let us consider macroscopic objects: Two metal spheres, one is positively charged, other one is negatively charged.

The spheres are floating in viscous oil and are being pulled towards each other by the coulomb force. The energy and the mass of the sphere-pair is decreasing, so the sphere-pair is doing work.

But the sphere on the right is not losing mass, so it is not doing any work, and same is true for the sphere on the left.

The sphere-pair is doing work - to be more specific the electric field of the sphere-pair is doing work, but not the electric field of the sphere on the left or the electric field of the sphere on the right.

As the sphere-pair loses mass, it floats a little bit higher in the oil. The sphere on the right has not lost any mass, but it floats higher.

That last sentence seems problematic to me. Did I make some error??
 
  • #30
jartsa said:
As the sphere-pair loses mass, it floats a little bit higher in the oil. The sphere on the right has not lost any mass, but it floats higher.

That last sentence seems problematic to me. Did I make some error??
Well I have thought about it, and the solution to the problem seems to be this: We do not say that the metal sphere on the left has a reduced mass, instead we say the left side of the system has a reduced mass. Somehow that seems to make the problem disappear. :smile:

We have the word nucleon, so how about if we do not call a free proton or neutron a nucleon, but instead we call them a proton and a neutron?

And those nucleon particles that can be thought to exist inside nuclei, how about if we do not say that they are protons and neutrons and have the same masses as protons and neutrons? Is it not a naive idea that nucleons would have the same masses as protons and neutrons?
 

Related to What causes mass defect in the nucleus?

1. What is mass defect in the nucleus?

Mass defect in the nucleus refers to the difference between the mass of an atom's nucleus and the sum of the masses of its individual protons and neutrons. This difference is caused by the conversion of some of the mass into energy during nuclear reactions.

2. What causes mass defect in the nucleus?

The main cause of mass defect in the nucleus is the strong nuclear force, which holds the protons and neutrons together in the nucleus. This force is stronger than the electrostatic repulsion between the positively charged protons, but it still requires energy to overcome it.

3. How is mass defect related to nuclear binding energy?

Mass defect and nuclear binding energy are directly related. The energy that is released during a nuclear reaction is equal to the mass defect multiplied by the speed of light squared (E=mc^2). This energy is known as nuclear binding energy and is what holds the nucleus together.

4. Can mass defect be observed in everyday objects?

Yes, mass defect can be observed in everyday objects, but it is usually on a very small scale. The most well-known example is nuclear power plants, where mass defect is harnessed to produce energy. However, it can also be observed in natural radioactive decay and in the formation of elements in stars.

5. How is mass defect calculated?

Mass defect is calculated by subtracting the mass of the individual protons and neutrons from the mass of the nucleus. This difference is then converted into energy using Einstein's famous equation, E=mc^2. The resulting energy is a measure of the strength of the nuclear binding force.

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