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Mass defect and binding energy

Nuclear binding energy is the energy required to split an atom’s nucleus into protons and neutrons. Mass defect is the difference between the predicted mass and the actual mass of an atom's nucleus. The binding energy of a system can appear as extra mass, which accounts for this difference. Created by Jay.

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  • duskpin tree style avatar for user Archit  Agarwal
    Why does a mass defect occur?
    (12 votes)
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  • aqualine ultimate style avatar for user Connor Stack
    Is it possible to determine how much mass is lost from the neutron and how much is lost from the proton individually?
    (25 votes)
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  • starky sapling style avatar for user Scott
    Why is the mass of a neutron greater than that of a proton?
    (7 votes)
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    • boggle blue style avatar for user Davin V Jones
      Protons are composed primarily of two up quarks and one down quark, while neutrons have two down quarks and one up quark. A down quark is more massive than an up quark, so that partly accounts for the difference in masses. However, quarks only account for a small portion of the total mass of nucleons. The rest of the mass is determined by the energy holding everything together.

      Here is a video that addresses this more in depth: https://www.youtube.com/watch?v=Ztc6QPNUqls
      (18 votes)
  • male robot donald style avatar for user Servaich
    Does this Nuclear binding energy releases photons?
    (9 votes)
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    • starky ultimate style avatar for user Collin
      From a quick Wikipedia search I found that the forming of a nucleus can release photons, but there are other forms of energy which can also be released. I suspect that a better answer may require more advanced quantum physics.
      (6 votes)
  • aqualine ultimate style avatar for user Haley
    I have two questions:
    1. How can mass be converted to energy? Does it relate to wavelength properties? Does this violate the conservation of mass or does it count as conversion to energy leading energy and mass to be more intrinsically tied?
    2. Forming the nucleus appears to be unfavorable in terms of entropy and charge therefore I would expect the reaction to be endothermic and require energy to proceed. Why does it not?
    (4 votes)
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    • male robot hal style avatar for user Charles LaCour
      The idea of conservation of mass is is based in the understanding of physics prior to understanding atomic interactions and the theory of special relativity. Einsteins well known equation E=m * c^2 shows that there is an equivalence between mass and energy.

      Besides the electromagnetic force being involved with the electric charges in the atomic nucleus there is also the residual strong nuclear force which is about 20 times stronger than the electromagnetic force at the distances in the atomic nucleus.
      (7 votes)
  • leaf blue style avatar for user Liroy Lourenco
    How is the actual mass of a nucleus determined experimentally?
    (5 votes)
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  • spunky sam blue style avatar for user Avi Perl
    I thought energy is released when you break the nucleus apart, not when you form it. What am I getting wrong here?
    (3 votes)
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    • male robot hal style avatar for user Charles LaCour
      It depends on the element you are dealing with. In general if it has an atomic number below iron then you energy out of fusing them together. If the atomic number is higher than iron's then you get energy out of breaking them up.

      There are exceptions where there are unstable atoms below iron and stable ones above but in general iron is the dividing line.
      (6 votes)
  • duskpin ultimate style avatar for user bookcrush
    When a nucleus is formed, only the electromagnetic force has to be overcome. But to break up a nucleus, the strong nuclear force (the strongest force) has to be overcome. So, why isn't more energy needed to break apart a nucleus than form it?
    (2 votes)
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    • male robot hal style avatar for user Charles LaCour
      The strong interaction has different effects at different ranges. At distances less than 0.8 * 10^-15 m or 0.8 fm (femtometer), the radius of a proton or neutron, the strong interaction is directly between quarks carried by particles gluons. This interaction is about 137 times stronger than the electromagnetic interaction.

      Distances greater than 0.8 fm are too far for the gluon interactions so the interaction is carried by particles call pi mesons or pions. These particles are able to carry this force to about 2.5 fm but it is not until about 1.7 fm that it becomes stronger than the electromagnetic force.

      The decrease of the strength of the strong interaction at distances greater than the radius of a proton is why more energy is not needed to break up a nucleus. Also because the strong interactions limited effect range it only affects adjacent protons/neutrons where the electromagnetic force is a long range force as the atomic nucleus gets larger the electromagnetic force builds up more that the strong interaction's force making them less well held together and once you get to the size of iron it takes more energy to force it together than is released by the binding so this is why stars can't fuse elements beyond iron in their normal fusion process.
      (4 votes)
  • aqualine tree style avatar for user hakeem
    I have a question regarding this unit. Based on the chemistry textbook. they also calculated the number of electrons which resulting a slightly different answer. So, do we have to calculate the number of electrons or not?
    (2 votes)
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    • male robot hal style avatar for user Andrew M
      These are nuclear reactions. They don't involve electrons.
      The only reason you might need to worry about the number of electrons is if you start with the mass of the entire atom you have to subtract the electrons to get the mass of the nucleus.
      (3 votes)
  • female robot grace style avatar for user ArslanAhmedAziz715
    why is the speed of light present in mass energy relationship ?what does it tell?
    (2 votes)
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Video transcript

- [Voiceover] Let's say you wanted to calculate the mass of a helium four nucleus, well first we need to figure out what's in the nucleus, so, with an atomic number of two, we know there are two protons in the nucleus. And, subtract the atomic number from the mass number, four minus two gives us two neutrons. And so, if we know the mass of a proton, and the mass of a neutron, we could easily calculate the expected mass of a helium four nucleus, and the mass of a proton in amus, atomic mass units, is equal to 1.00727647 and we have two protons, so we need to multiply this number by two. So let's go ahead and get out the calculator, and let's do that, so 1.00727647 times two gives us 2.01455294 so this is equal to 2.01455294, remember these are amus, atomic mass units. A neutron, one neutron, has a mass in amus of 1.00866490 and we have two neutrons, so we have to multiply this number by two. So let's go ahead and do that. So we have 1.00866490 times two, and this gives us 2.0173298. So 2.0173298 amus, so the mass of a helium nucleus, if we add those two numbers together, we should get that mass. So let's do that math, so this number plus our first number 2.01455294 gives us 4.03188274. So, 4.031882748 amus, so this is the predicted mass of the helium four nucleus. So let me go ahead and write this, this is the predicted mass. The actual mass of a helium four nucleus has been measured to be 4.00150608 amus. Let me go ahead and write this, this is the actual mass, so the actual mass. All right, there's a difference there. They're not the same number. The predicted number is higher than the actual mass. So let's calculate the difference between those two numbers. So if we subtract the actual from the predicted, we can see the difference between those numbers. So let's go ahead and do that. All right, so we have the predicted, and then we're going to subtract the actual from that. So 4.00150608 is going to give us .03037666 amus. So let's go ahead and write that. So, 0.03037666 amus, so this is the difference between those two numbers, and we call this the mass defect. So let me go ahead and write that down here. This is called the mass defect. The difference between the predicted mass of the nucleus and the actual mass of the nucleus. And it looks like we lost some mass here, and really what's happened is the mass, this mass right here, the mass defect, was converted into energy when the nucleus was formed. So that's pretty interesting, and we can calculate how much energy, according to Einstein's famous equation, which relates energy and mass. So this is the one that most people know. It's E is equal to MC squared. So E is equal to MC squared, where E refers to the energy in Joules, M is the mass in kilograms, and C is the speed of light, which is in meters per second, and so you'd be squaring that, so it would be meters squared over seconds squared. All right, so let's calculate the mass that we're dealing with here. So, using Einstein's equation, we see we need kilograms, and we've calculated the mass in amus and so the first thing we need to do, is convert the amus into kilograms. And I briefly mentioned in an earlier video, the conversion factor between amus and kilograms, so amus is just a different measure of mass. Let's get some more room down here. All right, so one amu is equal to 1.66054 times 10 to the negative 27 kilograms. And so the first thing we need to do, is convert that number, so let's go ahead and write it down. So 0.03037666 amus, so mathematically, how would I convert the amus into kilograms? Well, I need to cancel out the amu units. So my conversion factor is going to be 1.66054 times 10 to the negative 27, that's how many kilograms we have for every one amu, so I can put that in here. So there's my conversion factor. And notice what happens when we do this, our units for amus cancel, and this is going to give us kilograms. So let's go ahead and do this math. So what is this equal to? All right, so we get out the calculator, and we take this number, and we multiply it by, let's use some parentheses here, 1.66054 times 10 to the negative 27, so let's see what this gives us here, so this gives us 5.04417 times 10 to the negative 29, so let's round it like that. So this is going to give us, let's put it right here, 5.04417 times 10 to the negative 29 kilograms. All right, so now we have the mass in kilograms, so let's get some more room, and let's go ahead and calculate the energy. All right, so this is the mass that was lost when the nucleus was formed, so let's figure out how much energy was given off. So the total energy, energy is equal to that mass, so let's go ahead and plug that in. 5.04417 times 10 to the negative 29, times the speed of light, which is approximately three times 10 to the eighth meters per second, we'll use a more exact number, 2.99792 times 10 to the eighth, and we need to square that number. All right, so let's do our last calculation here. So let's start with the speed of light. 2.99792 times 10 to the eighth, and then we need to square that number. So we get this number, and we're going to multiply that by our mass, 5.04417 times 10 to the negative 29th, and so let's see what that gives us. 4.3346 times 10 to the negative 12, so let's write that down. 4 point, I think there was a five in there, 4.5334. So, 4.53346 times 10 to the negative 12, let me just double check that real fast here. So 4.55346 times 10 to the negative 12 is our answer. And the units should be Joules, so that's how much energy is given off. So here's our final calculation, so it took us awhile to get there. So remember, this is the energy that's released when the nucleus is formed. So let me go ahead and get some more space, and let me write this down. The energy released when the nucleus is formed. The energy released when the nucleus is formed. So let's draw a picture of what's happening. So we were talking about the helium nucleus, which had two protons, let me go ahead and draw those two protons in here, and two neutrons, so let me use neutrons here like that. And these came together, to form our nucleus. All right, so these two things come together. So we have our two positive charges in our nucleus. And then we have our neutrons as well. And this is supposed to represent our nucleus, our helium four nucleus here. So, energy is released when the nucleus is formed. So we could also put in this energy. So this energy is given off, so that's the number we just calculated. We spent several minutes getting this number, and this is the energy that's released when the nucleus is formed here. And so this is just a nice little picture to think about what's happening. So whatever this nucleus formed, energy was given off. The nucleus is stable because energy is given off here. And we can also think about going the opposite direction. So if we started with the nucleus, and you wanted to break it up into the individual components, so if you took this nucleus here, and you applied some energy, you could break it up and turn it back into protons and neutrons, and that energy that you would have to apply, is also equal to this energy. So this is also called the nuclear binding energy. So let me go ahead and write that. Nuclear binding energy, so again, that is the term for the energy that we just calculated here. So you can think about it two different ways, it's the energy that's released when the nucleus is formed, and that's also the amount of energy that's needed to break the nucleus apart. And so the nucleus is stable in this case. So we have a stable nucleus right? This is a stable nucleus, but that's a little bit weird because we have these positive charges, and we know that positive charges repel. So these two positive charges here are repelling each other right? We know that like charges repel, and so there must be some other force that's holding our nucleus together. And that's called the nuclear strong force, and we'll talk more about that in the next video.