Main content

## MCAT

### Course: MCAT > Unit 8

Lesson 22: Atomic nucleus- Atomic nucleus questions
- Radioactive decay types article
- Decay graphs and half lives article
- Atomic number, mass number, and isotopes
- Atomic mass
- Mass defect and binding energy
- Nuclear stability and nuclear equations
- Writing nuclear equations for alpha, beta, and gamma decay
- Types of decay
- Half-life and carbon dating
- Half-life plot
- Exponential decay formula proof (can skip, involves calculus)
- Introduction to exponential decay
- Exponential decay and semi-log plots
- More exponential decay examples
- Mass spectrometer

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# Mass defect and binding energy

Visit us (http://www.khanacademy.org/science/healthcare-and-medicine) for health and medicine content or (http://www.khanacademy.org/test-prep/mcat) for MCAT related content. These videos do not provide medical advice and are for informational purposes only. The videos are not intended to be a substitute for professional medical advice, diagnosis or treatment. Always seek the advice of a qualified health provider with any questions you may have regarding a medical condition. Never disregard professional medical advice or delay in seeking it because of something you have read or seen in any Khan Academy video.

## Want to join the conversation?

- Perhaps on future videos, can you guys show it without a calculator like it would be on the MCAT? That way we also get practice with good estimation techniques or rounding.(22 votes)
- These videos were created before making this MCAT playlist... So I doubt it. Also, I guarantee that you will not have to do calculations like this on test day.(7 votes)

- @6:20will we have to memorize kg/amu? the number? Or mass of a proton/neutron/electron?(3 votes)
- You should probably know the relative masses of protons, neutrons, and electrons, and how to get this info from the periodic table. I suspect that, at @6:20, you would be given this number if it was required.(1 vote)

- Binding energy is needed to make the nucleus more stable. Does it mean that more the binding energy, the more stable nucleus?(3 votes)
- Indeed! The more energy is released when the individual protons and neutrons combine to form the nucleus, the more stable that nucleus will be. Another way to think about it is that a more stable nucleus would require a larger energy input than a less stable one in order to be split into its constituent particles; this quantity of energy would be equal to the binding energy. :)(2 votes)

- I find this tutor's mic. to always be much quieter than others, so as progressing through videos I switch between hearing next to nothing, turning it up, then blasting my ears out! If you could sound mix it so his volume is on par with others it would be greatly appreciated!(3 votes)
- Honestly, do you believe we will need this for the MCAT? The non traditional student wants to know,(1 vote)
- If work done to overcome the strong nuclear force is the binding energy, I find the binding energy really small, why isn't fusion reaction possible to bring two nuclei close together?(1 vote)
- Why shouldn't we add the mass of electrons (for accurate binding energy)? Also, where does the actual mass of the Helium nucleus come from? Is this the weighted average of all He isotopes? In that case, I think I shouldn't be comparing that to find the binding energy?(1 vote)
- Here's what I don't get. When Nuclear fission of uranium occurs, the binding energy of the two products is higher and so the overall binding energy is higher than it was before. Im assuming that mass is converted to energy and the mass defect will increase. So, that means you would have to put in energy to seperate the atom since both of the products require more binding energy. How then, does fission release energy? Its not like the energy is escaping the products - its within the products as binding energy! Its completely contradictory to the storing and release of energy in chemical bonds such as those in ATP. Where has my logic gone wrong?(1 vote)

## Video transcript

- Let's say once you calculate the mass of a Helium four nucleus. First we have to figure out what's in the nucleus. With an atomic number of two, we know that 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 protron and the mass of a neutron, we could easily calculate the expected mass of the Helium four nucleus. And the mass of a proton in amu, 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, and remember these are amu, atomic mass units. A neutron, one neutron, has a mass in amu 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 amu. 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, it gives us 4.03188274. So 4.03188274 amu, 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 amu, let me go ahead and write this, this is the actual mass, so the actual mass. 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. Let's go ahead and do that. All right, so we have the predicted, and we're going to subtract the actual from that: 4.00150608 is going to give us .03037666 amu. Let's go ahead and write that: so, 0.03037666 amu, so this is the difference between those two numbers and we call this the mass defect. 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. It looks like we lost some mass here. What's happened is that 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 m c squared. So E is equal to m c squared, where "E" refers to the energy in jules, 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. Let's calculate the mass that we're dealing with here. Using Einstein's equation, we see we need kilograms, and we've calculated the mass in amu, so the first thing we need to do is convert the amu into kilograms, and I briefly mentioned in an earlier video, the conversion factor between amu and kilograms, so amu 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 ten to the negative 27 kilograms. So the first thing we need to do is convert that number, so let's go ahead and write it down: 0.03037666 amu. So, mathematically, how would I convert the amu into kilograms? Well, I need to cancel out the amu units, so my conversion factor is going to be 1.66054 times ten to the negative 27. That's how many kilograms we have for every one amu, so I can put that in here. There's my conversion factor. Notice what happens when we do this: our units for amu cancel, and this is going to give us kilograms. 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 parenthesis here, 1.66054 times ten to the negative 27, so let's see what this gives us here, so this gives us 5.04417 times ten 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 ten to the negative 29 kilograms. So now we have the mass in kilograms. Let's get some more room, and let's go ahead and calculate the energy. So, this is the mass that was lost when the nucleus was formed, so let's figure out how much energy was given off. The total energy, energy is equal to that mass, so let's go ahead and plug that in: 5.04417 times ten to the negative 29, times the speed of light, which is approximately three times ten to the eighth meters per second, we'll use a more exact number: 2.99792 times ten to the eighth, and we need to square that number. All right, so let's do our last calculation. Let's start with the speed of light, 2.99792 times ten 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 ten to the negative 29, and let's see what that gives us: 4.3346 times ten to the negative 12. Let's write that down. 4., I think there was a five in there, 4.53346 times ten to the negative 12, let me just double check that real fast here, so 4.53346 times ten 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, it took us a while to get there. Remember, this is the energy that's released when the nucleus is formed. So, let me 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, 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, 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. So this is just a nice little picture to think about what's happening. So, whenever 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 you started with the nucleus, and you wanted to break it up into the individual components, right? 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 you would have to apply, is also equal to this energy. So, this is also called the nuclear binding energy. 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.