Nuclear Binding Energy - Video Tutorials & Practice Problems
Get help from an AI Tutor
Ask a question to get started.
1
concept
Mass to Energy Conversion
Video duration:
2m
Play a video:
Now, nuclear binding energy is involved in our discussion of mass to energy conversions back and forth. We're gonna say if the mass defect is known, then its conversion to energy can be determined. We're gonna say nuclear binding energy which uses the variable E is the energy that is released during the formation of an isotope. We're gonna say recall the process can also be seen as energy being absorbed to break up the isotope. Now, here we're going to say that the higher the nuclear binding energy then the more stable the nucleus will be for any given isotope with nuclear binding energy. We have our own formula here. We're gonna say the formula for nuclear binding energy is per one mole of a radio isotope. And we're gonna say nuclear binding energy which is E equals MC squared. So our equation that we associate with Albert Einstein here E is our nuclear binding energy typically goods of joules from here. Once you know joules, you could change to kilojoules or even mega electron volts. Here, we're gonna say M is our mass defect. In this instance, it's not going to be in atomic mass units. It needs to be in kilograms because of the presence of jewels. That's because remember one jewel is equal to kilograms times meter squared over a second squared. Now the meter squared and the second squared come from squaring C which is our speed of light. Remember speed of light is equal to 3.00 times 10 to the 8 m per second. All of these variables together help us to calculate nuclear binding energy. Remember if you know nuclear binding energy, you can use it to determine your mass defect. And if you know your mass defect, you can use that to determine the nuclear binding energy they're connected to each other through this formula.
2
example
Nuclear Binding Energy Example
Video duration:
3m
Play a video:
Here, it says, calculate the nuclear binding energy in mega electron volts per mole of beryllium 10. Here we're told the atomic mass of beryllium 10 is 10.0135347 AM U. All right. So step zero is we'll repeat steps 1 to 3 of the previous topic to calculate the mass defect of the radio isotope. So beryllium 10. So beryllium has an atomic number four, it has four protons, four electrons and 10 minus 46 neutrons. Here, we would find out what the complete total mass of all of these subatomic particles are to help us determine our predicted mass. So here we'd multiply them by their masses in AM U. So we multiply them across and we add them all together. That's gonna give us our predicted mass. So our predicted mass equals 10.08324 A U. Here, they're giving us the atomic mass. The atomic mass is related to our nuclear mass, which is 10.0135347 emu subtracting these two gives us our mass defect, which is m the mass defect here will be 0.0697053 AM U. All right. So now that we have that what we would do next is we would have to convert AM U into kilograms. And here we have a conversion, that 1 a.m. U is equal to 1.66 times 10 to the negative 27 kg. All right. So here we're going to say 1 a.m. U is 1.66 times 10 to the minus 27 kg. So when we do that, we would get 1.1571 times 10 to the negative 28 kg. This is important to remember because re remember one jewel is equal to kilograms times meter squared over second squared. That's why we need to convert AM U into kilograms. At this point, we'd say that our nuclear binding energy is equal to mass defect times speed of light squared. So plug in what we just found for mass defect times our speed of light squared. When we do that, that's going to give me 1.041 times 10 to the negative 11 kg times meter squared over second squared or joules. Remember when we're using this formula for calculating nuclear binding energy, it is equal to one mole of that radio isotope. So this is joules per one mole. Now, in this question, they don't want joules per mole, they want mega electron volts per mole. So we use the other conversion factor that we have here where we have one mega electron volt equal to 1.60 times 10 to the negative 13 joules. Here, Jules would cancel out and we'd be left with mega electron volts per mole. Here, this would come out to b 65.087 mega electron volts per mole. So this would be our final answer.
3
Problem
Problem
Calcium-41 is commonly used radioisotope in the study of osteoporosis. If calcium-41 has a mass of 40.962278 amu, determine the nuclear binding energy per nucleon in MeV. (1 amu = 1.66 x 10-27 kg). (1 MeV = 1.60 x 10-13 J)
A
17.5537 MeV/nucleon
B
5.75533 MeV/nucleon
C
8.56276 MeV/nucleon
D
7.87683 MeV/nucleon
4
Problem
Problem
Calculate the mass defect (in g/mol) for the formation of a helium-6 nucleus, and calculate the binding energy in (MeV)/nucleon. (1 amu = 1.66 x 10-27 kg). (1 neutron = 1.00866 amu 1 proton = 1.00727 amu, & 1 electron = 0.00055 amu) (1 MeV = 1.60 x 10-13 J).
A
0.04267 g/mol, 2.122 MeV/nucleon
B
0.05318 g/mol, 7.050 MeV/nucleon
C
0.05219 g/mol, 8.017 MeV/nucleon
D
0.05138 g/mol, 7.996 MeV/nucleon
Do you want more practice?
We have more practice problems on Nuclear Binding Energy