The atomic masses of hydrogen-2 (deuterium), helium-4, and lithium-6 are 2.014102 amu, 4.002602 amu, and 6.0151228 amu, respectively. For each isotope, calculate

(c) the nuclear binding energy per nucleon.

Video transcript

Hello, everyone. Today, we have a following problem given the following atomic masses of hydrogen three, Brilli nine and lithium seven respectively, find the nuclear binding energy per nucleon and six significant figures for each of the following isotopes. First, we have hydrogen three, followed by Brilli nine and then lithium seven. So the first step is we want to find the nuclear mass for each of these isotopes. And we will do so by first subtracting the mass of the electrons from the atomic mass that we were given. So for example, for our hydrogen three, we have the given mass of 3.0160493 AM U or atomic mass units, we subtract that from the one electron present or the atomic number which is one. And then we multiply that by the mass of an electron which is 0.00055 AM U. This will give us a value of three point 015 or 993 AM U or atomic mass unit. Her beryllium nine, we will employ the same strategy. We will take the 9.0121821 AM U subtract that by the atomic number of Brilli which is three or which is four, multiplied by the mass of an electron which is 0.00055 AM U to give us a nuclear mass of 9.0099821 atomic mass units. And then for our lithium seven, we would take the given mass which is 7.01600 40 AM U and subtract that from the atomic number of lithium, which is three and then multiply by the mass of an electron. And this will give us a total nuclear mass for lithium seven of 7.143540 atomic mass units. Then we can find the nuclear binding energy. And to do this, we have to find a mass difference between the nucleus and the separate nucleus and then convert that mass difference into energy. And so we can do that. That's our second step is to find the nuclear binding energy. And we will use the equation that we have the energy released by particle release is equal to the change in mass associated with individual neutrons and protons multiplied by the speed of light squared. So before we do that though, we have to find the mass of individual neutrons and protons or we have to find the mass change. So for our hydrogen three to find our mass change, this is going to be equal to the energy released bari the mass of individual neutrons and protons subtracted by the nuclear mass. And so, for example, we have two because t was associated with the number of protons and neutrons that are in hydrogen, we will then multiply that by the one 0.00866 atomic mass units. And this value is from a nuclear mass, we will then add one because of the number or atomic number of hydrogen. And we multiply that by 1.00727 AM U which is essentially the mass of individual protons. And then lastly, we subtract that value by 3.015 or 993 AM U for a total of 0.0090907 atomic mass units. And we will employ the strategy for the rest of our molecules. So we have beryllium. Next, this will be five because uh that is the mass number of the neutrons plus the protons. We then multiply that by 1.00866 AM U, we add that to four times 1.00727 atomic mass units. And then we lastly subtract by the nuclear mass which is 9.0099821 atomic mass units or a total of 0.06 23979 atomic mass units. And then lastly, we have the mass change for lithium seven. This will be, or which is then multiplied by 1.00866 AM U. We add that to three times 1.00727 AM U. And then we subtract it from the nuclear mass which is 7.01 43540 AM U for a total mas change of 0.042096 AM U. Now we can plug this into our nuclear binding equation. So for our hydrogen three, we would have the change in energy is equal to our 0.0090907 atomic mass units. We divide that by the speed of light which is a 3.00 times 10 to the eighth. And this will be meters per second. And then this is a square. This gives us an answer of 1.358 times 10 to the negative 12. This can be in unit of kilograms times meters squared divided by second squared. We'll do the same thing for our burle nine or we will have our 0.0623979 atomic mass units multiplied by the speed of light. And then squared to give us a value of 9.322 times 10 to the negative 12 kg times a meter squared divided by a second squared. And then lastly, we have our lithium seven and this will be zero point 04, 2096 AM U multiplied by the speed of light. And then this is squared to give us a value our nuclear binding energy of 6.3289 times 10 to the negative 12 kg times meters squared divided by second squared. And then to find the nuclear binding binding energy per nucleon, we will simply just divide the nuclear binding energy by the total number of nucleons. So per hour first, we will actually do this down here. So we have for our hydrogen three, we can divide that value that nuclear binding energy by three nucleons. And this will give us an answer of 4.5 271. I just know it should be 27, 271 seven for the six significant figures times 10 to the negative 13th. And this can also be in units of jewels. We then divide our brilli nines nuclear binding energy by the nine nucleons which will give us a value of 1.03 58 one for our significant figures times 10 to the negative 12 joules. And then lastly, for our lithium seven, we can divide that nuclear binding energy by seven nucleons which will give us a total number or a total answer of 8.984 or nine times 10 to the negative 13th joules. And this will be Jes per nucleon. And with that, we have solved the problem overall. I hope this helped and hands on next time.