The nuclear masses of 7Be, 9Be, and 10Be are 7.0147, 9.0100, and 10.0113 amu, respectively. Which of these nuclei has the largest binding energy per nucleon?

Video transcript

Welcome back everyone in this example we have 9.1333 A. M. Use 10.1294 A. M. U. S. And 11.931 am use are all the nuclear masses for boron nine boron 10 and born 11 respectively, determined the nuclei with the highest nuclear binding energy from the given. So our first step is to recall the formula for binding energy. Recall that binding energy is represented by delta E. And it's calculated by taking our mass defect. So that's delta M. Multiplied by the speed of light squared. So C squared. So we can see clearly from this relationship that are binding energy is going to be directly related to our mast effect. So this is a direct relationship. And because we've outlined that, let's begin with our first isotope given as Boron nine. So when we find boron on the periodic table, we see that it corresponds to atomic number five. So we'll fill that in recall that the atomic number is written in the left hand subscript. And recall that our atomic number being our Z. Value here because it's typically represented by the term Z. Tells us that we have five protons. And if we know that we have five protons and we look at our mass number of nine given in the isotope name from the prompt recall that our mass number which is represented by the symbol A. Is equal to our number of protons plus our number of neutrons, meaning that we could say that our number of neutrons is equal to the difference between our mass number nine minus r five protons. And this is going to give us a total of four neutrons. And so now we can begin by calculating atomic mass and let's spell this clearly. So this is atomic mass. And we would calculate it by taking our five protons from our atomic number of moron. Multiplying this by The mass of a proton. Which we should recall from our textbooks or lectures is the value 1.072765 am use. And then we're going to add this to our four neutrons. I'm sorry this should be a four here for our four neutrons. So this is the number of neutrons here. So we would take our four neutrons and we're going to multiply it by the mass of a neutron which we should recall from our textbooks or lecture is the value one point oh 86649 A. M. Use And taking the product and some here we're going to have a atomic mass for our born nine equal to the value 9.07104-1. And this is zero here to one a.m. use. Alright now that we have this atomic mass, we want to get our total mass defect. So that's total delta M. So we're going to begin by recalling that to calculate mass defect. We're going to take our atomic mass which is what we just found subtracted from our nuclear mass which is given in the prompt. And so we're going to begin with that atomic mass. We just calculated as 9.07104-1 am use. And we're going to subtract this from the nuclear mass given in the prompt for boron nine as this first value here. So we have our nuclear mass being nine point oh 1333 am us. As given in the prompt for braun nine. And from this difference we calculate our total mass defect equal to the value 0. am. Us. Now that we have our total mass defect, we're going to go ahead and calculate our binding energy from our formula. So recall binding energy again is delta E. Equal to We're going to take that total mass defect that we calculated at 0.5771 to one a.m. U. S. And we're going to divide this by our mass number of boron are isotope boron which we stated is nine. So we'll interpret that as nine nucleons. So this mass total mass defect is per nine nucleons of our brawn nine isotope. Next we want to go ahead and get rid of that unit am use because as according to the prompt we want to calculate the binding energy for each isotope to find the highest binding energy. And recall that binding energy has units of jewels per nucleons. So these are the units for binding energy. And so we're going to recall that we can use avocados number to get rid of this AM U. Unit. So recall that avocados number tells us that we have six point oh 22 times 10 to the 23rd power AM use equivalent to one g. So we can now cancel out AM us and go from grams to kilograms. Where we recall that we have 10 to the third power grams for one of our kilograms. This allows us to get rid of grams. Now we will focus on incorporating that speed of light term from our binding energy formula where we plug in the speed of light as the value 2.99 times 10 to the eighth power with units of meters per second. And this is squared according to our formula for binding energy And just so everything is visible. We're going to scoot this calculation over for enough room. Okay now that we have enough room for our calculation, we're going to focus on canceling out units. So let's recognize that we have so far gotten rid of um us and grams right now what we have going on is kilograms multiplied by meters squared times seconds squared divided by nucleons. And if we recognize this we're going to realize that one jewel is actually equivalent to a kilogram times meters squared times meters squared divided by and sorry, this is a square power here, meter squared divided by seconds squared. So this conversion factor is equivalent to one Juul meaning that when we calculate our binding energy, we're going to be left with the units of joules per nucleons. And so we would have a value of 9. times 10 to the negative 29th power. And sorry, this is times 10 times 10 to the negative 29th power with units of jewels per new Cleon as our first binding energy for boron nine. So let's continue on to our next isotope of boron. Given in the prompt as boron 10. And then we have born 11 next. So now we have boron 10. Again, this corresponds to atomic number five. Because as we find brought on the periodic table, we see that we have a atomic number of five calculating our number of neutrons. We take the difference between our mass number of 10 minus r five protons from the atomic number which gives us a total of five neutrons. And now going back to calculate atomic mass for boron 10, we take our five protons multiplied by and let's use the color purple are five protons multiplied by the mass of a proton which from our textbooks is 1.72765 A. M. Use added to our five neutrons now multiplied by the mass of a neutron 1.86649 A. M. Use. And so taking the product and some here we're going to get a value equal to 10.0797 a. m. use. And actually just a correction here that's 10.0797. And then we have 07 again am us. So now calculating our total mass defect total delta M. We take the difference between our calculated atomic mass as 10.79707 A. M. Use subtracted from our given nuclear mass for boron 10 from the prompt of 10. A. M. Use. And so now when we take this difference here, we're going to have a total mass defect equal to the value of 0.66767 A. M. Use. And now finally moving in to calculate our delta or binding energy. We take our total mass defect of 0.66767 AM use which we are interpreting per hour mass number of 10 nucleons of our isotope of boron 10. This is multiplied by our first conversion factor. And let's just make some room here. So things are easily interpreted. So we have our first conversion factor to cancel out am us using a Vos number six point oh 22 times 10 to the 23rd power AM us which we understand is equivalent to one g. We can get rid of AM us and multiply by our conversion factor to get rid of graham's where we see that 10th of the third power grams is equal to one kg, canceling out grams. We plug in our last conversion factor which is C squared, which we recall is for the speed of light. A value of 2. times 10 to the eighth power units of meters per second. And this is squared and this we understand again is equivalent where we have kilograms times meters square divided by seconds squared, equivalent to one Juul as we stated earlier. And so from this calculation we're going to have a binding energy of 9.96466 times 10 to the negative 13th power. And we have units of joules per nucleons. And just to go back to our first calculation, I just realized I left out the unit for 10 to the eighth power in our speed of light. So this would change our binding energy for more on nine To the value of 9. times 10 to the negative 13th power jewels per nucleons. So sorry about that mistake. And let's keep going to our last isotope which we have from the prompt as Braun 11. So the mass number is born 11. From given in the isotope name again. We have an atomic number of five. And so calculating our number of neutrons. We take the difference of the mass number of 11 minus R five protons from the atomic number giving us a total of six neutrons. And so now moving on to our next step, we want to calculate our atomic mass. Now that we know our number of neutrons, we would say that we have from our five protons multiplied by the mass number of a proton from our textbooks. 1.72765 A. M. Use added to our six neutrons multiplied by the mass of a neutron from our textbooks. 1.86649 A. M. Use getting the product and some here we're going to have a value equal to 11. A. M. Use now that we have our atomic mass. We can go into calculating total mass defect total delta M. Equal to the difference between our calculated atomic mass of 11.883719 AM Use subtracted from our nuclear mass given in the prompt for born 11 as 11. am use from this difference we have a total mass defect equal to 0. A. M. Us. Now that we have this difference we're going to go into our binding energy or delta. The formula where we take that total mass effect. 0.790619 A. M. Us for born 11 which we interpret as per hour nucleons from our atomic mass of born 11. We're multiplying this to get rid of that am you term by using a Vos number six point oh 22 times 10 to the 23rd power AM Us. Which we are interpreting equivalent to one g canceling out Am Us. We're going to get rid of grams Now by recalling that 10 to the third power grams is equivalent to one kg and making some room here. We plug in to get rid of Now kilograms and sorry, this should be cancelation of grams here. So to get not to get rid of kilograms. Actually to plug in our speed of light term C squared. We recall that the speed of light is 2.99792458 times 10 to the eighth power units of meters per second. This is squared So we would now have again, kilograms times meters squared, divided by seconds squared. Which we understand is equivalent to one Juul. And this is going to yield a binding energy for boron 11 equal to 1.7269 times 10 to the negative 12 power with units of joules per nucleons. And so going back to the focus of our prompt which was to find the nuclei with the highest new Cleon binding energy. From our calculations we will see that for Brandi nine we have a nuclear binding energy sorry of 9.57 times 10 to the negative 13th power for boron 10. We have a binding energy of 9.96 times 10 to the negative 13th power. So so far that is a higher binding energy than Born nine. Moving on to our last binding energy for Born 11, we have a value of 1.7 times 10 to the negative 12 power jewels per nucleons. And because this is less negative with a power of 10 to the negative 12, we will conclude that again because of this power of 10 to the negative 12 we would say. Therefore born 11 has the highest binding energy again represented by delta E. And so for our final answer we have our isotope born 11 with the highest finding energy. This will complete this example as our final answer. If you have any questions leave them down below and I will see everyone in the next practice video.