Calculate the mass defect (in g/mol) and the binding energy (in MeV/nucleon) for the following nuclei. Which of the two is more stable? (a) 7Li (atomic mass = 7.016004) (b) 39K (atomic mass = 38.963706)

Video transcript

Welcome back everyone. We need to find the mass defects in atomic mass units. And binding energies in kg per mole for Molybdenum 84 with an atomic mass of 83.9401. And Zirconium 79 with an atomic mass of 78.949. We need to determine which isotope is more stable and use the speed of light equal to 299,792,458 m per 2nd. 1st focusing on our isotope of 84 molybdenum M. O. We want to begin by imagining our nucleus of molybdenum represented by this circle and within our nucleus recall that we have protons bounded to neutrons and as a unit are protons and neutrons bound together represent nucleons. So we're going to recall that our binding energy represented by delta E. Is describing the energy required to separate our nucleons. And recall that as our binding energy increases, this corresponds to an increase in nuclei stability, which also corresponds to a decrease in the potential energy of our nucleus, meaning that the nucleus would be more difficult to break for that atom. So, based on what we've outlined, we're going to need to find the mass of our nucleons in 84 molybdenum recall that we can calculate this by taking our number of protons multiplied by the mass of a proton, which is added to our number of neutrons in our nucleus added to. Or sorry, multiplied by the mass of a neutron. So focusing on our Molybdenum 84 isotope. Recall that our number given in our isotope name 84 represents our mass number represented by the symbol a or atomic mass. And recall that this number is calculated by taking the sum between our protons to neutrons in our nucleus. So next we want to note our Z value or atomic number of molybdenum. And on our periodic table we would find an atomic number for molybdenum of 42. So Z is equal to 40 to recall that our atomic number therefore tells us our number of protons. So we would therefore have 42 protons. And because we have a neutral atom of Molybdenum, there's no charge. We therefore also have an equivalent of 42 electrons. And so we can write out an equation to solve for our number of neutrons, where our mass number of 84 is equal to r protons plus N being our number of neutrons. So subtracting 42 from both sides of this equation, we would find that our number of neutrons equals of value of 42. Now that we know our number of neutrons and protons, we can calculate our mass of nucleons for molybdenum 84. So beginning with our number of protons, which we determined to be 42 protons, we're going to multiply by the mass of a proton, which we should recall is a value from our textbooks of 1.7 to 8 atomic mass units. This is added to our number of neutrons which we determined to be also 42 neutrons. And in this case we want to multiply by the mass of a neutron which from our textbooks we should recall is a value of 1.866 A M. U. S. So taking the sum here and products of our parentheses, we would find the mass of our nucleons equal to a value of 84.6695 atomic mass units. Now we need to find the mass of our nucleus For 80 for Molybdenum recall that we can find this by using the formula where we take our Atomic mass of Molybdenum 84 given in the prompt, subtracted from our number of electrons in this isotope multiplied by the mass of an electron. So plugging in our knowns, We can calculate the mass of our nucleus of Molybdenum 84 by plugging in its given atomic mass from the prompt, 83.9401 AM use. This is subtracted from our number of electrons which we determined for neutral molybdenum 84 being 42 electrons multiplied by the mass of an electron. Which we should recall is from our textbooks a value of 5.486 times 10 to the negative fourth power A. M use. And in our calculators we would find a value equal to 83.9171 AM use now that we have this information. Let's go back to our prompt and recall that mass defect is represented by delta M. That's what we're going to find next. And we want to recall that to find mass defect delta M. We would use the formula where we take the difference between our mass of our nucleons for our isotope subtracted from the mass of our isotopes nucleus. So in this case for the mast effect of Molybdenum, 84 we have the mass of nucleons which we determined to be 84.6695 AM use subtracted from our mass of mulet. Denham's 84. Um sorry molybdenum 80 four's nucleus which we determined to be the value 83.9171 atomic mass units. And actually before I continue, I'd like to make a correction for our calculation of our mass of the nucleus of Molybdenum 84 because we should keep as many decimals as possible. So we would have 83.9170588. So taking the difference, we have our mass of our nucleons 84.6695 minus 83.9170588 for the mass of molybdenum four's nucleus. And we would get a value of 0.752441 AM. Use which we can round to scientific notation as 7.52 atomic mass units. And sorry that's times 10 to the negative one Atomic mass units. Or as three significant figures, we would just have 7.52 times 10 to the negative first power AM us. And rather we'll keep our decimal as what we use for 66. So we'll have about 0. AM us as our mass defect. And this would be our first answer for our mass defect of molybdenum 84. Now we need to calculate its binding energy. So we're going to use our mass energy equivalence formula for the delta e binding energy of molybdenum 84 recalled that our mass energy equivalence formula states that we can find our binding energy by taking our mass defect delta M multiplied by the speed of light squared. So for our binding energy of molybdenum we would plug in our mass defect which we calculated above in scientific notation as 7.524412 times 10 to the negative first power am us. We want to use the long decimal value and actually note that our units should not be a M. U. S. For mass defect here. For our mass energy equivalence formula recall that mass effect should have units of grams per mole and we would recall that one a.m. U. Is actually equivalent to one g per mole. So we can interpret our mass defect in units of grams per mole here. This is then multiplied by the speed of light, in which from the prompt we're told to use the value 299,792, m/s squared. We want to recall that we need our unit of binding energy to be in killer joules per mole. So we're actually going to need to recall a second conversion factor in which one jewel is equivalent to one kg times meters squared, divided by seconds squared. And also a third conversion factor where one jewel is equivalent to one times 10 to the negative third power. Killing jules. So well actually expand our multiplier here. Where for our mass defect we want to convert from grams to mole two grams in the denominator, two kg in the numerator. Which is going to be our conversion factor. That we want to multiply by here. And so recall that for one g are prefix kilo tells us we have 10 to the negative third kilograms. So canceling out grams, we have kilograms per mole times meters squared, divided by seconds squared. So we're going to simplify this in our calculators by taking the products here and we would result in a value equal to 6.76 times to the positive 13th power. And our units are kilograms times meters squared divided by moles times seconds squared. So next we will worry about making that conversion of these units into killing joules per mole. So we're going to restart our formula for mass effect of We're sorry for binding energy delta E. Of molybdenum 84. Beginning with our calculation from before 6.76 times 10 to the positive 13th power kilograms times meters squared. Divided by moles times seconds squared. We're going to incorporate our first conversion factor where we were called that one kg times meters squared, divided by seconds squared is equivalent to one Juul in the numerator. And this allows us to cancel out kilograms meter squared and second squared. Leaving us with joules per mole. But we need kilograms per mole. So we're going to multiply by our next conversion factor where we recall that one jewel in the denominator is equivalent to 10 to the negative third power killer jewels in our numerator, canceling out jewels were left with killer joules per mole as our final units. Which is what we want for binding energy. And this will result in a binding energy equal to a value of 6.76 times 10 to the 10th power kila jules Permal as our second answer so far for our binding energy of our isotope molybdenum 84. So we'll separate our work here and focus on our next isotope zirconium 79. Based on its isotopes symbol, we have a mass number of 79 Which equals the sum between our protons and neutrons noting down the atomic number of Zirconium Z, we would find a value on our product table when we find zirconium equal to 40, meaning that therefore we have 40 protons. And because we have neutral zirconium we therefore have 40 electrons as well. So finding our some of or sorry number of neutrons. We're going to plug in our mass number of 79 equal to r 40 protons plus n. Subtracting 40 from both sides. We would find that our number of neutrons is equal to a value of neutrons. So beginning with finding the mass of our nucleons In Zirconium 79, We would take our number of protons which we determined to be 40 protons multiplied by the mass of a proton from our textbooks. 1.007-8 am use added to our number of neutrons, which we determined to be 39 neutrons multiplied by the mass of a neutron from our textbooks which we find is 1.866 A m U s. Taking the product and some here we would find a value equal 79.6- AM us as our mass of nucleons in Zirconium 79. Now moving on to find our mass Of our nucleus for Zirconium 79. Justice Before we take the atomic mass we'll just say a here for shorthand of Zirconium 79 which is subtracted from our number of electrons in zirconium 79 times the mass of an electron. So plugging in the given atomic mass from the prompt for zirconium 79. We have a value given as 78.949 atomic mass units. This is subtracted from our number of electrons which we determined for neutral Zirconium 79 being 40 electrons multiplied by the mass of an electron which we should recall from our textbooks is 5.486 times 10 to the negative fourth power atomic mass units. And this will simplify to our result for the mass of our nucleus of zirconium as 78.927056 AM us. And again keep as many decimals as possible for this calculation. So now we can go right into getting our mass defect. So delta M for zirconium 79 which as before we're going to take our mass of our nucleons Of Zirconium 79. Subtracted from our mass of our nucleus of Zirconium 79. So, plugging in our values for the mass of our nucleons above, we determined that to be a value of 79.6-894 AM us for our nucleons and then for the mass of our nucleus. We determined that to be above for zirconium 79 as 78.927056 AM Use. And this will give us this difference will result in our mass defect equal to a value of 0. am use in scientific notation. We can say that this is about 7. 884 times 10 to the negative first power am use and rounding our decimal two at least two sig figs. We would say that this is about 20.70 AM use so for our third answer we have our mass defect of zirconium 79 as our highlighted decimal 790.70 AM use. Now let's move into calculating its binding energy delta E For Zirconium 79. Just as before we take our mast effect delta m multiplied by the speed of light squared. And so plugging in our mass defect which we calculated above in scientific notation as 7.1884 times 10 to the negative first power instead of a muse because we know one a.m. U. Is equivalent to one g per mole. Our units are going to be grams per mole here just as before. We want to multiply by the conversion factor to go from grams in the denominator two kg in the numerator where recall that one g is equivalent to 10 to the negative third kilograms. And then just as before we plug in our speed of light term given in the prompt as 299,792, m per second. And this is squared, showing out our unit cancelation so far we can get rid of grams. We're left with kilograms times meters squared divided by moles times second squared. And in our calculators we will find a results for our binding energy equal to about 6.31 times 10 to the positive 13th power with units of kilograms times meters squared, divided by moles times seconds squared. And just as before recalled at our unit for binding energy should be in kila jules Permal. So we're going to need to convert this into killer jewels per mole. So just as before we're going to rewrite things so that we have our binding energy of zirconium 79 equal to our calculated value above 6.31 times 10 to the 13th power kilograms times meters squared, divided by moles times seconds squared. Multiplying by our first conversion factor where justice before we recalled that for one kg times meter squared, divided by seconds squared, we have an equivalent in the numerator of one jewel. So canceling out units of kilograms meter squared and second squared, we're left with joules per mole. And we need killer joules per mole. So we're going to multiply by our second conversion factor from before where one jewel in the denominator has an equivalence of 10 to the negative third power killer jewels. In our numerator, canceling out jewels were left with killer joules per mole as our final units for our binding energy. And in our calculators. This will simplify to a value of 6.31 times 10 to the 10th power kayla jules Permal as our binding energy of zirconium 79. And this would be our fourth final answer. So all of our final answers highlighted in yellow cover the mass defect and binding energies of each of our isotopes. And next we need to determine which isotope is more stable again, as we stated earlier, the greater the value of our binding energy delta E, the more stable the nucleus, meaning it's harder to break apart for our atom. And so because we have the following understanding where we can say that the binding energy delta E. For molybdenum 84 Which we calculated above as 6.76 times 10 to the 10th power killing joules per mole. This is less than our binding energy of zirconium 79 which we calculated above as 6.31 times 10 to the 10th power joules per mole. We can say that therefore this means that the greater binding energy associated with zirconium 79 being the more stable isotope. And this would be our 5th final answer. So all of our answers will correspond to choice. The in the multiple choice as the correct answer. I hope that this all made sense. And please let us know if you have any questions