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) 50Cr (atomic mass = 49.94605)
(b) 64Zn (atomic mass = 63.92915)

Question

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) 50Cr (atomic mass = 49.94605)
(b) 64Zn (atomic mass = 63.92915)

Accepted Answer

Welcome back everyone. We need to provide the mass defects in AM US and binding energies and killed jules per mole for rhodium 101 with an atomic mass of 100. and palladium 107 with an atomic mass of determine which is more stable. We're going to use the term for the speed of light, C as 299,792, m per second. Let's begin by imagining our nucleus of any atom and recall that within our nucleus are protons and neutrons are blinded to one another. They're bound together by a strong nuclear force and together they're known as nucleons within the nucleus. We next want to recall that binding energy is represented as delta E. And recall that are binding energy is our energy required to separate our nucleons. Next, we want to recall that in the development of a nucleus, the mass of our nucleons gets converted to energy, which is why there's a relationship defined by Einstein where mass is equal to energy. Lastly, we want to recall that with increasing binding energy of a nucleus, we have an increase in nuclear stability. And this is also associated with a decrease in potential energy of our nucleus because the nucleus will then be more difficult to break open or to break apart. So now with all of these facts outlined, let's begin by recalling how to calculate the mass of our nucleons within our isotopes. So in this case we want to recall that are massive, are nucleons is equal to our number of protons multiplied by the mass of a proton. And added to this, we would have our number of neutrons in our isotope, multiplied by the mass of a neutron. And so analyzing our isotope of rhodium 101. Recall that our mass number represented by the symbol a is equal to our number of protons plus our number of neutrons. We also want to recall that our atomic number represented by the symbol Z. For rhodium on the periodic table is equal to 45. And that would therefore tell us that we have 45 protons in rhodium one oh one. Notice that we also have a neutral atom of rhodium. And so that would also therefore tell us that we have 45 electrons since in neutral atoms atomic number will equal your number of protons and electrons. And so next we just need to solve for a number of neutrons. So because we know that our mass number is equal to the sum of protons and neutrons, we can write an equation where our mass number of 101 is equal to our number of protons 45 plus N, where N represents our number of neutrons. So solving for N, we would subtract 45 from both sides of our equation and we'll find that we have a total number of 56 neutrons in our atom of rhodium +101. So now we can go into calculating are Mass of Nucleons in Rhodium 101. And what we will find is that our number of protons, which we determined to be 45, multiplied by the mass of a proton, which in our textbooks is a value of 1.007-8 am use. And then adding to this, we have our number of neutrons which we determined to be 56 neutrons multiplied by the mass of a neutron in our textbooks as 1.866 A m U s. And so from this equation will find that are some will give us a value of 101.81256 am use as our massive nucleons and rhodium 101 Next we want to recall that to solve for our mass defect represented by Delta M. for Rhodium 101. We would take the mass of the nucleons in our isotope and this is subtracted from our mass of rhodium 101 nucleus, which we don't know. So we need to solve for the mass of rhodium one on ones nucleus. Next. So we'll do that here. Below the mass of rhodium 101 nucleus Where we would recall that we can find this by taking our mass number of rhodium 101 represented by a Which is subtracted from our number of electrons in Rhodium 101 times the mass of an electron. And so plugging in everything that we know, we're given our mass or mass number of rhodium in the prompt as 100.90616 AM use this is subtracted from our number of electrons and are neutral atom of rhodium 101 which we determined to be 45 electrons. Since we have an atomic number of 45 and then our mass of an electron we should recall in our textbooks is the value of 5. times 10 to the negative fourth power atomic mass units. And so taking the difference here, we would find a result of 100. A. M. Use for our mass of rhodium 101 nucleus. And so next we can go into calculating our mass defect delta M. For rhodium 101. Now that we know both of our variables are massive nucleons which we determined to be a value above as 101. We'll use the same color one oh 1. atomic mass units. Subtracted from our mass of rhodium one on one nucleus which we just determined above as 100.88 1473 atomic mass units. And so from this difference we'll find that our mass defective rhodium 101 is equal 2.931087 in scientific notation that's 9.31087 times 10 to the negative first power. Which we can round to about two sig figs as 20. AM. Use for our mass defect of rhodium 101. This will be our first answer here and now. We just need to calculate its binding energy. Recall that to calculate the binding energy of our atom Of Rhodium 101. We would take that and set it equal to our mast effect delta M multiplied by the speed of light C squared recall that our unit for mass defect in this formula should be in kilograms. Recall that for our binding energy our unit should be in joules per mole and recall that one a.m. U. Is equal to one g per mole. So going into our calculation for the binding energy delta E. For rhodium 101 plugging in our mass defect which above we found as 9.31087 times 10 to the negative first power. Instead of A. M. Us. We know that one a.m. U. Is equivalent to one g per mole. So we'll say that this is in units of grams per mole. But we're going to need to convert from grams in the denominator two kg in the numerator. Recall that for one g we have an equivalent of 10 to the negative third kilograms. So canceling out grams let's now plug in our speed of light, which in the prompt we're told to use a value of 299,792, m per second. And then our unit is squared here. And so what we will find is that this product gives us a value of 8.368, 19 times 10 to the 13th Power. Which we can round to about 8.47 times 10 to the 13th power. And our units that we are left with are kilograms times meters squared divided by seconds squared times moles. But as we stated earlier, we need this to be in units of killed jules per mole. So we're going to recalculate our mass defect for rhodium where we would take our value of 8.36819 times 10 to the 13th power kilograms times meters squared, divided by seconds squared times moles. And we want to recall our conversion factor where one kg times meter squared, divided by seconds squared is equivalent to one Juul in the numerator. And so we can go ahead and cancel out our unit of kilograms meters squared and second squared. And we're left with joules per mole in which we can multiply by our next conversion factor to go from jules and the denominator to kill a jewels in the numerator. Recall that for one jewel we have an equivalent of 10 to the negative third jewels. So canceling out, sorry, killer jewels and canceling out jewels were left with killer joules per mole as our final unit. And from this product will get a value of 8.47 times 10 to the 10th power kila jewels per mole as our binding energy for rhodium 101. So this is our second answer so far. So let's keep going and consider our second isotope palladium 107. So considering palladium 107, recall that are mass number of 107 tells us are some of protons and neutrons. And so if we know our mass number is 107, we can set up an equation where we just need to write down our atomic number. So our Z value for palladium, which on our periodic table will see is a value of 46 for the atomic number. Therefore telling us that we have 46 protons. So we would have 107 equal to 46 plus and for a number of neutrons, solving for N. And we subtract 46 from both sides of our equation and we'll find that we have a total number of 61 neutrons in palladium +107. So, going back to our atomic number because we know that we have a neutral atom Of palladium, we therefore know that we also have 46 electrons. So beginning the calculation to find our mass of our nucleons in palladium 107. So just like before we would take our number of protons times the mass of a proton. And added to this are number of neutrons multiplied by the mass of a neutron. I'm sorry, this is mn for mass of a neutron. So plugging in our information we have for our number of protons in palladium 107. We determined that to be 46, multiplied by the mass of a proton, which in our textbooks is a value of 1.007-8 am use. And adding to this are number of neutrons which we determined to be 61 for palladium 107 multiplied by the mass of a neutron in our textbooks as 1.866 A. M. Use. And from this some we would get a value equal to 107. AM us. So next we'll calculate our mass of palladium 107 nucleus Which recalled just like before we take our mass number which is given to us in our name for palladium 107. Subtracted from our number of electrons in our atoms multiplied by the mass of an electron. So plugging in our mass number which given in the prompt for palladium 107 is 106.90513 AM use. We are subtracting this from our number of electrons in platinum 107 which we determined to be 46. Multiplied by the mass of an electron which in our textbooks is 1.866 AM use. And so from this some we get a value for the mass of nucleons. Or sorry for the mass of our palladium 107 nucleus equal to a value of 106 point and apologies I made of air here. So plugging in our mass of an electron, I just want to correct myself and clarify that in our textbooks. That should be a value of 5.486 times 10 to the negative fourth power atomic mass units. And so now from this difference we get our mass of our nucleus of palladium 107 equal to 106. AM. Use now with this information we can finally calculate our mass defect of palladium 107 In which we again take our mass of our nucleons. Subtracted from our mass of palladium 107 nucleus. And this says nucleus. And so plugging in our values for the mass of nucleons, we determined that to be one oh 7. am Use. And subtracting this from our mass of platinum 107 nucleus which we determined to be 106.8798944 AM use from this difference we get a value equal to 0.983246 which in scientific notation is 9.832 46 times 10 to the negative first. Power and rounding our decimal to at least 266. We get 2660.98 AM use as our mass defect of palladium 107 which is going to be our third answer. So now we just need to find its binding energy. So just as before recall that the binding energy for this time palladium 107 is equal to our mass defect delta M multiplied by the speed of light squared. So for our mass defect of palladium 107 will plug in our value for the mass defect which above we calculated in scientific notation as 9.83246 times 10 to the negative first. Power and recall again that in this case we want to interpret our unit of am us as equivalent to one g Permal. So we have 9.83246 times 10 to the negative first power g per mole. And multiplying to convert from grams in the denominator two kg in the numerator, recall that for one g we have 10 to the negative third power kilograms. So canceling out grams we're left with kilograms per mole. And now we can plug in our speed of light given in the prompt as 299,792, m per second. And this is squared. So from the product here will get a value of 8.84 times 10 to the 13th power. And our units are kilograms times meters squared, divided by seconds squared times moles. And so we just need to ask before convert this to kill a jules Permal. And so taking our mass defect again of palladium we have 8.84 times 10 to the 13th power kilograms times meters squared, divided by second squared times most which we should recall. Our conversion factor. That we're going to multiply by where one kg times meters squared, divided by seconds squared is equivalent to one Juul. So canceling our kilograms meter squared and then second squared we have jewels per mole. And so we can multiply by our next conversion factor to go from jewels in the denominator To kill a jewels in the numerator. Where one jewelry recall is equivalent to 10 to the negative third power modules. So canceling out jewels were left with our final units of kg per mole for our finding energy. And we get a result equal to 8.84 times 10 to the 10th power killer, joules per mole. And this would be our fourth final answer. And now we just need our last final answer which is determined which isotope is more stable. And so we can note down that our binding energy for palladium 107, which we calculated to be 8.84 times 10 to the nine times 10 to the positive 10th power killer joules per mole is greater than our binding energy for rhodium 101 which we found above to be 8.47 times 10 to the 10th power killed jules Permal. And so because our binding energy for palladium 107 is greater than therefore palladium 107 is more or is the most stable isotope. And so this would be our fifth final answer to complete this example. So all of our final answers highlighted in yellow correspond to choice A as the correct answer to complete this problem. I hope this made sense and let us know if you have any questions.

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) 50Cr (atomic mass = 49.94605) (b) 64Zn (atomic mass = 63.92915)

Verified Solution

Key Concepts

Mass Defect

Binding Energy

Stability of Nuclei