Fraud in science is rare but does happen occasionally. In 1999, the creation of three superheavy elements (one new) was claimed when 208Pb was bombarded with 86Kr. The claim was subsequently found to be fraudulent and was with-drawn. Identify the isotopes X, Y, and Z that were claimed.

Video transcript

Hey everyone, we're told that three hypothetical super heavy elements were created when uranium 1 99 was bombarded with calcium 36. What are the X. Y. And Z. Isotopes? Note these isotopes do not appear in the periodic table. As we've learned. We know that an element can be shown as such with X. Being or element a being our mass number, nz being our atomic number. So first let's go ahead and determine our isotope X. We can do so by taking our calcium 36 And our uranium 1 99. And we can go ahead and balance out our atomic and our mass numbers. So to determine X, we can see that it captured a neutron. So we have that one neutron With a mass number of one and an atomic number of zero. So let's go ahead and balance out our atomic number and our mass number. So for our atomic number we have 36 plus 1 99 in our react inside and this is going to be equal to the mass number of X plus one. Solving for a mass number, we get a mass number of 2 34. Now let's go ahead and bounce out our atomic number. So we have 20 from our calcium 36 plus 92 from our uranium 1 99. This is going to be equal to the atomic number of X plus zero. So this means our atomic number of X comes up to 112. So our final symbol for X is going to be X With a mass number of 234 And an atomic number of 112. Now, let's go ahead and determine isotope why? To determine isotope why? We're going to take our isotope X with that mass number of 2 34 and that atomic number of 1 12. And we can see that we produced isotope Y, which underwent an alpha decay. So for alpha decay, we can write our alpha particle With a mass number of four and an atomic number of two. So let's go ahead and balance out our mass number and our atomic number, starting with our mass number, we have 2 34 is equal to the mass number of Why? Plus four, solving for the mass number, we get a mass number of 2:30. Now, let's go ahead and solve for the atomic number of Y. So we have 1 12 is equal to the atomic number of Y plus two, solving for the atomic number, we end up with 1 10. So our final symbol for isotope Y is going to be our element Y With a 2:30 as our mass number and a 1:10 as our atomic number. And this is going to be the final symbol for isotope Y. Now let's go ahead and determine isotope Z. Again, we're going to take the same steps as we did previously and we're going to take isotope Y with a mass number of 2 30 an atomic number of 1 10. And we ended up producing I stop Z, which also underwent alpha decay. So first let's go ahead and determine our mass number. So we have 2 30 from our isotope y. And this is going to be equal to the mass number of isotope C plus four, solving for the mass number. We end up with 2 26. Now, let's go ahead and solve for the atomic number of isotope Z. So we're going to take that 1 10 from isotope y. And this is going to be equal to the atomic number of isotope C Plus. To solving for the atomic number, we end up with a value of 108. So the final symbol for isotope Z is going to come up to A mass number of 1:10 on our top left and an atomic number of 108 at our bottom left. And this is going to be our final symbols. Now, I hope this made sense. And let us know if you have any questions