Radon-222 decays by a series of three a emissions and two b emissions. What is the final stable nucleus?

Video transcript

Welcome back everyone in this example, we need to determine the stable nucleus that will form from polonium 2 17. After it undergoes a series of two also emissions and three beta emissions. So let's go ahead and write out this expression. Were given polonium 2 17 as an isotope recalled that polonium is represented by uppercase P under case O where the 2 17 in the name represents our mass number written in the left hand exponents. And recall that on our periodic table. We can find polonium in group six a corresponding to the atomic number being 84. So let's represent atomic number in red. And let's actually use the color purple for our mass number. So this is going to be considered are reacting here and it's undergoing these emissions to produce our products where according to the prompt, we have two moles of our alpha particle produced. So this is our alpha particle which we should recall has a mass number of four and an atomic number of two. For our second product we form three beta emissions. So we form three moles of our beta emission, which we should recall. Our beta emission has the simple here or beta particle has the simple here and it has a mass number of zero and an atomic number of negative one. And then we have our third product which is our unknown nucleus, which we'll call X here where we recall that X represents our chemical symbol of our element where a is the atomic mass and Z is its atomic number. So let's fill that in atomic mass or mass number and then Z is our atomic number. So we're going to need to solve via two expressions for the atomic mass and atomic number of this unknown element. To figure out its identity, to figure out what our stable nucleus formed by Polonium 2 17 will be. So what we're going to have is solving first for, let's do atomic mass. We have our polonium 2 17 atomic mass, which we know is equal to 22 moles times the atomic mass of our alpha particle, which we know is four, which is then added to three moles of our atomic mass or times are atomic mass of our beta particle which we know is zero. And then this is going to be added to our unknown atomic mass for our unknown nucleus known as a here. And so solving for a what we're going to simplify too, is a equal to the difference between -8. And this is going to yield a atomic mass of our stable nucleus equal to 209. And so now we want to move forward and do the same steps but for this time, the atomic number of the stable nucleus. So let's go ahead and begin with what we know which is for polonium where we have that atomic number of 84, we know that this is equal to our products atomic number where we have two moles of our alpha particle multiplied by its atomic number being two added to three moles of our beta particle, multiplied by its atomic number being negative one. And then added to our unknown atomic number for our stable isotope Z. So solving for Z, what we should have is that Z is equal to the difference between 84 minus one. And that's because we see that two times two would give us four and then three times negative one would give us negative three. So four plus negative three would leave us with this minus one. And so 84 minus one is going to yield our atomic number of our stable isotope equal to the value 83. And now that we know this atomic number, we're going to go to our periodic table. And on our periodic table we would see that atomic number 83 corresponds to the atom bismuth, so be under case I. And so now we have identified our identity of our mass number of bismuth as well as as its atomic number 83 here for our stable isotope. And so we can complete our equation by saying that 2, 17 polonium. When it undergoes its submissions, we have two moles of our alpha particle formed with the mass number four and the atomic number two, we have three moles of our beta particle formed with the mass number zero and the atomic number negative one. And then our stable isotope that forms is our adam bismuth with the mass number 209 and the atomic number 83. And so our final answer is going to be this symbol here as our stable nucleus that is formed after polonium undergoes these two alpha and three beta emissions, so it's highlighted in yellow is our final answer. I hope everything I explained was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.