Naturally occurring uranium-238 undergoes a radioactive decay series and emits 8 a particles and 6 b particles. What is the stable nucleus at the end of the series?

Video transcript

Well everyone's in this video we're being told that they are in the radioactive decay chain of thorium 2 30 to 6 alpha particles. And for beta particles are being emitted. We're trying to determine the identity of the stable nucleus formed at the end. So again when we have emitted six alpha particles and four beta particles. So the story um 2 32 here this goes to 2 30 to 90 T. H. And this goes to a Z. X. So for my alpha particle this is just four to alpha. So we see here that we decrease the mass number by four and decrease the atomic number by two. As for my beta particle here it's just zero negative one. E. So we we increase the atomic number by one. So for my mass number here which is of 2 32 value, remind us four times six. And once you put that into the calculator, we see that the mass number Is a value of 208. As for the atomic number we have 90 minus two times six. Close one times four. Once you put that into the calculator we see that the atomic number is equal to 82. So filling I guess this value in is that we have 20882 two X. And taking a look at the pr table and looking for the number or the topic number of 82. This is going to be P. B. For the element symbol and this is just lead. So my final answer then is going to be. So we'll just put answer. The answer for this is PB. And this right here is going to be my final answer for this problem.