Write a nuclear equation for the indicated decay of each nuclide. c. Pb-214 (beta)

Video transcript

Hello. Everyone in this problem, we are trying to write the nuclear equation when this myth to 10 nucleotide undergoes beta to Cape. So whenever we have a radio active decay or emission reaction, the radioactive particles ejected from the nucleus and forms of product. So when we have a beta decay or a beta emission, it occurs when an unstable nucleus ejects a beta particle to create a new element. So a beta particle has no atomic bass and is represented by an electron. So it could be written as this right here. Or it can be written as 0-1. And we have the bait instead of our eight. Same thing for both sides. That zero is going to go ahead and represent the mass number and then over on the bottom are negative one. That's going to be our atomic number. An element can be represented in a very similar fashion. So the series of X. A. And Z. So our X. Is going to be the element symbol. Same thing on top. That's our mass number. And over to the bottom is going to be our atomic number. All right. So what we can say, we're given by looking at this right here, were given the atomic symbol of B. I. We have the mass number Given to us as to 10. And the atomic number by looking on the parent table, It's going to be 83. So our reacting. Then it's going to be 2 10, 83 B. I scrolling down a little. Then for the product we have that zero negative one beta plus R. A. Z. And X. So the mass number of reactant which is P. I will equal to the mass number of products. So focusing on the mass number of the B. I. Is going to be 2 10 And then the mass number of beta particles is zero. The mass number of the new element we don't know. So we just have as A. So basically then a. Is equal to 2 10 now for our Z value. So that's going to be the atomic number of erections which is B. I. That's gonna be 83. We're going to equal that to the atomic number of products. So first we have the atomic number of the beta particle but it's going to be negative one. We're adding the atomic number of the new element. Which we don't know as Z. So then that means that our z value is equal to 84. So looking at our atomic number we know is 84. Then looking at the para table that's going to be poli um The element symbol for polonium is P. O. Alright. Now finally running out our balanced nuclear equation with the products reactant and everything. Then we have the 2 10 B. I. To go ahead and make zero negative one and use a beta for this, adding the 2 10 P. O. All right, this is going to be my final answer for this question. Thank you all so much for watching.