Write balanced nuclear equations for the following processes.

(d) Positron emission of 165Ta

Video transcript

Welcome back, everyone consider the isotope mercury 201 provides the balanced nuclear equation when it undergoes positron emission. Now we are given mercury 201. So let's write down its symbol, it's HG 201, right? And if we look at the periodic table, it has an atomic number of 80. So we're going to write 80 as our subscript. If it undergoes positron emission, we have to recall that a positron is released. And what do we know about a positron? Well, it has a symbol of E which represents an electron. But the main difference is that even though it has the same mass number of zero, instead of having a negative one charge, it will have a positive one charge, hence the name positron, and then we are releasing an additional nucleus, we don't know what it is. So let's call it X. And let's say that the nucleus X has a mass number of M and an atomic number of C. So our goal is to identify M and Z, we can do that by using the laws of mass conservation and charge conservation. So first of all, if we use the law of mass conservation. The total mass on the reactant side must be equal to the total mass on the product side. Meaning if we take the top number as our superscript, that'd be 201 for mercury, it must be equal to the sum of M and zero. And we are going to use the same logic for the subscripts. Those are the charges, right, the atomic numbers. So 80 on the left side must be equal to the sum of Z and positive one. We want to solve these equations. Solving the first equation, we can just cross out the atomic mass of X is 201. And looking at the second equation, if we subtract one from both sides, then the atomic number Z is simply 79. Now, if we look at the periodic table, the atomic number of 79 represents gold and gold has a symbol of A U. So what we can do is just replace our X as the correct symbol of gold. That would be A U. We're going to replace the mass number M with 201. And we're going to replace Z the atomic number with 79. This gives us the balanced nuclear equation needed in this problem. Thank you for watching.