The radioactive isotope 100Tc decays to form the stable iso-tope 100Mo. (a) There are two possible pathways for this decay. Write balanced equations for both.

Video transcript

Hello everyone. So in this video being told that the new Clyde. So 2 30 or 2 53 FM undergoes radioactive decay in which the stable new Clyde to 53 E. S. Is formed. There are two possible nuclear decay pathways for this process. We're being asked to provide the balanced nuclear equation for the two processes. So from the PR table we can see that the atomic number For F. M. is equal to 100 and the atomic number for E. S. Is equal to 99. So where you see that we're going from 253, f. m. 22 53 99 E. S. So the atomic number decreased by one. Alright, so we have different types of radioactive decay. We have alpha decay beta decay gamma mission positron emission and electron capture. So the apostle radioactive decay that we have going on here. So let's write this in red. So we have the possible radioactive decays. We have maybe a beta decay we could have positron emission or we can have electron capture. So the reason why we said beta decay it is because we form a beta particle. The beta particle appears on the product side usually. And for the positron emission is because it forms a positron particle. And for the electron capture is that the initial nuclei captures in the electron and the electron appears on the reactive side. So let's go ahead actually test these possibilities out. Starting with of course debated decay I'll go and do this in green. So again we're doing the beta decay first, I'll just put beta for short so we have D 53. 100 F. M. Going to potential talk um a value Z. Value and we'll have this X. For some type of element there. And then we're having we have our beta decay so 0 - and beta. So let's fill these values in for A. So 2 52 equals two A plus zero. of course. We subtract zero on both sides and it's just that 252 is equal to now for a Z value, we have 100 you're going to Z plus are negative one value. So just adding one to both sides, we get 101 equaling two Z. Sonal, putting this all together. 2 53. 100 F. M. Goes to 2 53 X. And then the beta there. This is actually not possible. So this eliminated from our possible radioactive decays moving on to positron emission. So just simplify this to be just positron. So you're scrolling down here for more space. Again, we're starting off with 2 53. 100 F. M. Going to some a some Z. And some X december adding E. Again, settling for a we have 252 equal into a plus zero. So we just get that to 52 is equal to eight. Now for our Z value here is 100 equaling two Z plus one. Subtracting one from both sides. We get 99 equaling to this Z value. So putting this all together we get to 100 F. M. Going to 2 53 99 X plus zero one E. And this is possible. So this will not be eliminated from our list here. This is good to go. Next is going to be positron or electron capture rather. So I simplify this to be just electron here. So Again, starting off with the 2:53 100 FM. We're adding the zero negative one E. This gives us a Z. X. So now we're gonna go ahead and solve for eight. So this is 2 52 equaling 28 plus zero which is just 2 52 equaling 28. Now we're solving for our X value here, this is 100 plus this negative one to give us this Z value. Of course this is just 100 minus one which is just 91 99. Eagling to Z. So we put everything together. Now we have 2 53. 100 F. M adding zero negative one E. Which gives us 2 53 99 X. Of course this is possible as well. So we go ahead and check mark this here. So the balance reactions that we're gonna put as our final answers then Is that we have 2:53. f.m. Going to 2 53 99 E. S plus zero one. And the second balance reaction that will put is 2 100 f. M plus zero negative one E. And this gives us 2 53 99 e. S. So these right here are going to their final answers for this problem.