Each of the following transmutations produces a radionuclide used in positron emission tomography (PET). (a) In equations (i) and (ii), identify the species signified as 'X.' (i) 14N1p, a2X (ii) 18O1p, X218F (iii) 14N1d, n215O

Video transcript

Welcome back everyone in this example, we're told that the following trans mutations are used to produce trans uranium elements identify the new Clyde or the particle designated as X. So before we focus on parts one and two, we want to review something which is going to be the following expression where we have our target nucleus identified as our element X. With its corresponding mass number A and its corresponding atomic number Z. So this is mass number. This is our atomic number. This as a whole represents our target nucleus. And in parentheses we have the expression where we have our first particle which could be something like an alpha particle represented by this symbol. This is actually going to be considered our bombarding particle. Whatever it may be, is going to be our bombarding particle in this case are bombarding particle. Example is an alpha particle and then we would have a second particle listed which we can say for example is a beta particle. Maybe this particle that is listed second would be consider are ejected particle. So what is being produced as a product. And then on the outside of our parentheses on the right hand side of this expression, we would have a target nucleus which we have to identify the identity of. We'll just list it as X. For now with an unknown mass number and an unknown atomic number Z. So this is our product nucleus. So moving on to example one we are given the falling prompt where we have the atom curium with the mass number 42. This is going to be our target new glide or target nucleus. And recall that curium is located in our periodic table in the F block and corresponds to atomic number 96. So filling that in, we get that info from the periodic table. We're told from the prompt for part one that are bombarding particle is our alpha particle. So we're going to be bombarding this curium with our alpha particle which we should recall can be represented by the atom helium corresponding to the mass number four and atomic number two. So just to be clear, our alpha particle is equal to helium here, helium four. Now as our products, we're going to be having an ejected particle which according to part one of our prompt is going to be a neutron here, which we should recall corresponds to the mass number of one. An atomic number zero. So this would be our ejected particle in neutron according to part one of our prompt. And as our product new Clyde, that is what we need to determine because it's listed as X in part one. So we're going to determine the identity of X. We're going to need to determine its mass number and its atomic number Z. So setting up two expressions beginning with our solution for its mass number, we're going to plug in our knowns right now we know our mass number for curium 2 42. We know the mass number of our alpha particle helium being four. We know that this is set equal to our product side, where we have the mass number of a neutron being one. And the unknown mass number A that we need to solve for for our new Clyde. So simplifying this, we would have 2 46 equal to one plus are unknown mass number. And we would solve to say that our unknown mass number is equal to 2 46 minus one, which is equal to the value to 45 as our mass number. Following the same process. To solve for our atomic number Z of our new Clyde, We're going to plug in our notes. So from our equation, we know that our atomic number for Curium is 96. We know that our atomic number for our alpha particle helium is two. We know that it has set equal to our product side where our atomic number for a neutron is zero, added to our unknown atomic number Z. And so simplifying for Z. We would have 98 equal to Z. Meaning that when we look on our periodic table, We would see that the atomic number 98 corresponds to the atom California. And so that would be the identity of our new Clyde. So writing out our full equation, we would have to 42 curium With atomic number 96 being bombarded by our alpha particle helium, which ejects a ejecting particle being our neutron with atomic or atomic number zero. And mass number one and produces a target new Clyde or target nucleus being California with atomic number 98. And mass number 2 45. And this would be our final answer for part one of our prompt, which is our first new glide we've identified. Now let's move on to part two of our prompt. So going on to part two, we are given the Target. New Clyde being boron 10 were given in parentheses are bombarding particle which is our beta particle or a neutron. We're told that the ejected particle is going to be our new Clyde X. And sorry, this is a common here to be clear. And then our target or product nucleus rather is lithium seven given in the prompt. So let's write out our equation. We have boron 10. Recall that boron on our product table corresponds to atomic number five. We are bombarding this boron with a neutron beta particle which we recall has a mass number of one and an atomic number of zero. We are rejecting our unknown new Clyde X. Which we need to solve for its mass number Z. Or sorry, its atomic number Z. And its mass number A. And we are also having a product new Clyde which would be our lithium seven which we recognize on our product table corresponds to atomic number three. So solving first for our mass number A of our ejected particle X here. We're going to be plugging in our knowns, we know our mass number of boron 10 added to our mass number for our neutron one set equal to our product side. Where we have our unknown mass number A of our ejected particle added to our product nucleus mass number seven for lithium. So simplifying this, we would have 11 equal to a plus seven. And simplifying this further, we would have A is equal to four. So that's our mass number of our ejected particle. And now let's solve for our atomic number of the subjected particle. So Z. So solving for Z. Let's plug in our notes, we know that our atomic number of our known Target new Clyde Boron is five added to our bombarding particle, which is our neutron with the atomic number zero set equal to our product side where we have our ejected particle with the unknown atomic number Z that we have to sell for with our second product which is our product nucleus for lithium with atomic number of three. So simplifying for our symbol Z, we're going to have five minus three equal to Z. Meaning that our atomic number Z is equal to the number two. And on our periodic table we would see that this corresponds to our atom helium. And so to write out our full equation, we would have our Target nucleus boron being bombarded by our beta particle. The neutron with the atomic number zero. And mass number one, objecting our particle being an alpha particle being helium, with the atomic number of two and the mass number of four and our product nucleus being our lithium seven with the atomic number three. And so our second and final answer is going to be our identity of our ejected particle here, which is our alpha particle helium. So this is our second final answer. To complete this example. I hope everything I reviewed was clear. If you have any questions, leave them down below and I will see everyone in the next practice video.