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Ch.21 - Nuclear Chemistry

Chapter 21, Problem 16c

What particle is produced during the following decay processes: (c) iodine-122 decays to xenon-122?

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Welcome back everyone in this example, we need to determine which particle is going to be produced during the decay of a 17-22 rate on to 20. So let's write out our symbols for each of these isotopes. We know that their isotopes because they're listed with their names given as their mass numbers attached with the hyphen. So beginning with a 17 recall that that's represented by our symbol 80. Which we can find in Group seven A. On our periodic table corresponding to atomic number 85. And were given its mass number in the name of the isotope being 2 20. So we're told that this 19 to 20 is going to be decaying into the product radon 2 20. So we call that radon is represented by the symbol are under case N corresponding to the mass of 2 20 given in the name And the atomic number. Where we find radon in group A of our periodic table with atomic number 86. And we're also going to be producing a second particle product, which we need to identify the identity of. So we're going to call this unknown element X. With the atomic number designated by the symbol A. And the mass number. We're sorry, the mass number designated by the symbol A. And R. Atomic number designated by the symbol Z. So A. Is our mass number. And Z. We recall is our atomic number which defines our number of protons associated with the molecule or atom. So let's go ahead and figure out how to determine this identity of this particle. What we're going to have to do is set up to expressions to solve for a our mass number and then for ZR atomic number which will ultimately lead us into the identity of our element which is X. For now. So solving for a first, we're going to have the following expression where we know that our Constantine mass number is 2 20. We know that this is our reactant. So this is equal to our product side where we have also the mass number 2 20 for Reagan plus our second reactant mass number which is unknown as the symbol. A. And so solving for a we would say that A should equal the value zero. Moving forward to our atomic number Z, solving for this, we are going to begin with our atomic number of A. 17 which we are or which we determined from our periodic tables as 85. This is set equal to our product side where we have the atomic number for Reagan being 86. Which we got from our periodic table added to Z are unknown atomic number. And sorry, this is a Z. Here added to Z. Our atomic number of our unknown particle. And so solving for Z, we would say that Z is going to equal 85 minus 86. And this difference gives us the value of negative one as our atomic number. And so to write in our full reaction, what we should have is our 2 20 Aston teen With Atomic # 85. Which The case to produce 220 Reagan With atomic number 86. And as the identity we've secured of our unknown element. We now know that it has a mass number of zero and an atomic number of negative one. And we should recall that this is the description of our particle known as our electron. And so for our final answer, the particle produced during our decay of a 17 to 20 is our electron here with the mass number zero and the atomic number minus one. So what's highlighted in yellow is our final answer as our particle produced during the decay of a 17. If you have any questions, leave them down below and I will see everyone in the next practice video.