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Ch.21 - Nuclear Chemistry

Chapter 21, Problem 14a

Write balanced nuclear equations for the following transformations: (a) bismuth-213 undergoes alpha decay.

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Hey everyone, we're asked, what is the balanced nuclear equation? When the isotope act any um to 25 decays by alpha emission first, let's go ahead and write out our incomplete nuclear equation. So we have actively um to 25. And when we look at our periodic table, we find that it has an atomic number of 89. So this is going to decay by alpha emission. So we will have that daughter knew Clyde plus our alpha particle first. We want to go ahead and find the mass number of our daughter knew Clyde in order to do so we can label it as X. And we can go ahead and add that four from our alpha particle. And this will both be equal to 2 25. Using algebra, we simply subtract four from both sides And we end up with a mass number of 221. Now let's go ahead and calculate the atomic number to calculate our atomic number. Again, we can label this as X. And we're going to add that to from our alpha particle And this will be equal to 89, which is the atomic number of act any um subtracting two from both sides. We end up with an x. 87, Which is going to be the atomic number of our daughter new glide. And when we look at our periodic table, the element with the atomic number 87 is going to be francie. Um so when we complete our nuclear equation, we will have francie. Um And it will have an atomic mass of 221 And it has an atomic number of 87. So this is going to be our final balanced nuclear equation. Now, I hope that made sense and let us know if you have any questions.