Write a partial decay series for Rn-220 undergoing the sequential decays: a, a, b, b.

Video transcript

Hey everyone, we're asked to identify the final new glide. If polonium 2 14 undergoes the partial decay series, alpha, beta beta alpha First, let's go ahead and start off with our Polonium 2 14. Looking at our periodic table, we find that it has an atomic number of 84. Since this will undergo an alpha decay, we would have 4, 2 and an Alpha Particle. Now, in order to determine the missing particle, We can subtract four from our 2 14 and this will get us to An atomic mass of 2 10 And we subtract two from 84 And this will get us to an atomic number of 82. This means our element is going to be led. So we have led to 10, Taking our lead to 10 where then Going to undergo a beta decay, which is a beta particle with a zero and a -1 to find our missing particle. We're going to do the same steps. We will take our 210 from our lead And subtract R0 from our beta decay, which will get us to to 10 as well. Taking our 82 And subtracting our -1. We end up with an atomic number of 83, which means our element is going to be bismuth. Next taking Bismuth to 10. We then undergo another beta decay will do the same exact steps as the previous step 2, 0 comes up to 2, 10 and 83 minus negative one comes up to 84 which means our element is going to be polonium Taking our Polonium 2, 10. we then undergo an alpha decay and to find the missing particle. We would then take our 2 10 and subtract our four, Which will get us to an atomic mass of 206. Next, we're going to take our atomic number of 84 from our polonium and subtract R two from r alpha particle And we end up with a value of 82, Which means that we have led 206. So the final new glide is going to come up to be lead 206. So I hope this made sense and let us know if you have any questions.