What nuclide is produced in the following radioactive decays? (a)...

Question

What nuclide is produced in the following radioactive decays? (a) a decay of 239 94 Pu; (b) b- decay of 24 11Na; (c) b+ decay of 15 8O.

Accepted Answer

Hi, everyone. In this practice problem, we're being asked to determine the daughter isotope that is going to be produced by the following reaction. First negative beta decay of phosphorus, 32 15 p, second alpha decay of radium 226 88 radium and third positive beta decay of carbon, 11 6 carbon. The options listed the different daughter isotopes for different decay equations. So in this practice problem, the way we want to tackle this problem is to remember that the left side and the right side of the A and the C or the total mass on each side of the decay equation, the total charge and the nucleon number should be conserved. So in this case, let A Z X B, the daughter isotope for each decay equation. And let's start by actually tackling the first part which is going to be the negative decay or negative beta decay of phosphorus. 32 15 B. For an uh negative beta decay, we have an emission of an electron, the total charge and the nucleon number should be conserved. So 32 15 P will decay into zero minus one E plus AC X So on the left side, we have 15, which will also uh be, well, which will then be equal to, on the right side minus one plus C for the Z. And for the A, on the left side, we have 32 which will be equal to zero plus A. So with this two equation, we have the Z to then be equal to 16 and the A to then be equals to 32. And then we can notice that the atomic number of the daughter nude increases by one while its atomic mass remains constant. So then with the uh the A of 32 the Z of 16, we have the daughter nucleus to be 16 sulfur or s So that will be the answer for the first part. And then moving on to the second part, we have the alpha decay of radium, which will have the uh decay equation of 226 radium decay into a Z X plus four to helium because it is an alpha decay. So looking into the Z, we have 88 on the left side, which will be equals to Z plus two. So then our Z for the daughter isotope will then be equal to 86 of uh looking at the A, on the left side, we have 226 A plus four. And then that will give us an A of 22 on the daughter isotope. So the total charge and the nuclear number are conserved through this ac uh values. And the atomic number of the daughter nude decreases by two and its, and its atomic mass decreases by four. So according to this, we can uh determine that the daughter nucleus is going to be redden, which is going to be 222 86. Er And lastly looking into the third equation or the third part of our problem, which will be the uh positive beta decay beta plus of carbon 11. For this decay, we will have an emission of positron. So the decay equation for carbon 11 will be 11 6 C or carbon 11 decaying into zero plus one E or positron plus AC X. Looking at the Zs, we have six on the left side plus one on the right side plus C. So then our Z will then uh Z $40 isotope will then be equals to five. And then looking at the ace, we have 11 on the left side, zero plus A and the A will still remains to be 11 of on the daughter aceto. So the total charge and the nuclear number are conserved through this and the atomic number of the daughter newly decreased by one while its atomic mass remains constant. Therefore, we can determine that the daughter nucleus is going to be 5 boron or 11 5 B. So there we have it for this practice problem. The first uh decay equation will have the daughter isotope 32 16 S or sulfur. And then for the second one, it will have 222 86 Redden or R N. And for the last part, it will have 11 5 boron or B. So all of this will correspond to option A in our answer choices. So option A will be the answer to this particular practice problem. So that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this one. Thank you.

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What nuclide is produced in the following radioactive decays? (a) a decay of 239 94 Pu; (b) b- decay of 24 11Na; (c) b+ decay of 15 8O.

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