In neutron activation analysis (NAA), stable isotopes are bombarded with neutrons. Depending on the isotope and the energy of the neutron, various emissions are observed. What are the products when the following neutron-activated species decay? Write an overall equation in shorthand notation for the reaction starting with the stable isotope before neutron activation.(c) ²⁸₁₃Al* → [β⁻ emission]

When the nucleus undergoes decay, it produces a more stable nucleus. There are different types of decay in this problem. We're looking at the idea of beta decay, which means that this particle is produced another form of decay that we may see is positron or be positive, which means that this particle is produced. Both decays are initiated by a neutron bomb. Birdman.

So one of the react Ince's always a neutron. Chazz, this symbol here. And we can predict the products of each reaction by remembering that the mass numbers an atomic numbers must be equal. So upon neutron bombardment, an unstable nucleus is produced. We just symbolized with little star next to it. So to figure out how that formed, we know that we had some isotope plus a neutron to identify what the nucleus is.

We used the atomic numbers, the bottom number. So what? Plus zero makes 23 or 23. 23 has the same symbol. The the mass number is the different number. What? Plus one makes 52. So we're starting with vanadium 51. This quickly decays through beta decay. So we formed this particle plus to figure out the other product. We know that the numbers on the bottom have to equal 23.

We're X equals 24 which, using the periodic table we see is chromium and the mass number has to equal 52. So it's just 52. We can write this in shorthand notation, which is the reactant nuclear I'd and then in parentheses, the particle going in, followed by the particle going out and then the product. New climb. So for this one, the shorthand notation his vanadium 51 with the neutron going in, beta particle coming out, forming chromium 52.

So to identify the initial element, we know that the mass number has to be 20. The atomic number has to be 29 and the atomic number has to be 63 because 63 plus one is 64. The new element formed has to have an atomic number of 28 because 28 plus one is 29. This is nickel and the mass number is 64 because 64 plus zero is 64 in shorthand notation.

Final product

It's the starting reactant, followed by the neutron in in the beta particle out, followed by the final product when we could in fact right this using our symbols for a neutron is an N and for positron is B plus. Finally, receive the formation of the unstable nucleus a l 23 which forms from the bombardment of a neutron and produces a new particle through beta decay.

So the initial particle is still aluminum, the mass number of 22 and the final product has a mass number 14 which is s I in a mass number of 23 atomic number of 14 mass number of 23 in shorthand notation It's 22 a l and Data s I 23

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