# Video: Deducing the Nuclide Symbol of the Parent Element of a Given Nuclide Formed After a Series of 𝛼 and 𝛽 Decays in a Set of Nuclide Symbols

What is the parent isotope of an isotope with a mass number of 102 and an atomic number of 27 if it is formed as a result of 2 𝛼 and 3 𝛽⁻ decays?

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### Video Transcript

What is the parent isotope of an isotope with a mass number of 102 and an atomic number of 27 if it is formed as a result of two 𝛼 and three 𝛽 minus decays? And as options, we’ve been given five nuclide symbols.

We use nuclide notation to indicate the identity of a particle, be it an atom of an isotope, a 𝛽 particle, an 𝛼 particle, and so on. When referring to an isotope, the root element symbol is used. But in this question, the letter X and the Greek letters 𝜋, Ω, and 𝜑 have been used in place of standard elements symbols indicating we’re not supposed to pay too much attention to the elements involved. What’s more important are the numbers to the left-hand side. The number to the top left is the mass number. And that indicates the number of protons and neutrons in total in the nucleus. The number to the bottom left is the atomic number. And that’s simply the number of protons. The nuclide referenced in the question has an atomic number of 27. So you know in its nucleus, there are 27 protons.

We also know there is 75 neutrons because the mass number is 102. And 102 minus 27 is 75. The question tells us we’re forming this nuclide from a parent isotope. So we have an isotope that’s decaying and our final product is what’s in the question, our X 102 nuclide. So for this question it’s very important we get that the right way round. It’s not our X isotope that’s decaying. It’s another isotope that’s decaying into our X isotope. The question tells us that our decay chain involves two 𝛼 and three 𝛽 minus decays. In an 𝛼 decay, a nucleus will release a high-energy 𝛼 particle, which is the equivalent of a helium nucleus. That’s two protons and two neutrons. So what happens when our parent isotope undergoes two 𝛼 decays?

Well, to start off with, the mass number will decrease by eight because we’re losing two protons and two neutrons per 𝛼 particle. And we’re losing two 𝛼 particles. And since each 𝛼 particle contains two protons, we’re going to lose four from the atomic number. 𝛽 minus decay involved the release of a high-energy electron. But this electron isn’t released from the cloud of electrons around the nucleus. It’s released from the nucleus itself when a neutron decays to form a proton. So what happens to the mass number when our nuclide undergoes three 𝛽 minus decays? The answer is nothing. The mass number is simply the number of protons and neutrons. So if a neutron turns into a proton, the mass number isn’t affected.

However, the atomic number increases by three. So we go from 𝑍 minus four to 𝑍 minus one. This is because the atomic number is the number of protons and each 𝛽 minus decay turns a neutron into a proton. We’ve accounted for all our decays, two 𝛼 decades and three 𝛽 decays. It doesn’t matter what order we did them in. The result would be the same. All that remains is to work out the value of 𝐴 and 𝑍, the mass number and atomic number of our parent isotope. We know that if we take eight away from the mass number of our parent isotope, we’ll get the mass number of the final isotope, 102. So the mass number of our parent isotope is 110.

We also know that the atomic number of our final isotope is only one less than the atomic number of our parent isotope. So the atomic number of our parent isotope is 28. This corresponds with answer A. If we look up on a periodic table, we know the element with atomic number 28 is nickel. But the heaviest isotope I could find for nickel only has a mass number of 80. And it’s highly radioactive with a half-life of only 24 milliseconds. So the parent isotope would actually be incredibly unstable and unlikely to form in the first place. The important thing with this question was to recognize how to manipulate the mass number and the atomic number to move from nuclide to another.