In problems 5.51 to 5.54, a nucleus is shown with protons and neutrons.
Draw the nucleus that emits a beta particle to complete the following:

Question

In problems 5.51 to 5.54, a nucleus is shown with protons and neutrons.

Draw the nucleus that emits a beta particle to complete the following:

Accepted Answer

Welcome back, everyone illustrate the change that occurred after a nucleus emits a beta particle. In the following diagram, we are given a legend where the orange particles represent protons. The gray particles represent neutrons and the purple particles represent a beta particle. And on the reactant side of this nuclear reaction, we have a nucleus which is going to be the parent nu glide. And we observe illustrated inside of its nucleus, a total of seven gray particles which represent seven neutrons. And we would observe a total of five orange colored particles which represent five protons. So writing the parent nu glide symbol, we would have X and then we would place for an isotope symbol, the mass number A which is equal to the sum of protons plus neutrons. And that is written in the left-hand superscript. So that is the mass number. And then we've got z the atomic number written in the left-hand subscript which corresponds to the number of protons. Since we counted five orange particles that corresponds to an atomic number of five equaling five protons. We would fill that in the left hand subscript of the apparent nudes isotope symbol. And for its mass number, we would take the five protons added to the seven neutrons that we counted. And that would yield a mass number of 12. So we would fill in 12 for the mass number in the left-hand superscript of the isotope symbol of the parent nu glide. Now with our parent nu glide determined, we're going to observe that because our product emits a beta particle where the beta particle is the purple particle. And recall represents an electron with a mass number of zero and an atomic number of negative one. So ultimately, what we have occurring is the parent new glide undergoing a beta decay. Recall that in a beta decay, a neutron is converted to a proton and an electron. And so that is why we have a beta particle representing an electron as one product. And then the daughter new glide, which is the nucleus of the product of decay of the parent New glide. The daughter new glide is going to be represented by its isotope symbol as Y. And we're going to need to determine its mass number. A and atomic number Z where its atomic number is actually going to be Z plus one because as we stated, a neutron from the parent new glide is converted to a proton and an electron. And so the atomic number of the daughter new glide should be the atomic number of the original parent new glide plus one extra proton. And so showing the balanced nuclear reaction, we would have 12 X for the parent new glide with its atomic number of five which forms as a product, the do new glide Y with a mass number of a, an atomic number of Z plus one. And then this is added to our second product, the beta particle which is an electron with a mass number of zero and an atomic number of negative one. Solving for a, we would determine that we have 12, equal to a plus zero, which simplifies to a equal to 12. So the mass number stays the same in the daughter new glide because as we stated, a neutron is just converted to a proton. And so ultimately, we have the same number of particles as we had in the parent new glide. It's just that and actually what I mean is we have the same number of nucleons making up the nucleus of both the parent nu glide and daughter new glide being a total of 12 nucleons. However, it's just the number of individual particles changed to where we now have one less neutron, which is converted to an additional proton in the daughter nu glide. And so it makes sense that the mass number stays the same for the daughter New glide. Now determining the atomic number of the daughter new glide, we would be solving for Z and so we would have five which should equal Z plus one minus one. And that would mean that when simplifying Z is equal to five. And so that means that we would have for the daughter new glide an atomic number of now six. So we can fill in that in the left hand subscript and its mass number of 12. So illustrating the nucleus of the daughter nu glide, we would need to draw a total of only six neutrons. Now, since one of the original seven neutrons in the parent nu glide was converted to a proton. So that leaves us with six neutrons now, and then we would draw in a total of six protons or six orange particles. Since again, a neutron was converted to gain an additional proton moving us from five in the parent new glide to a total of six in the daughter new glide. And so our final answer is going to be the illustration of the daughter new glide that results once the parent new glide undergoes a beta decay to convert a neutron into a proton and an electron. So our final answer is the illustration where we have six neutrons and six protons illustrated. I hope that this made sense. And let us know if you have any questions.

In problems 5.51 to 5.54, a nucleus is shown with protons and neutrons.
Draw the nucleus that emits a beta particle to complete the following:
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Key Concepts

Beta Decay

Nuclear Composition

Particle Emission