Question Video: Determining Strangeness in a Xi Baryon Decay | Nagwa Question Video: Determining Strangeness in a Xi Baryon Decay | Nagwa

Question Video: Determining Strangeness in a Xi Baryon Decay Physics

The following equation shows a Xi baryon decaying into a lambda baryon and a pion, which is an interaction that does not conserve strangeness. 𝛯⁻ (dss) ⟶ ⋀⁰ (uds) + 𝜋⁻ (ud). What is the total strangeness before the interaction takes place? What is the total strangeness after the interaction takes place?

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Video Transcript

The following equation shows a Xi baryon decaying into a lambda baryon and a pion, which is an interaction that does not conserve strangeness. What is the total strangeness before the interaction takes place? What is the total strangeness after the interaction takes place?

Looking at our equation, we see this particle here, called a Xi baryon, decaying into a lambda baryon and a pion. Along with the symbol for each of these particles, in parentheses, we’re told what quarks make up each one. So, for example, for our pion, this is made up of an up antiquark and a down quark.

Our question is focused on the strangeness of this interaction. And we can recall that this has to do with how many strange quarks and strange antiquarks are present. Specifically, every strange quark in a particle has a strangeness of negative one, while every strange antiquark has a strangeness of positive one. This means, for example, that if we had a strange quark all by itself, then the strangeness of that particle would be negative one.

The first part of our question asks about the total strangeness before the interaction takes place. In other words, what is the strangeness of the Xi baryon before it decays into the lambda baryon and the pion? We see that this particle is made up of a down quark and two strange quarks. The down quark doesn’t contribute anything to strangeness. Only strange quarks and strange antiquarks can. But each one of the strange quarks, according to our rule, will contribute negative one to the overall strangeness of the particle. So if we call 𝑆 sub b the total strangeness before our interaction takes place, that’s equal to zero, that’s the contribution of the down quark in the Xi baryon, minus one minus one. Those are the contributions of the two strange quarks. So that equals negative two. And that’s our answer to the first part of our question.

Part two asks us, what is the total strangeness after the interaction takes place? Now, if strangeness was conserved in this equation, then our answer here would be the same as our answer earlier. But our problem statement tells us that, in this case, strangeness is not conserved. So let’s take a look at the product side of this interaction.

First, we have our lambda baryon. And we see this is made of an up quark, a down quark, and a strange quark. Both the up and the down quark don’t contribute anything to the strangeness of this particle overall, while the strange quark, according to our rule, contributes a strangeness of negative one. We can say then that the overall strangeness of our lambda baryon is negative one.

And next, we look at the pion, which we see is made up of an up antiquark and a down quark. Since neither of these is a strange quark or a strange antiquark, they contribute nothing to the strangeness of the pion. And so we can say that the strangeness of the pion overall is zero. This means that the total strangeness of this interaction after the interaction takes place is negative one plus zero or just negative one. And we see now that indeed strangeness is not conserved in this interaction. Before the interaction, it was negative two. And after, it’s negative one.

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