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(a) Show that baryon number and charge are conserved in the following reactions of a pion with a proton:

$$

\begin{array}{l}{\text { (1) } \pi^{+}+\mathrm{p} \rightarrow \mathrm{K}^{+}+\Sigma^{+}} \\ {\text { (2) } \pi^{+}+\mathrm{p} \rightarrow \pi^{+}+\Sigma^{+}}\end{array}

$$

(b) The first reaction is observed, but the second never occurs. Explain these observations. (c) Could the second reaction happen if it created a third particle? If so, which particles in Table 30.2 might make it possible? Would the reaction require less energy or more energy than the reaction of Equation (1)? Why ?

a. Consider the following reaction:

$\pi^{+}+\mathrm{p} \rightarrow \mathrm{K}^{+}+\Sigma^{+}$

Conservation of charge:

$(+1)+(+1) \rightarrow(+1)+(+1)$

2$\rightarrow 2$

Therefore, the charge is conserved.

Conserve baryon number:

$(0)+(+1) \rightarrow(0)+(+1)$

1$\rightarrow 1$

Therefore, the baryon number is conserved.

b.Conservation of strangeness of first reaction:

$(0)+(0) \rightarrow(-1)+(+1)$

$\quad 0 \rightarrow 0$

$\quad 0 \rightarrow 0$

The strangeness is conserved and hence it is observed via strong interaction.

Conservation of strangeness of first reaction:

$(0)+(0) \rightarrow(0)+(+1)$

The strangeness is not conserved and hence it is not observed via either strong interaction or

electromagnetic interaction.

c. The third particle emitted in the second reaction should conserve all conservation laws. The

possibility of such a particle must be a neutral Kaon, which conserves the strangeness of the

reaction. The total energy of product particles of second reaction is greater than that of first

reaction. Therefore, the total energy of the reactant particles of the second reaction should be

greater than that of the first reaction.

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in this first part off the question. Looking at whether these two reactions conserve charge and barrel number. So let's go through one of them first. So for the first reaction, we can t charge that is plus one plus two. So it's plus two on both left and right. So charges concert. If you look at the burial number. So pipe classes Amy's on so that zero Putin is a very on. So you have one. Just put it reet one plus one very one number The K on a same isn't so that zero and the signal. Plus, it's a very on. So we get a plus one as well, Right? So DeBare anomalies conserved. Next for the other equation symptoms. It is pi plus first brought on. So rest water off. Plus two charge on the 11 side. Eddie, I had sight looking at a very low number. Proton is it, Baron? The signal pluses Burial in my keep I hypersea safeties on. So there's cereal so total barrel number concepts s well, but given that t first reaction, actually a curse, but not t second. So there's something wrong going on for a second reaction. Some other conservation laws that me be violating. So let us check. Ah, when I d auto possible conservation laws. Right. So left on number. Not very possible since there's no left turns involved over here. Maybe strangeness. So let us check the quite content and look for strange clocks. All right, So for the first equation, I wish here the strange clock on left inside. There's no strange cook. All right, So thats zero put debts here, so there's zero strangeness. We're here, Bonnie, right? The signal Plus actually has or contains strange clock anti strange and the que on also contents a streetcar. All right? And so what it get right? It's actually minus one plus one. Sorry. Distribute the other way. Wrong. Single plus has a clock as a street clock, which keeps strangeness off minus one. He keeps us k on contains a anti strange and anti straight quirk that gives us a striations value off plus one. So the total strangeness is through zero. Right? So it's concert 40. Other reaction. Frustration is is there a way to start? But finally, we have t signal, plus the pie Pluses contains no stretch cook. So that's zero but a single plus, remember? Yes, a strange quirk. So strange yourself. Minus one as well. The left inside the right hand side, actually to not have the same strangeness value I saw. Strange dress is violated. So how do we actually allow this reaction to proceed while we allowing or the conservation laws to be conserved? Right. So the question actually hinted that possibly we could introduce a particle in our product over here product over here. But in order to introduce these products in must allow a must to conserve the conservation laws which is charge para number and also helped do cons oaf our strangeness. So first off charge for you to not affect the conservation of charges, it must be neutral. And in order for it to not affected Baron number, it must be a muse on on not Barry on you carry me zone. Left own etcetera. Strangeness. It must contain a strangeness value off plus one so that it can counter contract e minus one to give us a total off zero. Right? And so it cannot be a left on anymore, since they left on do not have a strangeness value. So it must be Amy's on if a stranger's off plus one and a charge off Syria, and one of it that's possible would be adding in neutral. Kill, like a neutral cheon contains an anti strange and which has a stranger's off plus one. And this will help us to get a total strangers off zero so that our strangers is conserved. No. As a result of this, this was more like the require more energy to proceed. And why is that so? That's because the total mess off the protects the pi plus the signal plus e. K. On we have a greater mass than just e k plus at the signal Plus. And that means debut on eat More energy to be inputted Tohti initial rectums richest pi plus anti proton in order to allow them to collide with greater energy such that they would have more energy to produce more particles. And so definitely this reaction would require more energy

National University of Singapore