Chapter 32, Problem 22P

To determine

Is ${\pi }^{-}+P\to n+\eta °$ possible reaction.

Answer to Problem 22P

Solution:

Yes, ${\pi }^{-}+P\to n+\eta °$ is possible reaction?

Explanation of Solution

Given info.:

Given reaction is ${\pi }^{-}+P\to n+\eta °$

Concept used:

In a reaction charge, Baryon number, lepton number and strangeness number are conserved.

Baryon number: proton and neutrons both have baryons.

$\begin{array}{l}\text{Lepton number Le}\\ {\overline{V}}_e-1\\ e^{+}-1\\ {\overline{V}}_{\mu }-1\\ {\mu }^{+}-1\\ {\overline{V}}_{\Sigma }-1\\ {\Sigma }^{+}-1\\ V_e\text{ 1}\\ {\text{e}}^{-}\text{ 1}\\ V_{\mu }\text{ 1}\\ {\mu }^{-}1\\ V_{\Sigma }1\\ {\Sigma }^{-}1\end{array}$

Strangeness number:

Strangeness number of quarks up, down and strange.

$\begin{array}{l}\text{ strangeness}\\ \text{U 0}\\ \text{d 0}\\ \text{s }-\text{1}\end{array}$

Calculation:

${\pi }^{-}+P\to n+\eta °$

Charge on reactant side: $-1+1=0$

Charge on product side: $0+0=0$

$\left(\text{charge on reactant side}\right)=\left(\text{charge on product side}\right)=0$

Therefore, charge is conserved.

Baryon number on reactant side $=0+1=1$

Baryon number on product side $=1+0=1$

${\left(\text{Baryon number}\right)}_{\text{Reactant}}={\left(\text{Baryon number}\right)}_{\text{product}}$

Therefore Baryon number is conserved.

Lepton number on reactant side $=0+0=0$

Lepton number on product side $=0+0=0$

${\left(\text{Lepton number}\right)}_{\text{Reactant side}}={\left(\text{Lepton number}\right)}_{\text{product}}$

Therefore, lepton number is conserved.

Strangeness number on reactant side $=0+0=0$

Strangeness number on product side $=0+0=0$

${\left(\text{Strangeness number}\right)}_{\text{Reactant side}}={\left(\text{Strangeness number}\right)}_{\text{product}}$

Therefore, strangeness number is conserved.

Therefore, reaction ${\pi }^{+}+P\to n+\pi °$ is possible.

To determine

(b)

Is ${\pi }^{+}+P\to n+\pi °$ possible?

Answer to Problem 22P

Solution:

Reaction ${\pi }^{+}+P\to n+\pi °$ is not possible.

Explanation of Solution

Given info.:

${\pi }^{+}+P\to n+\pi °$

Concept used:

In a reaction charge, Baryon number, lepton number and strangeness number are conserved.

Baryon number: proton and neutrons both have baryons.

$\begin{array}{l}\text{Lepton number Le}\\ {\overline{V}}_e-1\\ e^{+}-1\\ {\overline{V}}_{\mu }-1\\ {\mu }^{+}-1\\ {\overline{V}}_{\Sigma }-1\\ {\Sigma }^{+}-1\\ V_e\text{ 1}\\ {\text{e}}^{-}\text{ 1}\\ V_{\mu }\text{ 1}\\ {\mu }^{-}1\\ V_{\Sigma }1\\ {\Sigma }^{-}1\end{array}$

Strangeness number:

Strangeness number of quarks up, down and strange.

$\begin{array}{l}\text{ strangeness}\\ \text{U 0}\\ \text{d 0}\\ \text{s }-\text{1}\end{array}$

Calculation:

Charge on Reactant $=1+1=2$

Charge on product $=0+0=0$

Charge on Reactant $\ne$ charge on product

Here, charge is not conserved.

Therefore, the reaction ${\pi }^{+}+P\to n+\pi °$ is not possible.

To determine

(c)

Is ${\pi }^{+}+P\to p+e^{-}$ possible reaction?

Answer to Problem 22P

Solution:

${\pi }^{+}+P\to p+e^{-}$ Is not possible reaction.

Explanation of Solution

Given info.:

${\pi }^{+}+P\to p+e^{-}$

Concept used:

In a reaction charge, Baryon number, lepton number and strangeness number are conserved.

Baryon number: proton and neutrons both have baryons.

$\begin{array}{l}\text{Lepton number Le}\\ {\overline{V}}_e-1\\ e^{+}-1\\ {\overline{V}}_{\mu }-1\\ {\mu }^{+}-1\\ {\overline{V}}_{\Sigma }-1\\ {\Sigma }^{+}-1\\ V_e\text{ 1}\\ {\text{e}}^{-}\text{ 1}\\ V_{\mu }\text{ 1}\\ {\mu }^{-}1\\ V_{\Sigma }1\\ {\Sigma }^{-}1\end{array}$

Strangeness number:

Strangeness number of quarks up, down and strange.

$\begin{array}{l}\text{ strangeness}\\ \text{U 0}\\ \text{d 0}\\ \text{s }-\text{1}\end{array}$

Calculation:

Charge on reactant $=1+1=2$

Charge on product $=1+1=2$

Therefore, charge is conserved.

Lepton number on reactant $=0+0=0$

Lepton number on product $=0+1=1$

Therefore, Lepton number is not conserved.

Therefore, Reaction is not possible.

To determine

(d)

Is $p\to e^{+}+V_e$ a possible reaction?

Answer to Problem 22P

Solution:

No, $p\to e^{+}+V_e$ is not a possible reaction.

Explanation of Solution

$p\to e^{+}+V_e$

Concept used:

In a reaction charge, Baryon number, lepton number and strangeness number are conserved.

Baryon number: proton and neutrons both have baryons.

$\begin{array}{l}\text{Lepton number Le}\\ {\overline{V}}_e-1\\ e^{+}-1\\ {\overline{V}}_{\mu }-1\\ {\mu }^{+}-1\\ {\overline{V}}_{\Sigma }-1\\ {\Sigma }^{+}-1\\ V_e\text{ 1}\\ {\text{e}}^{-}\text{ 1}\\ V_{\mu }\text{ 1}\\ {\mu }^{-}1\\ V_{\Sigma }1\\ {\Sigma }^{-}1\end{array}$

Strangeness number:

Strangeness number of quarks up, down and strange.

$\begin{array}{l}\text{ strangeness}\\ \text{U 0}\\ \text{d 0}\\ \text{s }-\text{1}\end{array}$

Calculation:

Charge on reactant $=1$

Charge on product $=1+0=1$

Therefore, charge is conserved.

Baryon number on reactant $=1$

Baryon number on product $=0+0=0$

Therefore, Baryon number is not conserved.

Hence, reaction is not possible.

To determine

(e)

Is ${\mu }^{+}\to e^{+}+\overline{V}\mu$ a possible reaction?

Answer to Problem 22P

Solution:

No, ${\mu }^{+}\to e^{+}+\overline{V}\mu$ is not a possible reaction.

Explanation of Solution

Given info.:

${\mu }^{+}\to e^{+}+\overline{V}\mu$

Concept used:

In a reaction charge, Baryon number, lepton number and strangeness number are conserved.

Baryon number: proton and neutrons both have baryons.

$\begin{array}{l}\text{Lepton number Le}\\ {\overline{V}}_e-1\\ e^{+}-1\\ {\overline{V}}_{\mu }-1\\ {\mu }^{+}-1\\ {\overline{V}}_{\Sigma }-1\\ {\Sigma }^{+}-1\\ V_e\text{ 1}\\ {\text{e}}^{-}\text{ 1}\\ V_{\mu }\text{ 1}\\ {\mu }^{-}1\\ V_{\Sigma }1\\ {\Sigma }^{-}1\end{array}$

Strangeness number:

Strangeness number of quarks up, down and strange.

$\begin{array}{l}\text{ strangeness}\\ \text{U 0}\\ \text{d 0}\\ \text{s }-\text{1}\end{array}$

Calculation:

Charge on reactant $=1$

Charge on product $=1+0=1$

Therefore, charge is conserved.

Baryon number on reactant $=0$

Baryon number on product $=0$

Therefore, Baryon number is conserved

Lepton number on reactant $=0$

Lepton number on product $=-1+0=-1$

Therefore, Lepton number is not conserved.

Therefore, Reaction is not possible.

To determine

(f)

Is $p\to n+e^{+}+V_e$ not possible reaction?

Answer to Problem 22P

Solution:

NO, $p\to n+e^{+}+V_e$ is not possible reaction.

Explanation of Solution

Given info.:

$p\to n+e^{+}+V_e$

Concept used:

In a reaction charge, Baryon number, lepton number and strangeness number are conserved.

Baryon number: proton and neutrons both have baryons.

$\begin{array}{l}\text{Lepton number Le}\\ {\overline{V}}_e-1\\ e^{+}-1\\ {\overline{V}}_{\mu }-1\\ {\mu }^{+}-1\\ {\overline{V}}_{\Sigma }-1\\ {\Sigma }^{+}-1\\ V_e\text{ 1}\\ {\text{e}}^{-}\text{ 1}\\ V_{\mu }\text{ 1}\\ {\mu }^{-}1\\ V_{\Sigma }1\\ {\Sigma }^{-}1\end{array}$

Strangeness number:

Strangeness number of quarks up, down and strange.

$\begin{array}{l}\text{ strangeness}\\ \text{U 0}\\ \text{d 0}\\ \text{s }-\text{1}\end{array}$

Calculation:

Charge on reactant side $=1$

Charge on product side $=0+1+0=1$

Therefore, charge is conserved.

Baryon number on reactant side $=1$

Baryon number on product side $=1+0+0=1$

Therefore, Baryon number is conserved

Lepton number on reactant $=0$

Lepton number on product $=0+\left(-1\right)+1=0$

Therefore, Lepton number is conserved.

Strangeness number on reactant $=0$

Strangeness number on product $=0+0+0=0$

Strangeness number in a reaction is conserved.

Mass energy: $938.3MeV/c^2<939.6MeV/c^2+0.511MeV/c^2$

Mass energy is not conserved.

Therefore, reaction is not possible.