Chapter 32, Problem 23P

To determine

Which of the following reaction will not happen.

$\begin{array}{l}p+p\to p+p^{-}\\ p+p\to p+p+p^{-}\\ p+p\to p+p+p+p^{-}\\ p+p\to p+e^{+}+e^{+}+p^{-}\end{array}$

Answer to Problem 23P

Solution:

List of reaction which will not happen.

$\begin{array}{l}p+p\to p+p^{-}\\ p+p\to p+p+p^{-}\\ p+p\to p+e^{+}+e^{+}+p^{-}\end{array}$

Explanation of Solution

Given info:

$\begin{array}{l}p+p\to p+p^{-}\\ p+p\to p+p+p^{-}\\ p+p\to p+p+p+p^{-}\\ p+p\to p+e^{+}+e^{+}+p^{-}\end{array}$

Formula 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:

For reaction:$p+p\to p+p^{-}$

Charge on reactant $=1+1=2$

Charge on product $=1-1=0$

Since, charge is not conserved. Therefore,

Reaction $p+p\to p+p^{-}$ is not possible.

For reaction:

$p+p\to p+p+p^{-}$

Charge on reactant $=1+1=2$

Charge on product $=1+1-1=1$

Since, charge is not conserved. Therefore, this reaction will not happen.

For Reaction, $p+p\to p+p+p+p^{-}$

Charge on reactant $=1+1=2$

Charge on product $=1+1+1-1=2$

Since, charge is conserved.

Brayon number on reactant $=1+1=2$

Brayon number on product $=1+1+1-1=2$

Brayon number is conserved.

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

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

Therefore, Lepton number is conserved.

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

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

Strangeness number is conserved.

Since, Reaction obeys all conserved laws and $p+p\to p+p+p+p^{-}$ is a possible reaction.

For Reaction, $p+p\to p+e^{+}+e^{+}+p^{-}$

Charge on reactant $=1+1=2$

Charge on product $=1+1+1-1=2$

Charge is conserved.

Brayon number on reactant $=1+1=2$

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

Brayon number is not conserved.

Therefore, this reaction is not possible.