Chapter 32, Problem 8Q

To determine

Verify that change and baryon number are conserved in each decay.

Answer to Problem 8Q

Solution:

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

In this piondecay charge and baryon number are conserved.

Explanation of Solution

Given info:

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

Formula used:

Proton is positively, neutron is neutral changed particle and ${\pi }^{-}$ is negatively changed particle.

Calculation:

Charge conserved.

Charge on Reactant side

$\begin{array}{l}=1+0\\ =1\end{array}$

Charge on Product side

$\begin{array}{l}=1+1-1\\ =1\end{array}$

Here,

${\left(\text{Charge}\right)}_{\left(\text{Reactant}\right)}={\left(\text{Charge}\right)}_{\text{Product}}=1$

Therefore, charge is conserved.

Proton and neutron both are baryons.

Number of baryons in reactant side $=2$

Number of baryons in product side $=2$

Hence, baryon number is also conserved.