Chapter 12, Problem 30P

(a)

To determine

The explanation of nuclear decay of the nuclear reaction ${\Sigma }^{-}\to n+{\pi }^{-}$

Explanation of Solution

Given:

Nuclear reaction,

  ${\Sigma }^{-}\to n+{\pi }^{-}$

Formula used:

Following particles can be written in terms of quarks as,

  $\begin{array}{l}{\Sigma }^{-}=dds\\ n=udd\\ {\pi }^{-}=\overline{u}d\end{array}$

Calculation:

  $\begin{array}{l}{\Sigma }^{-}\to n+{\pi }^{-}\\ \cancel{d}\cancel{d}s\to u\cancel{d}\cancel{d}+\overline{u}d\\ s\to u+\overline{u}d\end{array}$

Conclusion:

Thus, s quark is converted into one u quark, one $\overline{u}$ and one d quark.

(b)

To determine

The explanation of nuclear decay of the nuclear reaction, ${\Lambda }^0\to p+{\pi }^{-}$

Answer to Problem 30P

  $s\text{quark}$ converts into $d$ and one pair of $u\overline{u}$ .

Explanation of Solution

Given:

Nuclear reaction

  ${\Lambda }^0\to p+{\pi }^{-}$

Formula used:

Given particles in the nuclear reaction can be written in terms of quark as,

  $\begin{array}{l}{\Lambda }^0=uds\\ p=uud\\ {\pi }^{-}=\overline{u}d\end{array}$

Calculation:

  $\begin{array}{l}{\Lambda }^0\to p+{\pi }^{-}\\ uds\to uud+\overline{u}d\end{array}$

Conclusion:

In this decay, $s$ quark converts into d quark and one pair of $u\overline{u}.$

(c)

To determine

The explanation of nuclear decay of the nuclear reaction $K^{*+}\to K^0+{\pi }^{+}$

Answer to Problem 30P

  $u\text{converts into u and one pair of d<mover accent="true"><mo>d</mo><mo>¯</mo></mover>}\text{.}$

Explanation of Solution

Given:

Nuclear reaction,

  $K^{*+}\to K^0+{\pi }^{+}$

Formula used:

Following mesons can be written in terms of quark as,

  $\begin{array}{l}{K}^{*+}=u\overline{s}\\ K^0=d\overline{s}\\ {\pi }^{+}=u\overline{d}\end{array}$

Calculation:

  $\begin{array}{l}{K}^{*+}\to K^0+{\pi }^{+}\\ u\overline{s}\to d\overline{s}+u\overline{d}\end{array}$

Conclusion:

In this nuclear reaction, u converts into u and one pair of $d\overline{d}.$

(d)

To determine

The explanation of nuclear decay of the nuclear reaction ${\Omega }^{-}\to {\Lambda }^0+K^{-}$

Answer to Problem 30P

  $s\text{quark converts into d and one pair of u<mover accent="true"><mo>u</mo><mo>¯</mo></mover>}\text{.}$

Explanation of Solution

Given:

Nuclear reaction,

  ${\Omega }^{-}\to {\Lambda }^0+K^{-}$

Formula used:

Given particles in the nuclear reaction can be written as,

  $\begin{array}{l}{\Omega }^{-}=sss\\ {\Lambda }^0=uds\\ K^{-}=\overline{u}s\end{array}$

Calculation:

  $\begin{array}{l}{\Omega }^{-}\to {\Lambda }^0+K^{-}\\ sss\to uds+\overline{u}s\end{array}$

Conclusion:

  $\text{Thus},\text{one }s\text{ quark converts into }d\text{ and one pair of }u\overline{u}\text{.}$