Chapter 12, Problem 29P

To determine

(a)

The explanation of the nuclear reaction ${\pi }^{+}+p\to {\Sigma }^{+}+K^{+}$ in terms of constituent of quarks.

Answer to Problem 29P

The quarks $d\text{and}\overline{d}$ annihilate and recreate a new quark $s$ and $\overline{s}$.

Explanation of Solution

Given:

Nuclear reaction,

${\pi }^{+}+p\to {\Sigma }^{+}+K^{+}$.

Formula used:

$\begin{array}{l}{\pi }^{+}=u\overline{d}\\ p=uud\\ {\Sigma }^{+}=uus\\ +K^{+}=u\overline{s}\end{array}$.

Calculation:

$\begin{array}{l}{\pi }^{+}+p\to {\Sigma }^{+}+K^{+}\\ u\overline{d}+uud\to uus+u\overline{s}\\ \cancel{u}\overline{d}+\cancel{uu}d\to \cancel{uu}s+\cancel{u}\overline{s}\\ d+\overline{d}\to s+\overline{s}\end{array}$.

Conclusion:

$d\text{and}\overline{d}$ annihilate and energy released during the process is used to recreate another new particle $s\text{ and }\overline{s}$.

To determine

(b)

The explanation of the nuclear reaction $\gamma +n\to {\pi }^{-}+p$ in terms of constituent of quarks.

Answer to Problem 29P

The energy of gamma radiation converts into mass of the particles $u\text{and}\overline{u}.$.

Explanation of Solution

Given:

Nuclear reaction,

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

Formula used:

$\begin{array}{l}n=udd\\ {\pi }^{-}=d\overline{u}\\ p=uud\end{array}$.

Calculation:

$\begin{array}{l}\gamma +n\to {\pi }^{-}+p\\ \gamma +udd\to d\overline{u}+uud\\ \gamma \to u+\overline{u}\end{array}$.

Conclusion:

The energy of the gamma radiation is converted into the mass as $u\text{and}\overline{u}.$.

To determine

(c)

The explanation of the nuclear reaction $p+p\to p+p+p+\overline{p}$ in terms of constituent of quarks.

Answer to Problem 29P

The kinetic energy of two protons converts into two $(d+\overline{d})$ pairs and a $(u+\overline{u})$ pair.

Explanation of Solution

Given:

Nuclear reaction,

$p+p\to p+p+p+\overline{p}$.

Formula used:

$\begin{array}{l}p=uud\\ \overline{p}=\overline{u}\overline{u}\overline{d}\end{array}$.

Calculation:

$\begin{array}{l}p+p\to p+p+p+\overline{p}\\ \text{(kinetic energy colliding of protons) + }uud+uud\to uud+uud+uud+\overline{u}\overline{u}\overline{d}\\ \text{kinetic energy of colliding protons }\to 2(u+\overline{u})+(d+\overline{d})\end{array}$.

Conclusion:

Kinetic energy of colliding particle converts into masses of two pairs of $(u+\overline{u})$ and one pair of $(d+\overline{d})$.

To determine

(d)

The explanation of the nuclear reaction $K^{-}+p\to K^{+}+K^0+{\Omega }^{-}$ in terms of constituent of quarks.

Answer to Problem 29P

$\begin{array}{l}u\text{}\text{and}\overline{u}\text{annihilate and energy of this particles is used to produced two pairs of particles}\\ (s+\overline{s}).\end{array}$.

Explanation of Solution

Given:

Nuclear reaction,

$K^{-}+p\to K^{+}+K^0+{\Omega }^{-}$.

Formula used:

$\begin{array}{l}{K}^{-}=s\overline{u}\\ p=uud\\ K^{+}=u\overline{s}\\ K^0=d\overline{s}\\ {\Omega }^{-}=sss\end{array}$.

Calculation:

$\begin{array}{l}{K}^{-}+p\to K^{+}+K^0+{\Omega }^{-}\\ s\overline{u}+uud\to u\overline{s}+d\overline{s}+sss\\ (u+\overline{u})\to 2(s+\overline{s})\end{array}$.

Conclusion:

Thus, $u$ and $\overline{u}$ annihilate and convert into $2(s+\overline{s}).$

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