Chapter 32, Problem 36P

(a)

To determine

To Show: $B^{-}$ meson is a $b\overline{u}$ quark is consistent for all quantum number.

Explanation of Solution

Introduction:

  $B$ meson is made from the bottom antiquark and $B^{+},B^0,B_s^0$ or charm quark ( $B_c^{+}$ ).

1. According to cheek charge conservation

  $\begin{array}{l}b\overline{u}=\frac{e}{3}-\frac{2e}{3}\\ b\overline{u}=-e\end{array}$

2. According to Baryon Number conservation

  $\begin{array}{l}B=\frac{1}{3}-\frac{1}{3}\\ B=0\end{array}$

3. According to strangeness conservation

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

4. Charm conservation

  $0=0+0$

5. According to topness, $0=0+0$
6. According to Spin, $0=\frac{1}{2}-\frac{1}{2}$
7. According to bottomness: $-1+0=-1$

Conclusion:

  $B^{-}$ meson is a $b\overline{u}$ quark is consistent for all quantum number.

(b)

To determine

Quark combinations of $B^{+},B^0,\overline{B^0}$ .

Answer to Problem 36P

  $B^{+}=\overline{b}u$

  $B^0=b\overline{d}$

  $\overline{B^0}=\overline{b}d$

Explanation of Solution

The $B^{+}$ is the antiparticle of $B^{-}$ ,therefore, $B^{+}=\overline{b}u$

The $B^0$ have some bottom quark but it is neutral. So $B^0=b\overline{d}$

The $\overline{B^0}$ is the anti-particle to $B^0$ , therefore, $\overline{B^0}=\overline{b}d$