Chapter 32, Problem 41P

To determine

The quark combination would produce by a ${\text{D}}_S^{\text{+}}$ meson.

Explanation of Solution

Introduction:

The quark combination would produce by a ${\text{D}}_S^{\text{+}}$ meson is .

Mesons consist of a quark–antiquark pair. Quarks may be so tightly bound together that

they may not ever exist singly in the free state. But quarks can be detected indirectly when they turn into narrow jets of other particles.

To produce a ${\text{D}}_S^{\text{+}}$ meson, the total charge on the quark combination must be $1$. The value of strangeness $S$ must be $1$ and the value of charm $c$ must be $+1$ Also the value of the baryon number $B$ is $0$ as per given condition. Assume that there is no topness and bottomness. Since to obtain a charm $c$ the quark must be $c$ with a charge of $+\frac{2}{3}e$. The net charge on the ${\text{D}}_S^{\text{+}}$ meson is $+1$ so there will be another quark with a charge of $+\frac{1}{3}e$

Here the baryon number is $0$ so, another quark must be an antiquark. To obtain the value of strangeness is $1$, the antiquark must be an anti-strange. So, the given combination of the quarks to produce a ${\text{D}}_S^{\text{+}}$ meson is .