5.4.2 So­lu­tion spin-b

Ques­tion:

Delta par­ti­cles have spin $\leavevmode \kern.03em\raise.7ex\hbox{\the\scriptfont0 3}\kern-.2em /\kern-.21em\lower.56ex\hbox{\the\scriptfont0 2}\kern.05em$. What is their square spin an­gu­lar mo­men­tum?

An­swer:

$$s (s+1)\hbar^2 = {\textstyle\frac{15}{4}}\hbar^2.$$