Question

Use the conservation laws and Tables 44-3 and 44-4 to identify particle$\mathit{x}$in each of the following reactions, which proceed by means of the strong interaction: (a)$\mathit{p}\mathbf{+}\mathit{p}\mathbf{\to }\mathit{p}\mathbf{+}{\mathit{\Lambda }}^{\mathbf{0}}\mathbf{+}\mathit{x}$(b)$\mathit{p}\mathbf{+}\overline{\mathbf{p}}\mathbf{\to }\mathit{n}\mathbf{+}\mathit{x}$(c).${\mathit{\pi }}^{\mathbf{-}}\mathbf{+}\mathit{p}\mathbf{\to }{\mathit{{\rm E}}}^{\mathbf{0}}\mathbf{+}{\mathit{K}}^{\mathbf{0}}\mathbf{+}\mathit{x}$

Short answer

(a) The value of$x$ is.$k^{+}$

(b) The value of$x$ is.$\overline{\nu }$

(c) The value of$x$ is.$K^0$

Step-by-step solution

(a) Strangeness of.p+p→p+Λ0+x

The given reaction is written as:

$\begin{array}{l}p+p\to p+{\Lambda }^0+x\\ p\to {\Lambda }^0+x\end{array}$

	p	${\Lambda }^0$	$x$
Charge	e	0	e
Baryon number	+1	+1	0
Strangeness	0	-1	+1

As the unknown particle is meson with spin zero, charge $+e$and strangeness.$+1$ The spin zero meson with charge of$+e$ and a strangeness of$+1$ is$K+$ particle. Thus, the value of$x$ is.$k^{+}$

(b) Strangeness of.p+p¯→n+x 

The given reaction is written as follows;

	p	$\overline{p}$	$n$	$x$
Charge	e	-e	0	0
Baryon number	+1	-1	+1	-1
Strangeness	0	0	0	0
Spin	$\frac{1}{2}$	$-\frac{1}{2}$	$\frac{1}{2}$	$-\frac{1}{2}$

From the above table, it is clear that the unknown particle is anti baryon with spin $-\frac{1}{2}$, charge and strangeness both zero. Therefore the unknown particle should be anti neutron.

Thus, the value of$x$ is.$\overline{\nu }$

(c) Strangeness of .π−+p→Ε0+K0+x

The given reaction is written as follows:

	${\pi }^{-}$	$p$	${{\rm E}}^0$	$K^0$	$x$
Charge	-e	e	0	0	0
Strangeness	0	0	-2	+1	+1

As the unknown particle has zero spin, zero charge and strangeness of.$+1$Thus, the value of$x$is.$K^0$$K^0$