Use the conservation laws to identify the particle labelled x in the following reactions, which proceed by means of the strong interaction. (a) ; (b) ; (c) .

Solution:             Click For PDF Version

(a)

In strong interactions charge, baryon number and strangeness are conserved. In the reaction

,

As the baryon number of a proton is 1, the total baryon number of the initial particles is 2. As the strangeness of a proton is zero, the strangeness of the initial particles is zero. As the charge of a proton is e, the total charge of the final particles has to be +2e. Now we note that of the final particles are baryons and each has a baryon number 1, so the particle x has to be a meson. As the strangeness of is , the missing particle is a meson with strangeness +1 and charge e. From the octet of mesons, we note that the particle that has strangeness +1, and charge +e is , the particle x is .

(b)

In the reaction

,

the total baryon number is 0 as proton and antiproton have baryon number of 1 and , respectively. As one of the two final particles is a neutron, the particle x is an antineutron, .

(c)

In the reaction

,

the total charge and strangeness of the initial particles are zero, and the total baryon number is +1, as is a meson and proton is a baryon and its baryon number is 1.

Of the final particles is a baryon with baryon number 1 and strangeness , and is a meson with strangeness and baryon number 0, therefore, the particle x has to have strangeness , baryon number 0, and is uncharged. Therefore, it is another .