Question

Consider the semi leptonic weak decays 

(a) Σ⁻ → n + e⁻ + bar ν_e 

(b) Σ⁺ → n + e⁺ + ν_e 

Explain why the reaction (a) is observed but (b) is not.

Answer 1

The reaction (a) can go via the mechanism of the diagram shown in Fig. 10.12a. However, no diagram with single W exchange can be drawn for the reaction 

(b) which at the quark level implies the transformation 

uus → udd + e⁺ + ν_e 

as in Fig. 10.12b and would require two separate quark transitions which involve the emission and absorption of two W bosons – a mechanism which is of higher order and therefore negligibly small. 

The above conclusion can also be reached by invoking for the selection rule for semi leptonic decays. 

Reaction 

(a) obeys the rule ΔS = ΔQ = +1 and is therefore allowed, while in reaction 

(b) we have ΔS = +1, but ΔQ = −1, and therefore forbidden.