One of the decay modes of the omega minus is \({\Omega ^ - } \to {\Xi ^0} + {\pi ^ - }\).

(a) What is the change in strangeness?

(b) Verify that baryon number and charge are conserved, while lepton numbers are unaffected.

(c) Write the equation in terms of the constituent quarks, indicating that the weak force is responsible.

In particle physics, there are six standard conservation laws that must be followed by all the processes. These are mass, charge, momentum, angular momentum, lepton number, and baryon number.

Given reaction is,\({\Omega ^ - } \to {\Xi ^0} + {\pi ^ - }\;\; \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left( 1 \right)\)

Strangeness number for\({\Omega ^ - } = - 3\)

Strangeness number for\({{\rm{\Xi }}^{\rm{0}}}{\rm{ = - 2}}\)

Strangeness number for\({{\rm{\pi }}^{\rm{ - }}}{\rm{ = 0}}\)

Baryon number for\({\Omega ^ - } = + 1\)

Baryon number for\({{\rm{\Xi }}^{\rm{0}}}{\rm{ = + 1}}\)

Baryon number for\({{\rm{\pi }}^{\rm{ - }}}{\rm{ = 0}}\)

Charge for\({\Omega ^ - } = - 1\)

charge for\({{\rm{\Xi }}^{\rm{0}}}{\rm{ = 0}}\)

charge for\({{\rm{\pi }}^{\rm{ - }}}{\rm{ = - 1}}\)

Quark structure of\({{\rm{\Omega }}^{\rm{ - }}}{\rm{ = sss}}\)

Quark structure of\({{\rm{\Xi }}^{\rm{0}}}{\rm{ = }}\)uss

Quark structure of\({{\rm{\pi }}^{\rm{ - }}}{\rm{ = \bar ud}}\)

(a)Considering reaction (1), for the given values of strangeness,

Total strangeness on the right is: \( - 2 + 0 = - 2\)

Total strangeness on the left is:\( - 3\)

\(\therefore \)Change in strangeness will be:\( - 2 - \left( { - 3} \right) = 1\)

Hence, the required change in strangeness for the reaction \({\Omega ^ - } \to {\Xi ^0} + {\pi ^ - }\) is \({\rm{ + 1}}\).

(b) Considering the given reaction and the given values of baryon number,

Total baryon number on the right is:\( + 1 + 0 = + 1\)

Total baryon number on the left is:\( + 1\)

\(\therefore \)Change in baryon number will be:\( + 1 - \left( { + 1} \right) = 0\)

Hence, baryon number is conserved in reaction.

Similarly for the given values of charge and the known value of elementary charge(e),

Total charge on the right side of the reaction is:\( - 1e + 0 = - 1e\)

Total charge on the left side of the reaction is:\( - 1e\)

\(\therefore \)Change in charge will be:\( - 1e - \left( { - 1e} \right) = 0\)

So, charge is also conserved in the reaction.

Because there are no leptons involved in this reaction, lepton number is unaffected.

Therefore, the conservation of charge and baryon number for the reaction \({\Omega ^ - } \to {\Xi ^0} + {\pi ^ - }\) is verified and also, the lepton numbers are unaffected.

(c)Considering the given reaction and the composition of particles:

The reaction in terms of quarks, can be written as:

\(\begin{array}{l}{\Omega ^ - } \to {\Xi ^0} + {\pi ^ - }\\sss \to uss + \bar ud\end{array}\)

The reaction alters the flavor of the quarks and only weak reaction can modify the flavor of a quark, whereas a strong reaction cannot modify the flavor of a quark. The above equation indicates that a weak force is to blame for the reaction.

Therefore, the required equation \({\Omega ^ - } \to {\Xi ^0} + {\pi ^ - }\) in terms of quarks is given by \(sss \to uss + \bar ud\). As the flavor of the quark is changing, the weak forces are only responsible for the reaction.