Question

Repeat the previous problem for the decay mode

\({{\rm{\Omega }}^{\rm{ - }}} \to {{\rm{\Lambda }}^{\rm{0}}}{\rm{ + }}{{\rm{K}}^{\rm{ - }}}{\rm{. }}\)

Short answer

a. The change in strangeness for the reaction \({\Omega ^ - } \to {\Lambda ^0} + {K^ - }\) is \({\rm{ + 1}}\).

b. The conservation of charge and baryon number for the reaction \({\Omega ^ - } \to {\Lambda ^0} + {K^ - }\) is verified and also, lepton numbers are unaffected in the reaction

c. The equation \({\Omega ^ - } \to {\Lambda ^0} + {K^ - }\) in terms of quarks is given by \(sss \to uss + \bar ud\).

Step-by-step solution

Concept

Only weak nuclear forces have the ability to change the flavor of the quark in a decay process. Thus, if the change in the flavor of the quark is observed in the decay process, then weak nuclear forces are responsible for the decay of the particle.

Find the change in strangeness

(a)

Consider the given information:

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

Strangeness number for\({{\rm{\Lambda }}^{\rm{0}}}{\rm{ = - 1}}\)

Strangeness number for\({{\rm{K}}^{\rm{ - }}}{\rm{ = - 1}}\)

For the given reaction:\({\Omega ^ - } \to {\Lambda ^0} + {K^ - }\)

Total strangeness on the right side of the reaction is:\( - 1 - 1 = - 2\)

Total strangeness on the Left side of the reaction is:\( - 3\)

the change in strangeness is:

\(\Delta S = - 2 - ( - 3) = + 1\)

Therefore, the required change in strangeness for the reaction \({\Omega ^ - } \to {\Lambda ^0} + {K^ - }\) is \({\rm{ + 1}}\).

Verify the baryon number and charge are conserved

(b)

Considering the given information:

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

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

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

charge on\({K^ - } = - 1e\)

charge on\({\Omega ^ - } = - 1e\)

charge on\({\Lambda ^0} = 0\)

For the given reaction:\({\Omega ^ - } \to {\Lambda ^0} + {K^ - }\)

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

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

the change in charge is:

\(\begin{array}{c}\Delta q = - 1 - ( - 1)\\ = 0\end{array}\)

Total Baryon number on the right side of the reaction is:\( + 1 + 0 = + 1\)

Total Baryon number on the Left side of the reaction is:\( + 1\)

the change in Baryon number is:

\(\begin{array}{c}\Delta B = 1 - (1)\\ = 0\end{array}\)

\(\therefore \)charge and baryon number are conserved in the reaction\({\Omega ^ - } \to {\Lambda ^0} + {K^ - }\).

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

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

Find the equation

(c)

Considering the given information:

Given reaction is,\({\Omega ^ - } \to {\Lambda ^0} + {K^ - }\)

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

Quark structure of\({{\rm{\Lambda }}^{\rm{0}}}{\rm{ = uds}}\)

Quark structure of\({{\rm{K}}^{\rm{ - }}}{\rm{ = \bar us}}\)

As a result, the equation in terms of quarks is as follows:

\(\begin{array}{l}{\Omega ^ - } \to {\Lambda ^0} + {K^ - }\\sss \to uds + \bar us\end{array}\)

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

Therefore, the given reaction \({\Omega ^ - } \to {\Lambda ^0} + {K^ - }\) in terms of quarks is given as \(sss \to uds + \bar us\). As the flavor change of quark is experienced, weak nuclear forces must be responsible for the decay.