Question

Is the decay \({\rm{n}} \to {{\rm{e}}^{\rm{ + }}}{\rm{ + }}{{\rm{e}}^{\rm{ - }}}\)possible considering the appropriate conservation laws? State why or why not.

Short answer

The decay \({\rm{n}} \to {{\rm{e}}^{\rm{ + }}}{\rm{ + }}{{\rm{e}}^{\rm{ - }}}\)is not possible as per the conservation laws.

Step-by-step solution

Definition of Concept

A decay which does not satisfy anyone of the conservation laws of mass, charge, momentum, angular momentum, baryon number or lepton number, does not exist.

Explain is the decay \({\rm{n}} \to {{\rm{e}}^{\rm{ + }}}{\rm{ + }}{{\rm{e}}^{\rm{ - }}}\) possible considering the appropriate conservation laws

Considering the given information:

Given reaction is\(n \to {e^ + } + {e^ - }\)

We can check the following conservation laws to see if the decay\(n \to {e^ + } + {e^ - }\)is possible.

Decay:\(n \to {e^ + } + {e^ - }\)

Charge:\(0 \to + 1 - 1\)

\(\therefore \)charge is conserved.

Baryon Number\(\left( B \right): + 1 \to 0 + 0\)

\(\therefore \)B is not conserved.

Electron Lepton Number\(\left( {{L_e}} \right):0 \to - 1 + 1\)

\(\therefore \)\({L_e}\)is conserved.

Strangeness\(\left( S \right):0 \to 0 + 0\)

\(\therefore \)\(S\)is conserved

While charge, strangeness, and lepton number are all conserved, the baryon number is not conserved for the given decay. As a result, according to conservation laws, decay is not possible.

Therefore, the required decay\({\rm{n}} \to {{\rm{e}}^{\rm{ + }}}{\rm{ + }}{{\rm{e}}^{\rm{ - }}}\)is not possible as per the conservation laws.