Discussion
In the decay, \(\mu^{+} \rightarrow e^{+} + \nu_{e} + X\), what is \(X\)?
Answer: D
\(X\) is the muon antineutrino, \(\bar{\nu}_{\mu}\).
The decay is \(\mu^{+} \to e^{+} + \nu_{e} + \bar{\nu}_{\mu}\). Check the lepton family numbers. On the left, \(\mu^{+}\) has muon number \(L_{\mu} = -1\) and electron number \(L_{e} = 0\). On the right, \(e^{+}\) has \(L_{e} = -1\) and \(\nu_{e}\) has \(L_{e} = +1\), cancelling, so the remaining particle must carry \(L_{\mu} = -1\) and no charge: the muon antineutrino.
A photon carries no lepton number, so \(\gamma\) leaves \(L_{\mu}\) unbalanced. \(\bar{\nu}_{e}\) would make \(L_{e} = -1\) and leave \(L_{\mu}\) unbalanced. \(\nu_{\mu}\) has \(L_{\mu} = +1\), the wrong sign; it is the partner in the \(\mu^{-}\) decay.
Post your comment
Please log in or register to post a comment.
Examination Categories
- SSC Exams
- Banking Exams
- Teaching Exams
- Civil Services Exams
- Railways Exams
- Engineering Recruitment Exams
- Defence Exams
- State Government Exams
- Police Exams
- Insurance Exams
- Nursing Exams
- Other Government Exams
- NRA CET
- PG Entrance Exams
- Campus Placement
- MBA Entrance Exams
- CUET and UG Entrance
- Judiciary Exams
- Regulatory Body Exams
- Software Development