Question

For the following reaction, the particle ' x ' is \({}_6\text{C}^{11} \longrightarrow {}_5\text{B}^{11} + \beta^+ + \text{X}\)

• A. proton
• B. neutrino
• C. anti neutrino
• D. neutron

Hint:

$\beta^+$ (Positron) $\rightarrow \nu$ (Neutrino).
$\beta^-$ (Electron) $\rightarrow \bar{\nu}$ (Anti-neutrino).

Answer 1

The Correct Option is B

Step 1: Concept
In $\beta^+$ decay (positron emission), a proton converts into a neutron, releasing a positron and a specific neutral particle to conserve lepton number.

Step 2: Meaning
$\beta^+$ is a positron ($e^+$). To balance the lepton number (which is 0 initially), the emission of a positron (lepton number $-1$) must be accompanied by a neutrino (lepton number $+1$).

Step 3: Analysis
If the decay was $\beta^-$ (electron), an anti-neutrino would be emitted. Since this is positron ($\beta^+$) emission from Carbon-11 to Boron-11, the accompanying particle is a neutrino ($\nu$).

Step 4: Conclusion
The particle 'X' is a neutrino. Final Answer: (B)