For the given reaction, the particle X is _6C | vedclass

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(D) In $\beta^+$ decay (positron emission), a proton inside the nucleus is converted into a neutron, a positron $(\beta^+)$, and a neutrino $(
u)$.Th

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$A$ neutrino

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(D) In $\beta^+$ decay (positron emission), a proton inside the nucleus is converted into a neutron, a positron $(\beta^+)$, and a neutrino $(
u)$.The reaction is given by: $_6C^{11} 	o _5B^{11} + _1e^0 + 
u$.According to the law of conservation of lepton number, the lepton number must be conserved. Since the positron $(\beta^+)$ has a lepton number of $-1$, a neutrino $(
u)$ with a lepton number of $+1$ must be emitted to balance the equation.Therefore, the particle $X$ is a neutrino.

For the given reaction, the particle $X$ is $_6C^{11} \to _5B^{11} + \beta^+ + X$.

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