$A$ nuclear reaction given by $_Z{X^A} \to {_{Z+1}}{Y^A} + _{-1}{e^0} + \bar{\nu}$ represents:

The radionuclide $^{11} C$ decays according to
$_{6}^{11} C \rightarrow_{5}^{11} B + e^{+} + \nu: \quad T_{1/2} = 20.3 \; min$
The maximum energy of the emitted positron is $0.960 \; MeV$. Given the mass values:
$m(_{6}^{11} C) = 11.011434 \; u$ and $m(_{5}^{11} B) = 11.009305 \; u$
Calculate $Q$ and compare it with the maximum energy of the positron emitted.

Assertion: Radioactive nuclei emit $\beta^-$ particles.
Reason: Electrons exist inside the nucleus.

Who discovered radioactivity?

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

$A$ nucleus $X$ undergoes the following transformation:
$X \xrightarrow{\alpha} Y$
$Y \xrightarrow{2\beta} Z$
Then: