$A$ nucleus of atomic mass $A$ and atomic number $Z$ emits a $\beta^-$ particle. The atomic mass and atomic number of the resulting nucleus are

An element $A$ decays into element $C$ by a two-step process:
$A \to B + {\;_2}He^4$
$B \to C + 2e^-$
Then:

When a radioactive substance emits an $\alpha$-particle,its position in the periodic table is lowered by

In the following nuclear reaction,$D \xrightarrow{\alpha} D_{1} \xrightarrow{\beta^-} D_{2} \xrightarrow{\alpha} D_{3} \xrightarrow{\gamma} D_{4}$. The mass number of $D$ is $182$ and the atomic number is $74$. The mass number and atomic number of $D_{4}$ respectively will be:

$A$ free neutron decays into a proton, an electron and

In $\beta^{-}$ decay,a neutron transforms into a proton within the nucleus according to the equation:
$\text{neutron} \rightarrow \text{proton} + \beta^{-} + x$
In this equation,the particle represented by '$x$' is: