The ${ }_{92}^{238} U$ atom disintegrates to ${ }_{84}^{214} Po$ with a half-life of $4.5 \times 10^9$ years by emitting $6$ $\alpha$-particles and $n$ electrons. Here,$n$ is:

What happens to the mass number and atomic number of an element when it emits $\gamma$-radiation?

In the given nuclear reaction,how many $\alpha$ and $\beta$ particles are emitted in the decay $_{92}X^{235} \to _{82}Y^{207}$?

The nucleus $_{10}^{23} Ne$ decays by $\beta^{-}$ emission. Write down the $\beta^{-}$-decay equation and determine the maximum kinetic energy of the electrons emitted. Given that:
$m(_{10}^{23} Ne) = 22.994466 \; u$
$m(_{11}^{23} Na) = 22.989770 \; u$

$A$ radioactive nucleus undergoes $\alpha$-emission to form a stable element. What will be the recoil velocity of the daughter nucleus if $V$ is the velocity of $\alpha$-emission and $A$ is the atomic mass of the radioactive nucleus?

$A$ nucleus $X$ emits a beta particle to produce a nucleus $Y$. If their atomic masses are $M_{x}$ and $M_{y}$ respectively,the maximum energy of the beta particle emitted is (where $m_{e}$ is the mass of an electron and $c$ is the velocity of light):