The ${ }_{92}^{235} U$ atom disintegrates to ${ }_{82}^{207} Pb$ with a half-life of $10^9 \ yr$. In the process,it emits $7 \ \alpha$ particles and $n \ \beta^{-}$ particles. Here,$n$ is

The atomic mass number of the element thorium is $232$ and its atomic number is $90$. The end product of this radioactive element is an isotope of lead (atomic mass $208$ and atomic number $82$). The number of alpha and beta particles emitted is:

$A$ nucleus of lead $Pb_{82}^{214}$ emits two electrons followed by an $\alpha$-particle. The resulting nucleus will have

Choose the correct nuclear process from the below options $[p : \text{proton}, n : \text{neutron}, e^{-} : \text{electron}, e^{+} : \text{positron}, v : \text{neutrino}, \overline{v} : \text{antineutrino}]$.

After one $\alpha$ and two $\beta$ emissions,what happens to the nucleus?

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:
$m(_{10}^{23} Ne) = 22.994466 \; u$
$m(_{11}^{23} Na) = 22.989770 \; u$