$A$ radium $_{88}Ra^{224}$ isotope,on emission of an $\alpha$-particle,gives rise to a new element whose mass number and atomic number will be:

After the emission of an $\alpha$-particle from the atom $_{92}X^{238}$,the number of neutrons in the resulting atom will be:

The number of $\alpha$ and $\beta$-particles emitted in the nuclear reaction $_{92}U^{238} \to _{90}Th^{234} \to _{91}Pa^{234}$ are respectively:

An atom has a mass number of $232$ and an atomic number of $90$. How many $\alpha$-particles should it emit after the emission of two $\beta$-particles,so that the resulting atom has a mass number of $212$ and an atomic number of $82$?

What is the symbol for the nucleus remaining after $_{20}Ca^{42}$ undergoes $\beta$-emission?

$^{27}_{13}A$ is a stable element. $^{29}_{13}A$ is expected to disintegrate by: