The atomic number of a nucleus after a $\beta$-emission from a nucleus with an atomic number of $40$ will be:

After the emission of a $\beta$-particle followed by an $\alpha$-particle from ${ }_{83}^{214} Bi$, the number of neutrons in the resulting atom is:

After losing a number of $\alpha$ and $\beta$ particles,$_{92}U^{238}$ changed to $_{82}Pb^{206}$. The total number of particles lost in this process is

In the reaction $_{93}Np^{239} \xrightarrow{} _{94}Pu^{239} + (?)$,the missing particle is

$_{11}^{23}Na$ is the more stable isotope of $Na$. Find out the process by which $_{11}^{24}Na$ can undergo radioactive decay.

During a $\beta$-decay,the mass of the atomic nucleus