In the nuclear reaction $_4^9Be(p, \alpha)X$,the $X$ is

Artificial radioactivity was first discovered by

The isotope $_{92}U^{235}$ decays in a number of steps to an isotope of lead $_{82}Pb^{207}$. The groups of particles emitted in this process will be

The missing particle in the reaction,$_{92}^{235}U + _{0}^{1}n \to _{56}^{146}Ba + ... + 3_{0}^{1}n$ is

In the nuclear transmutation $_4^9 Be + X \longrightarrow _4^8 Be + Y$,the pair $(X, Y)$ is (are):

For the nuclear reaction,$_{12}^{24}Mg + _{1}^{2}D \to _{2}^{4}He + ?$,the missing nuclide is