When $_{90}^{228}Th$ transforms to $_{83}^{212}Bi$,the number of emitted $\alpha$- and $\beta$- particles is,respectively:

Who discovered radioactivity?

$A$ nucleus of mass $218\, amu$ in a free state decays to emit an $\alpha$-particle. The kinetic energy of the emitted $\alpha$-particle is $6.7\, MeV$. The recoil energy (in $MeV$) of the daughter nucleus is:

In a radioactive decay chain,the initial nucleus is ${}_{90}^{232}Th$. At the end,$6$ $\alpha$-particles and $4$ $\beta$-particles are emitted. If the final nucleus is ${}_{Z}^{A}X$,then $A$ and $Z$ are given by:

$A$ radioactive nucleus of mass number $A$,initially at rest,emits an $\alpha$-particle with a speed $v$. The recoil speed of the daughter nucleus will be

The end product of the decay of $ { }_{90} Th^{232} $ is $ { }_{82} Pb^{208} $. The number of $ \alpha $ and $ \beta $ particles emitted are respectively: