A nucleus of an element $_{84}{X^{202}}$ emits an $\alpha-$ particle first, a $\beta-$ particle next and then a gamma photon. The final nucleus formed has an atomic number

When a radioactive substance emits an $\alpha$- particle, its position in the periodic table is lowered by

Suppose a ${ }_{88}^{226} Ra$ nucleus at rest and in ground state undergoes $\alpha$-decay to a ${ }_{56}^{22} Rn$ nucleus in its excited state. The kinetic energy of the emitted $\alpha$ particle is found to be $4.44 MeV$. ${ }_{86}^{22} Rn$ nucleus then goes to its ground state by $\gamma$-decay. The energy of the emitted $\gamma$-photon is. . . . . . . .$keV$,

[Given: atomic mass of ${ }_{ gs }^{226} Ra =226.005 u$, atomic mass of ${ }_{56}^{22} Rn =222.000 u$, atomic mass of $\alpha$ particle $=4.000 u , 1 u =931 MeV / c ^2, c$ is speed of the light $]$

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

Consider a $\beta$ decay reaction

${}_1^3H \to {}_2^3He + {e^{ - 1}} + \bar v$

Atomic mass of ${}_1^3H$and ${}_2^3He$ are $3.016050\,u$ and $3.016030\,u$. Find the maximum possible energy of electron ....... $MeV$

The isotope ${ }_5^{12} \mathrm{~B}$ having a mass $12.014 \mathrm{u}$ undergoes $\beta$-decay to ${ }_6^{12} \mathrm{C} .{ }_6^{12 .}$ has an excited state of the nucleus $\left({ }_6^{12} \mathrm{C}^*\right)$ at $4.041 \mathrm{MeV}$ above its ground state. If ${ }_5^{12} \mathrm{~F}$ decays to ${ }_6^{12} \mathrm{C}^*$, the maximum kinetic energy of the $\beta$-particle in units of $\mathrm{MeV}$ is ( $1 \mathrm{u}=931.5 \mathrm{MeV} / c^2$, where $c$ is the speed of light in vacuum).