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

A nucleus decays by ${\beta ^ + }$ emission followed by a gamma emission. If the atomic and mass numbers of the parent nucleus are $Z$ and $A$ respectively, the corresponding numbers for the daughter nucleus are respectively.

In the following nuclear rection, $D \stackrel{\alpha}{\longrightarrow} D _{1} \stackrel{\beta-}{\longrightarrow} D _{2} \stackrel{\alpha}{\longrightarrow} D _{3} \stackrel{\gamma}{\longrightarrow} D _{4}$ Mass number of $D$ is $182$ and atomic number is $74$ . Mass number and atomic number of $D _{4}$ respectively will be

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 $]$

The particles emitted by radioactive decay are deflected by magnetic field. The particles will be

Consider the following radioactive decay process 

${ }_{84}^{218} A \stackrel{\alpha}{\longrightarrow} A_1 \stackrel{\beta^{-}}{\longrightarrow} A_2 \stackrel{\gamma}{\longrightarrow} A_3 \stackrel{\alpha}{\longrightarrow} A_4 \stackrel{B^{+}}{\longrightarrow} A_5 \stackrel{\gamma}{\longrightarrow} A_6$

The mass number and the atomic number $A _6$ are given by