A nucleus  $_n{X^m}$ emits one $\alpha$ particle and two $\beta$ particles. The resulting nucleus is

In the equation ${ }_{13}^{27} Al +{ }_2^4 He \longrightarrow{ }_{15}^{30} P + X ,$ The correct symbol for $X$ is

A certain radioactive material $_ZX^A$ starts emitting $\alpha $ and $\beta $ particles successively such that the end product is $_{Z-3}Y^{A-8}$. The number of $\beta $ and $\alpha $ particles emitted are

A nuclear decay is possible if the mass of the parent nucleus exceeds the total mass of the decay particles. If $M(A, Z)$ denotes the mass of a single neutral atom of an element with mass number $A$ and atomic number $Z$, then the minimal condition that the $\beta$ decay $X_Z^A \rightarrow Y_{Z+1}^A+\beta^{-}+\bar{v}_e$ will occur is ( $m_e$ denotes the mass of the $\beta$ particle and the neutrino mass $m_v$ can be neglected)

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 $]$

Assertion: ${}_Z{X^A}$ undergoes a $2\alpha  -$ decays, $2\beta  -$ decays and $2\gamma  - $ decays and the daughter product is ${}_{Z - 2}{X^{A - 8}}$

Reason : In $\alpha  - $decays the mass number decreases by $4$ and atomic number decreases by $2$. In $2\beta  - $ decays the mass number remains unchanged, but atomic number increases by $1$ only.