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 $]$

$_6^{12}C$ absorbs an energetic neutron and emits beta particles. The resulting nucleus is

In an $\alpha -$ decay, the kinetic energy of $\alpha -$ particle is $48\, MeV$ and $Q$ value of the reaction is $50\, MeV$. The mass number of the mother nucleus is [assume that daughter nucleus is inground state]

In which radioactive disintegration, neutron dissociates into proton and electron

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.

Three $\alpha - $ particles and one $\beta - $ particle decaying takes place in series from an isotope $_{88}R{a^{238}}$. Finally the isotope obtained will be