When $_3Li^7$ nuclei are bombarded by protons, and the resultant nuclei are $_4Be^8$, the emitted particles will be

A radioactive nucleus with $Z$ protons and $N$ neutrons emits an $\alpha - $ particle, $2\beta$- particles and $2$ gamma rays. The number of protons and neutrons in the nucleus left after the decay respectively, are

$_1{H^1}{ + _1}{H^1}{ + _1}{H^2} \to X + {\;_{ + 1}}{e^0} + $energy. The emitted particle is

A nucleus $_Z{X^A}$ emits $9 \alpha$ - particles and $5 \beta$ particle. The ratio of total protons and neutrons in the final nucleus is

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

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 $]$