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 radioactive nucleus undergoes $\alpha$- emission to form a stable element. What will be the recoil velocity of the daughter nucleus if $V$ is the velocity of $\alpha$-emission and $A$ is the atomic mass of radioactive nucleus

${ }_{82}^{290} X \xrightarrow{\alpha} Y \xrightarrow{e^{+}} Z \xrightarrow{\beta^{-}} P \xrightarrow{e^{-}} Q$

In the nuclear emission stated above, the mass number and atomic number of the product $Q$ respectively, are

$_{90}^{232}Th$ an isotope of thorium decays in ten stages emitting six  $\alpha$-particles and four $\beta$-particles in all. The end product of the decay is

Before the neutrino hypothesis, the beta decay process was thought to be the transition, $n \to p + {e^ - }$ If this was true, show that if the neutron was at rest, the proton and electron would emerge with fixed energies and calculate them. Experimentally, the electron energy was found to have a large range. 

In the nuclear reaction $_{85}{X^{297}} \to Y + 4\alpha ,\;Y$ is