The mass of a nucleus ${ }_Z^A X$ is less that the sum of the masses of $(A-Z)$ number of neutrons and $Z$ number of protons in the nucleus. The energy equivalent to the corresponding mass difference is known as the binding energy of the nucleus. A heavy nucleus of mass $M$ can break into two light nuclei of masses $m_1$ and $m_2$ only if $\left(m_1+m_2\right)M^{\prime}$. The masses of some neutral atoms are given in the table below:
| ${ }_1^1 H$ |
$1.007825 u$ |
${ }_2^1 H$ |
$2.014102 u$ |
${ }_3^1 H$ |
$3.016050 u$ |
${ }_2^4 He$ |
$4.002603 u$ |
| ${ }_3^6 Li$ |
$6.015123 u$ |
${ }_7^3 Li$ |
$7.016004 u$ |
${ }_70^30 Zn$ |
$69.925325 u$ |
${ }_{34}^{82} Se$ |
$81.916709 u$ |
| ${ }_{64}^{152} Gd$ |
$151.919803 u$ |
${ }_{206}^{82} Gd$ |
$205.974455 u$ |
${ }_{209}^{83} Bi$ |
$208.980388 u$ |
${ }_{84}^{210} Po$ |
$209.982876 u$ |
$1.$ The correct statement is:
$(A)$ The nucleus ${ }_3^6 Li$ can emit an alpha particle
$(B)$ The nucleus ${ }_{84}^{210} P_0$ can emit a proton
$(C)$ Deuteron and alpha particle can undergo complete fusion.
$(D)$ The nuclei ${ }_{30}^{70} Zn$ and ${ }_{34}^{82} Se$ can undergo complete fusion.
$2.$ The kinetic energy (in $keV$ ) of the alpha particle, when the nucleus ${ }_{84}^{210} P _0$ at rest undergoes alpha decay, is:
$(A)$ $5319$ $(B)$ $5422$ $(C)$ $5707$ $(D)$ $5818$
Give the answer question $1$ and $2.$
