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 $2.$

• A

  $(C,A)$

• B

  $(B,C)$

• C

  $(B,D)$

• D

  $(A,D)$