If there is a mass defect of $0.1\%$ in a nuclear fission process,how much energy will be released in the fission of $1\, kg$ of mass?

Energy in the sun is generated mainly by

In an atomic reactor,the kinetic energy of fast-moving neutrons can be reduced by colliding them with:

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

The binding energy of deuteron $_1^2H$ is $1.112 \, MeV$ per nucleon and an $\alpha$-particle $_2^4He$ has a binding energy of $7.047 \, MeV$ per nucleon. Then in the fusion reaction $_1^2H + _1^2H \to _2^4He + Q$,the energy $Q$ released is ........ $MeV$.

The mass of a nucleus ${ }_Z^A X$ is less than the sum of the masses of $(A-Z)$ neutrons and $Z$ protons. The energy equivalent to this mass difference is the binding energy. $A$ heavy nucleus of mass $M$ can break into two light nuclei of masses $m_1$ and $m_2$ only if $M > (m_1+m_2)$. The masses of some neutral atoms are given in the table below:

${ }_1^1 H$: $1.007825 u$	${ }_1^2 H$: $2.014102 u$	${ }_1^3 H$: $3.016050 u$	${ }_2^4 He$: $4.002603 u$
${ }_3^6 Li$: $6.015123 u$	${ }_3^7 Li$: $7.016004 u$	${ }_{30}^{70} Zn$: $69.925325 u$	${ }_{34}^{82} Se$: $81.916709 u$
${ }_{64}^{152} Gd$: $151.919803 u$	${ }_{82}^{206} Pb$: $205.974455 u$	${ }_{83}^{209} Bi$: $208.980388 u$	${ }_{84}^{210} Po$: $209.982876 u$

$1.$ The correct statement is:
$(A)$ The nucleus ${ }_3^6 Li$ can emit an alpha particle.
$(B)$ The nucleus ${ }_{84}^{210} Po$ can emit a proton.
$(C)$ Deuteron $({ }_1^2 H)$ and alpha particle $({ }_2^4 He)$ 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 ${ }_{84}^{210} Po$ at rest undergoes alpha decay, is:
$(A)$ $5319$ $(B)$ $5422$ $(C)$ $5707$ $(D)$ $5818$