Energy released in the fusion of $1\, kg$ of deuterium nuclei is:

$113$. Energy released in the fission of a single uranium nucleus is $200 \text{ MeV}$. Then the number of fissions per second to produce $5 \text{ mW}$ power is

$A$ $^{235}U$ nuclear reactor generates energy at a rate of $3.70 \times 10^7 \text{ J/s}$. Each fission liberates $185 \text{ MeV}$ of useful energy. If the reactor has to operate for $144 \times 10^4 \text{ s}$, the mass of the fuel needed is (Assume Avogadro's number $= 6 \times 10^{23} \text{ mol}^{-1}$, $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$) (in $\text{ kg}$)

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:
$(A)$ $5319$ $(B)$ $5422$ $(C)$ $5707$ $(D)$ $5818$

Fusion reaction is initiated with the help of

What will be the energy released in joule, in the process of fission by $1 \text{ mg}$ of ${ }_{92}^{240} U$? Assume energy release per fission is $200 \text{ MeV}$. [Use Avogadro's number as $6 \times 10^{23}$ and $1 \text{ eV} = 1.6 \times 10^{-19} \text{ J}$]