$113$. 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

The control rod in a nuclear reactor is made of

The average energy released per fission for the nucleus of $_{92}^{235} U$ is $190 \text{ MeV}$. When all the atoms of $47 \text{ g}$ pure $_{92}^{235} U$ undergo fission process, the energy released is $\alpha \times 10^{23} \text{ MeV}$. The value of $\alpha$ is . . . . . . . . . . . (Avogadro Number $= 6 \times 10^{23} \text{ per mole}$)

The binding energy per nucleon of ${}_3^7Li$ and ${}_2^4He$ nuclei are $5.60 \, MeV$ and $7.06 \, MeV$, respectively. In the nuclear reaction ${}_3^7Li + {}_1^1H \to 2 \, {}_2^4He + Q$, the value of energy $Q$ released is ............ $MeV$.

Energy released when two deuterons $\left({ }_1 H ^2\right)$ fuse to form a helium nucleus $\left({ }_2 He ^4\right)$ is $:$
(Given $:$ Binding energy per nucleon of ${ }_1 H ^2=1.1 \ \text{MeV}$ and binding energy per nucleon of ${ }_2 He ^4=7.0 \ \text{MeV}$) (in $\text{MeV}$)

$U^{235}$ 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 $144 \times 10^4 \text{ s}$, then 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 $\text{ kg}$)