What is the amount of energy released by deuterium and tritium fusion (in $,MeV$)?

The energy released in the fusion of $2 \ kg$ of hydrogen deep in the sun is $E_{H}$ and the energy released in the fission of $2 \ kg$ of ${ }^{235} U$ is $E_U$. The ratio $\frac{E_H}{E_U}$ is approximately :
(Consider the fusion reaction as $4{ }_1^1 H + 2 e^{-} \rightarrow { }_2^4 He + 2 \nu + 6 \gamma + 26.7 \ MeV$,energy released in the fission reaction of ${ }^{235} U$ is $200 \ MeV$ per fission nucleus and $N_{A} = 6.023 \times 10^{23}$ )

Fast neutrons can be easily slowed down by:

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 ............ $MeV$.

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 isotope ${}_{5}^{12}B$ having a mass $12.014 \text{ u}$ undergoes $\beta$-decay to ${}_{6}^{12}C$. ${}_{6}^{12}C$ has an excited state of the nucleus $({}_{6}^{12}C^*)$ at $4.041 \text{ MeV}$ above its ground state. If ${}_{5}^{12}B$ decays to ${}_{6}^{12}C^*$, the maximum kinetic energy of the $\beta$-particle in units of $\text{MeV}$ is ($1 \text{ u} = 931.5 \text{ MeV}/c^2$, where $c$ is the speed of light in vacuum).