$1 \, g$ of hydrogen is converted into $0.993 \, g$ of helium in a thermonuclear reaction. The energy released is

The binding energy of a deuteron is $2.2 \, MeV$ and the binding energy of $_2^4He$ is $28 \, MeV$. If two deuterons fuse to form a $_2^4He$ nucleus,the energy released is ...... $MeV$.

The binding energies of nuclei $X$ and $Y$ are $E_1$ and $E_2$ respectively. Two atoms of $X$ fuse to give one atom of $Y$ and an energy $Q$ is released. Then:

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 $= 6 \times 10^{23} \text{ per mole}$)

Fast neutrons can be easily slowed down by:

$A$ star initially has $10^{40}$ deuterons. It produces energy via the processes:
$_1H^2 + _1H^2 \to _1H^3 + p$
$_1H^2 + _1H^3 \to _2He^4 + n$
The masses of the nuclei are as follows:
$M(H^2) = 2.014 \, amu; \, M(p) = 1.007 \, amu;$
$M(n) = 1.008 \, amu; \, M(He^4) = 4.001 \, amu$
If the average power radiated by the star is $10^{16} \, W$, the deuteron supply of the star is exhausted in a time of the order of: