In the proton-proton cycle in the Sun,the energy released when an electron and its antiparticle combine is $\qquad$ .

Complete the following nuclear reaction:
${ }_{0}^{1} n+{ }_{92}^{235} U \rightarrow{ }_{92}^{236} U \rightarrow$ $\qquad$ $+{ }_{41}^{99} Nb+$ $\qquad$.

The minimum frequency of a photon required to break a particle of mass $15.348 \ amu$ into $4 \ \alpha$ particles is . . . . . . $kHz$.
[Mass of $He$ nucleus = $4.002 \ amu$, $1 \ amu = 1.66 \times 10^{-27} \ kg$, $h = 6.6 \times 10^{-34} \ J \cdot s$ and $c = 3 \times 10^8 \ m/s$]

If three helium nuclei combine to form a carbon nucleus,then the energy released in this reaction is $\times 10^{-2} \text{ MeV}$. (Given $1 \text{ u} = 931 \text{ MeV}/c^2$,atomic mass of helium $= 4.002603 \text{ u}$,atomic mass of carbon $= 12.000000 \text{ u}$)

If the number of uranium nuclei required per hour to produce a power of $64 \text{ kW}$ is $7.2 \times 10^{18}$, then the energy released per fission is

What is nuclear fission and nuclear fusion?