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Q.
Four atoms of hydrogen combine to form an ${ }_{2}^{4} He$ atom with a release of energy of
Nuclei
Solution:
The fusion reaction in the sun is a multi-step process in which hydrogen is burned into helium, hydrogen being the fuel and helium the ashes. The proton-proton $(p, p)$ cycle by which this occurs is represented by the following sets of reactions:
${ }_{1}^{1} H +{ }_{1}^{1} H \rightarrow{ }_{1}^{2} H +e^{+}+v+0.42 \,MeV \dots$(i)
$e^{+}+e^{-} \rightarrow \gamma+\gamma+1.02\, MeV\dots $(ii)
${ }_{1}^{2} H +{ }_{1}^{1} H \rightarrow{ }_{2}^{3} He +\gamma+5.49 \,MeV \dots$(iii)
${ }_{2}^{3} He +{ }_{2}^{3} He \rightarrow{ }_{2}^{4} He +{ }_{1}^{1} H +{ }_{1}^{1} H +12.86 \,MeV \dots$(iv)
For the fourth reaction to occur, the first three reactions must occur twice, in which case two light helium nuclei unite to form ordinary helium or nucleus. If we consider the combination $2( i )+2( ii )+2( iii )+( iv )$, the net effect is
$4_{1}^{1} H +2 e^{-} \rightarrow{ }_{2}^{4} He +2 v+6 \gamma+26.7 \,MeV$