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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 $$= 4.002603 \text{ u}$$)
Correct Answer: 727
Three helium nuclei combine to form carbon: $$3 \, ^4_2He \rightarrow \, ^{12}_6C$$
We calculate the mass defect as follows:
$$ \Delta m = 3 \times 4.002603 - 12 = 12.007809 - 12 = 0.007809 \text{ u} $$Using this mass defect, the energy released is:
$$ E = \Delta m \times 931 \text{ MeV} = 0.007809 \times 931 = 7.27 \text{ MeV} $$ $$ = 727 \times 10^{-2} \text{ MeV} $$Therefore, the final answer is 727.
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