Physics
When four hydrogen nuclei combine to form a helium nucleus in the interior of sun, the loss in mass is 0.0265 a.m.u. How much energy is released?
Radioactivity
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Answer
Given,
loss in mass (∆m) = 0.0265 a.m.u.
We know that,
1 a.m.u = 931.5 MeV
Therefore, E = 0.0265 a.m.u. × 931.5 MeV = 24.7 MeV
Hence, the energy released = 24.7 MeV
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