When four hydrogen nuclei combine to form a helium nucleus in the interior of sun, the loss in mass in $0.0625$ amu. How much energy is released?

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Solution

Energy equivalent of 1 atomic mass unit $= 931.5 M e V = 1.492 \times 10^{- 10} J$
Energy released $= 0.0625 \times 931.5 = 58.22 M e V$
$= 0.0625 \times 1.492 \times 10^{- 10} = 9.325 \times 10^{- 12} J$

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When four hydrogen nuclei combine to form a helium nucleus in the interior of sun, the loss in mass in 0.0625 amu. How much energy is released?

Energy equivalent of 1 atomic mass unit =931.5MeV=1.492×10−10JEnergy released =0.0625×931.5=58.22MeV =0.0625×1.492×10−10=9.325×10−12J

When four hydrogen nuclei combine to form a helium nucleus in the interior of sun, the loss in mass in $0.0625$ amu. How much energy is released?

Energy equivalent of 1 atomic mass unit $= 931.5 M e V = 1.492 \times 10^{- 10} J$
Energy released $= 0.0625 \times 931.5 = 58.22 M e V$
$= 0.0625 \times 1.492 \times 10^{- 10} = 9.325 \times 10^{- 12} J$