Question

When four hydrogen nuclei combine to form a helium nucleus in the interior of the sun, the loss in mass is 0.0265 amu?

Answer 1

Answer:

0.0265 amu is 4.4x10^(-29) kg multiply by c^2 (in m/sec) and obtain 3.955*10^(-12) Joules or 24.68 MeV. Multiply by Avogadro’s Number to obtain 2.382*10^12 Joules for one mole, or 4 grams, of hydrogen. Divide by 0.004 to obtain 595.4 trillion joules per kilogram of hydrogen. 634.0 million tons of hydrogen is converted to 629.8 million tons of helium and 4.2 million tons of energy each second in the Sun to produce sufficient energy to maintain the solar surface at 5772 Kelvin producing 62.94 MW/m2 which drops to 1,362 W/m2 at 1 AU distance from the Solar surface.

Answer 2

Answer:

The energy released is 265

Explanation:

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