The equation E = m_0 c^2 tells us how much energy is required create a certain a
ID: 1997316 • Letter: T
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The equation E = m_0 c^2 tells us how much energy is required create a certain amount of mass, or equivalently how much energy is released when a certain amount of mass is destroyed. In this equation, m_0 represents the rest mass (what you are used to thinking of as the mass). The Sun's energy comes from nuclear fusion, fusing four hydrogen atoms together to make one helium atom. The mass of a helium atom is slightly less than the mass of four hydrogen atoms. This "missing" mass is released as energy. Look up the masses of hydrogen and helium atoms and use E = m_0 c^2 to calculate how much energy is released in a single fusion reaction in the Sun. The Sun's power output is approximately 4* 10^26 Watts. How many fusion reactions happen each second in the Sun? When the Sun formed it was about 75% hydrogen by mass. 10% of this hydrogen is available for fusion (the temperature is only hot enough for fusion in the core). How many hydrogen atoms were available for fusion when the Sun formed? (You should look up the Sun's mass online.) Based on your answers above, how long will the Sun shine before all of its available hydrogen is converted to helium? How does your result in (d) compare to the accepted lifetime of our Sun? If your result differs from the accepted value by more than 50% then you have an error in your calculation.Explanation / Answer
a) Mass of 4 H atoms: 4.03130 AMU
Mass of 1 He atom: 4.00268 AMU
difference in mass = 0.02862 AMU = REST MASS = 4.75 * 10-29 kg
ENERGY LIBERATED = E = mc2
E = (4.75 * 10-29 kg) x c2
E = (4.75 * 10-29 kg) x (3x108m/s)2
E = 5.278x10-14 J
b) no of fusions per second = 4* 1026 / 5.278x10-14= 7.58* 1039
c) mass of sun = 1.99 * 1030 kg
mass of sun comprised of hydrogen = 75 % of 1.99 * 1030 kg
= 1.49 * 1030 kg
Mass of hydrogen = 1.67 * 10-27 kg
no of hydrogen atoms in sun = 1.49 * 1030 / 1.67 * 10-27 = 8.92 * 1056
no of hydrogen used in fusion per second = no of fusion * no of H used in 1 fusion
= 7.58* 1039 *4 = 3.03 *1040
lifetime of sun = no of hydrogen atoms present in sun / no of Hydrogen atoms used in 1 s
= 8.92 * 1056 / 3.03 *1040 (in seconds) = 9.33* 108 years
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