Nuclear fusion involves the formation of larger nuclei from smaller nuclei. Duri
ID: 521458 • Letter: N
Question
Nuclear fusion involves the formation of larger nuclei from smaller nuclei. During fusion, mass is converted to energy according to Einstein's equation: Delta E = (Delta m)c^2 where Delta E is the energy produced in joules, Delta m is the change in mass in kilograms, and c is the speed of light (3.00 times 10^8 m/s). In the core of the Earth's sun, fusion involves hydrogen atoms (^1_1H) forming helium atoms (^4_2H) accompanied by the release of tremendous amounts of energy. The mass of our sun is about 2 times 10^30 kg. About 80% of this mass is hydrogen, and about 25% of the hydrogen is located in the sun's core, where it is available for fusion. Proposed solar fusion reaction Spectroscopic studies have shown that the fusion reaction in the sun's core proceeds by a complicated mechanism. One proposed overall equation for the solar fusion reaction is 4^1_1 H + 2^0_-1 e rightarrow^4_2 He which is also accompanied by the release of "massless" neutrinos and photons in addition to energy. Precise masses of the reactants and product are given below. The sun produces energy via nuclear fusion at the rate of 4 times 10^26 J/s. Based on the proposed overall fusion equation, how long will the sun shine in years before it exhausts its hydrogen fuel? (Assume that there are 365 days in the average year.) Express your answer to one significant figure and include the appropriate units.Explanation / Answer
4(1.007825) + 2(0.00549) - 4.002603 = 0.029795 amu is the energy from hydrogen gas.
4.0313 amu H+ > 0.029795 amu = ratio of 1 to 0.00739
dE = ((2 x 1030 kg) x 0.8 x 0.25)(3.00x108 m/s)2 = 3.6x1046 kg x m2/s2
3.6x1046 kg x m2/s2 x 0.00739 = 2.6604 x 1044 kg x m2/s2
4x1026 J/s x s = 2.6604 x 1044 kg x m2/s2
Solving for s.
s = 6.651x1017 seconds
6.651x1017 seconds x 1 min/60 s x 1 hour/ 60 min x 1 day/ 24 hour x 1 year / 365 days = 2.1x1010 years
2X1010 years will be the answer (approximately).
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