practice problem 1
4 11H → 42He + 2(0+1e + 00γ + 00ν)
The mass of the sun is 1.99 × 1030 kg, 91% of which is hydrogen. Its power output is 3.85 × 1026 W. Determine…
- the mass of four hydrogen atoms
- the mass defect when four hydrogen atoms fuse into one helium atom (in atomic mass units and megaelectronvolts)
- the rate at which the sun's mass is decreasing
- the total mass destroyed if all the sun's hydrogen were converted into helium
- the expected lifetime of the sun (assuming its power output will remain constant)
The mass of four hydrogen atoms is…
4 × 1.007825 u = 4.031300 u
The mass difference between four hydrogen atoms and one helium atom is…
4 × 1.007825 u
− (4.00260 u) = 0.02870 u
(0.02870 u)(931 MeV/u) = 26.72 MeV
The rate at which mass is destroyed is related to the rate at which energy is produced.
∆m = P t c2 ⇐ P = E = ∆mc2 t t ∆m = 3.85 × 1026 W t (3.00 × 108 m/s)2 ∆m = 4.29 × 109 kg/s t
About four million tons of the sun vanishes every second.
The fraction of the sun's mass lost if every hydrogen nucleus was used to produce helium is the same as the the ratio of the mass defect to the original mass.
destroyed mass = m = 0.0287 u initial mass (0.91)(1.99 × 1030 kg) 4.0313 u
m = 1.29 × 1028 kg
Setting up another proportion finishes the problem.
∆m = 4.29 × 109 kg = 1.29 × 1028 kg t 1 s t
t = 3.01 × 1018 s = 95 × 109 years
This is a not a good estimate, however. Although extremely hot in human terms, most of the sun is just too cool for hydrogen to fuse into helium. Only the central core is hot enough and dense enough. Thus, it is estimated that not more than ten percent of the sun's total hydrogen will ever be available for thermonuclear fusion. Ten percent of 95 billion years 9.5 billion years, which is still a good long time. The earth is some 4.5 billion years old already, placing us somewhere in the middle of the sun's life. The sun is a middle-aged star.
practice problem 2
practice problem 3
practice problem 4