Assuming that solar energy is due to the thermonuclear reaction 4 1 H 1 — → 2He 4 + 2 +1 e 0 and 1 e 0 + – 1 e 0 — → γ , calculate the rate at which the hydrogen is consumed in the sun. Also calculate the time in which there would be no sun.
Given:
(i) Energy reaching the earth = 1400 Wm –2 , on the surface normal to sun rays.
(ii) Distance between earth and sun, D se = 1.5 × 10 8 km.
(iii) M( 1 H 1 ) = 1.0078 amu, m( 2 He 4 ) = 4.0028 amu, m( +1 e 0 ) = 0.005 amu.
(iv) Radius of sun = 7 × 10 8 m.
Assume density of sun = d kg/m 3 .
Text Solution
Verified by ExpertsCHECK THE SOLUTION.
Sol. Let m kg of hydrogen be consumed in one second.
Number of hydrogen atoms burnt = 
× m
When 4 atoms are burnt, energy released because of mass to energy conversion is
(4 × 1.0078 – 4.0028) amu × 931 MeV = 26.63 MeV
When two positrons are released, they combine with two electrons producing two photons.
Photon energy released = 0.005 × 4 amu
or 0.002 × 9.31 MeV = 1.862 meV
Total energy released by burning of 4 hydrogen atoms = 26.63 + 1.862 = 28.5 MeV
Total energy released by burning of m kg hydrogen
= 
× (6.03 × 10 26 m) MeV = 6.87 × 10 14 mJ
Energy/unit area = 
= 1400 (given)
M = 5.76 × 10 11 kg/s
Rate of burning of hydrogen = 5.76 × 10 11 kg/sec
Let ‘t’ be the time in which there would be no sun, i.e., all hydrogen will get consumed. Then
T = 
⇒ = 
= 
= 
= 2.5 × 10 15 sec = 7.9 × 10 7 years
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