Nuclear Physics And Radioactivity Question 477
Question: The proton -proton mechanism that accounts for energy production in the sun releases $ 26.7MeV $ energy for each event. In this process, protons fuse to form an alpha particle $ {{(}^{4}}He) $ . At what rate $ \frac{dm}{dt} $ is hydrogen being consumed in the core of the sun by the p-p cycle? Power of sun is $ 3.90\times 10^{26}W $ .
Options:
A) $ 1.6\times 10^{10}kg/s $
B) $ 2.3\times 10^{9}kg/s $
C) $ 6.2\times 10^{11},kg/s $
D) $ 5.5\times 10^{10},kg/s $
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Answer:
Correct Answer: C
Solution:
[c] The rate $ dm/dt $ can be calculate as; Power, $ P=\frac{dE}{dt}=\frac{dE}{dm}\times \frac{dm}{dt}=\frac{\Delta E}{\Delta m}\times \frac{dm}{dt} $
$ \therefore \frac{dm}{dt}=\frac{\Delta m}{\Delta E}P $ ?..(i) We known that $ 26.2,MeV=4.20\times {{10}^{-12}}J $ of thermal energy is produced when four protons are consumed. This is $ \Delta E=4.20\times {{10}^{-12}}J $ for $ \Delta m=4\times (1.67\times {{10}^{-27}}kg). $ Substituting these values in equation (i), we have $ \frac{dm}{dt}=\frac{\Delta m}{\Delta E}P=\frac{4(1.67\times {{10}^{-27}})}{4.20\times {{10}^{-12}}}\times (3.90\times 10^{26}) $ $ =6.2\times 10^{11}kg/s $
BETA
