InterviewSolution
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InterviewSolution
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A star initially has `10^40` deuterons. It produces energy via the processes `_1H^2+_1H^2rarr_1H^3+p` and `_1H^2+_1H^3rarr_2He^4+n`. If the average power radiated by the star is `10^16` W, the deuteron supply of the star is exhausted in a time of the order of (a) `10^6s` (b) `10^8s` (c) `10^12s` The masses of the nuclei are as follows `M(H^2)=2.014` amu, `M(n)=1.008` amu, `M(p)=1.007` amu,`M(He^4)=4.001`amu |
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Answer» Correct Answer - C The given reactions are `_1H^2+_1H^2rarr_1H^3+p` `_1H^2+_1H^3rarr_2He^4+n` implies `3_1H^2rarr_2He^4+n+p` Mass defect, `Deltam=(3xx2.014-4.001-1.007-1.008)` amu `=0.026` amu Energy released `=0.026 xx 931MeV` `=0.026xx931xx1.6xx10^-13J` `=3.87xx10^-12J` This is the energy produced by the comsumption of three deuteron atoms. :. Total energy released by `10^40` deuterons `=10^40/3xx3.87xx10^-12J=1.29xx10^28J` The average power radiated is `P=10^16W` or `10^16J//s`. Therefore, total time to exhaust all deuterons of the star will be `t=(1.29xx10^28)/(10^16)=1.29xx10^12s~~10^12s` :. The correct option is (c). |
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