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`amuA. `10^(6)s`B. `10^(8)s`C. `10^(12)s`D. `10^(16)s` |
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Answer» Correct Answer - c The given ractions are `{:(._(1)H_(2)+._(1)H^(2)rarr._(1)h^(3)+p),(._(1H^(2)+._1H2)rarr._2He^4_n),(3.h^3 rarr ._2He^4+n+p):}` Mass defect, `Delta m=(3xx2.014 -4.001 -1.007 -1.008)a.m.u.` = `0.026 a.m.u.` Energy released `=0.026 xx 931 MeV` `=0.026xx931 xx1.6 xx10^(-13) J` `=3.87 xx 10^(-12) J` This is the energy produced by the consumption of three deutron atoms. Therefore, total energy released by `10^(40)` deuterons is `(10^(40))/(3) xx 3.87 xx 10^(-12) J=1.29 xx 10^(28) J` The average power radiated is `P=10^(6)W` or `10^(16) J s^(-1)`. Therefore, total time is exhaust all deutrons of the star will be `t=1.29 xx(10^(28))/(10^(16))=1.29 xx 10^(12) s ~~ 10^(12)s`. |
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