InterviewSolution
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InterviewSolution
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It is proposed to use the nuclear fusion reaction, `_1^2H+_1^2Hrarr_2^4He` in a nuclear reactor 200 MW rating. If the energy from the above reaction is used with a 25 per cent efficiency in the reactor, how many grams of deuterium fuel will be needed per day?(The masses of `_1^2H` and `_2^4He are 2.0141 atommic mass units and 4.0026 atomic mass units respectively.) |
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Answer» Correct Answer - A::B Mass defect in the given nuclear reaction, `Deltam=2(mass of deuterium)-(mass of helium)` `=2(2.0141)-(4.0026)=0.0256` Therefore, energy released `DeltaE=(Deltam)(931.48)MeV=23.85MeV` `=23.85xx1.6xx10^-13J=3.82xx10^-12J` Efficiency is only 25%, therefore, `25% of DeltaE=(25/100)(3.82xx10^-12)J` `=9.55xx10^-13J` i.e. by the fusion of two deuterium nuclei, `9.55xx10^-13J` energy is available to the nuclear reactor. Total energy required in one day to run the reactor with a given power of 200 MW, `E_(Total)=200xx10^6xx24xx3600=1.728xx10^13J` :. Total number of deuterium nuclei required for this purpose, `n=(E_(Total))/(DeltaE//2)=(2xx1.728xx10^13)/(9.55xx10^-13)` `=0.362xx10^26` :. `Mass of deuterium required = (Number of g-moles of deuterium required)xx2g` `=((0.362xx10^26)/(6.02xx10^23))xx2=120.26g` |
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