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Assuming that about 20 MeV of energy in released per fusion in the reaction ""_(1)H^(2)+""_(1)H^(3) to ""_(0)n^(1)+ ""_(2)He^(4). then the mass of ""_(1)H^(4) consumes per day in a bowiem reactor of puwwer Imawat will styroximately be: |
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Answer» 0.001gm `=10^(6) XX 24 xx 60 xx 60 J//day` Energy/fusion `=20 xx 10^(6)eV` `=20 xx 10^(6) xx 1.6 xx 10^(-19)J` In each fusion, MASS of `""_(1)H^(2)` used =2 a.m.u. `=2 xx 1.6 xx 10^(-27)kg ..........(1)` Number of fusion/day `=(10^(6) xx 24 xx 60 xx 60)/(20 xx 10^(6) xx 1.6 xx 10^(-19))` Mass consumed//day of `""_(1)H^(2)` `=(10^(6) xx 24 xx 60 xx 60)/(20 xx 10^(6) xx 1.6 xx 10^(-19)) xx 2 xx 1.6 xx 10^(-27)kg` `approx 10^(-4)kg=0.1gm` |
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