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A nuclear reactor generates power at 50% efficiency, by fusion ""_(92)U^(235) into two canal fragments of ""_(6)Pd^(116) with the emission of two gamma rays and three neutrons. The average B.E. per particle of U^(235) and Pd^(116) is 7.2 MeV and 8.2 MeV respectively. 75. Amount of U235 consumed per hour to produce 1600 mega watt power is |
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Answer» 0.02kg `=1600 xx 10^(6) xx 3600` `n=50% and n=("output energy")/("input energy")` Input energy `=2 xx 1600 xx 10^(6) xx 3600` `=1.152 xx 10^(13)J` No. of FISSION per hour `=(1.152 xx 10^(13))/(200 xx 10^(6) xx 1.6 xx 10^(-19))` No. of atoms in 1 gm of `U^(235)=(6.023 xx 10^(23))/(235)` Mass of `U^(235)` consumed per hour `=("no. of fission/hr")/("no. of atoms in 1 gm of "U^(235))` `=(1.152 xx 10^(32) xx 235)/(200 xx 10^(6) xx 1.6 xx 6.023 xx 10^(23))` `=(1.152 xx 235 xx 10^(3)gm)/(200 xx 1.6 xx 6.023)` `=0.14 xx 10^(3)gm=0.14kg`. |
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