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20 MeV energy is released per fusion reaction ""_1H^2 +""_1H^2 to ""_2He^4 +""_0n^1 Calculate the mass of ""_1H^2 consumed in a fusion reactor of power 1 MW in 1 day . |
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Answer» Solution :`P =1 MW =10^6 W=10^6Js^(-1)` `t=1 "DAYS" =24xx60xx60=86400 s` `therefore` Energy released in one DAY `=pt=86400xx10^6J` Energyreleased per fusion =20 MeV `=20 xx1.6 xx10^(-13)=3.2 xx10^(-12)J` Mass of `""_1H^2` consumed in one fusion `(""_1H^2+""_1H^2)` =4u `=4xx1/66 xx10^(-27)` kg `=6.64 xx10^(-27)` kg `=(6.64xx10^(-27))/(3.2 xx10^(-12))xx86400xx10^6` `=1.79 xx10^(-4)` kg |
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