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Assuming that about 20 MeV of energy is ...

Assuming that about 20 MeV of energy is released per fusion reaction
`1^(H^(2))+1^(H^(2))to2^(He^(4))`
then the mass `1^(H^(2))`consumed per day in a jfusion rector of power 1 MW wll apporximately be :

Text Solution

Verified by Experts

`P =1 MW =10^6 W=10^6 Js^(-1)`
`t=1 "days" =24xx60xx60=86400 s`
`therefore` Energy released in one day `=pt=86400xx10^6J`
Energy released 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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