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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 day = 24 x 60 x 60 =86400 s
`therefore` Energy released in one day = Pt = `86400 xx 10^6` J
Energy released per fusion
=20 MeV
`=20xx1.6xx10^(-13)`
`=3.2xx10^(-12) J`
Mass of `._1H^2` consumed in one fusion `(._1H^2+._1H^2)`
=4u
`=4xx1.66xx10^(-27) kg`
`=6.64xx10^(-27)` kg
`=(6.64xx10^(-27))/(3.2xx10^(-12))xx86400xx10^6`
`=1.79xx10^(-4)` kg
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