Assuming that about `20 M eV` of energy is released per fusion reaction `._(1)H^(2)+._(1)H^(3)rarr._(0)n^(1)+._(2)He^(4)`, the mass of `._(1)H^(2)` consumed per day in a future fusion reactor of powder `1 MW` would be approximately

Options

A. `0.001g`
B. `0.1g`
C. `10.0g`
D. `1000 g`

Answer

b

Explanation

`P=10^(6)W`
Time `= 1 day =24 xx 36 xx 10^(2)s`
Energy produced,
`U=Pt=10^(6) xx24 xx 36xx 10^(2)`
`=24 xx 36 xx10^(8) J`
Energy released per fusion reaction is
`20 MeV =32 xx 10^(-13) J`
Energy released per atom of `._(1)H^(2)` is
`32 xx 10^(-13) J`
Number of `._1H^2` atoms used is
`(24 xx 36 xx10^(8))/(32 xx 10^(-12)=22 xx 10^(21))`
Mass of `6 xx10^(23)` atom `=2g`
Mass of `27 xx10^(21)` atoms` =(2)/(6) xx 10^(23) xx 27xx 10^(21) =0.1g.`