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

Answer

B

Explanation

Energy produced,
`U=Pt=10^(6)xx24xx36xx10^(2)`
`=24xx36xx10^(8)J`
Enegy released per fusion reaction
`=20MeV`
`=20xx10^(6)xx1.6xx10^(-19)`
`=32xx10^(-13J`
Energy released per atom of `._(1)H^(2)`
`=32xx10^(-13)J`
Number of `._(1)H^(2)` atoms used
`=(24xx36xx10^(8))/(32xx10^(-13))=27xx10^(21)`
Mass of `6xx10^(23)` atoms=2g
`therefore`Mass of `27xx10^(21)` atoms
`=(2)/(6xx10^(23))xx27xx10^(21)`
`=0.1g`