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Assuming that about 20 M eV of energy is...

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

A

`0.1 gm`

B

`0.01 gm`

C

`1 gm`

D

`10 gm`

Text Solution

Verified by Experts

The correct Answer is:
A
no of moles of `._(1)H^(2)` consumed
`=(1MWxx(24xx3600)sec//day)/((20 MeVxx6.023xx10^(23)))=0.05`
`:. m=0.1 g`
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