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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 :

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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