It is propssed to use the nucles fasion
`_(1)^(2) H +_(1)^(2) H rarr _(2)^(4) He`
in a nucleas of `200MW` rating if the energy from the above reaction is used with a `25` per cast effecincy in the rector , low maney game of deuterium fiel will be needed per day (The masses of `_(1)^(2)H and _(4)^(2) He are 2.0141`atomic mass unit and `4.0028`atyomic mass uniot repertively)

Answer

A

Explanation

energy required per day
` E = p xx 1 = 200 xx 10^(9) xx 24 xx 60 xx 60`
`= 1.728 xx 10^(13)J`
energy released per fusion reaction
`= [ 2 (2.0141)-4.0026] xx 931.5 MeV`
`= 23.15 xx 10^(-13)J`
`:. ` No of fusion reactions required
`= (1.728 xx 10^(13))/(38.15 xx 10^(-13)) = 0.045 xx 10^(26)`
:.` No of deuterium atoms required
`= 2 xx 0.045 xx 10^(26) = 0.09 xx 10^(26)`
number of deterium atom
`= (0.09 xx 10^(26))/(6.02 xx 10^(23)) = 14.95`
`:.`Mass in gram of deterium atom
`= 14.95 xx 2 = 29.9 g`
But the efficiency is `25%`
Therefore, the efficiency mass required` = 199.6 g`