It is proposed to use the nuclear fusion reaction,
`._1^2H+_1^2Hrarr_2^4He`
in a nuclear reactor `200 MW` rating. If the energy from the above reaction is used with a `25` per cent efficiency in the reactor, how many grams of deuterium fuel will be needed per day?(The masses of `._1^2H` and `._2^4He` are `2.0141` atomic mass units and `4.0026` atomic mass units respectively.)

Explanation

Energy released in the nuclear fusion is `Q = Deltamc^(2) = Deltam(931) MeV` (where `Deltam` is in amu)
`Q = (2 xx 2.0141 - 4.0026) xx 931 MeV = 23.834 MeV = 23.834 xx 10^(-13) J`
Since efficient of reactor is `25%`
So effective energy used `= 25/100 xx 23.834 xx 10^(6) xx 1.6 xx 10^(-19) J = 9.534 xx 10^(-13) J`
Since the two deuterium nuclei are involved in a fusion reaction,
therefore, enrgy released per deuterium is , `(9.534 xx 10^(-13))/(2)`
For `200 MW` power per day, number of deutrium, nuclei required `= (200 xx 10^(6) xx 86400)/((9.534 xx 10^(-23))/(2)) = 3.624 xx 10^(25)`
Since `2g` of deuterium constitute `6 xx 10^(232)` nuclei, therefore amount of deuterium requried is
`= (2 xx 3.624 xx 10^(25))/(6 xx 10^(23)) = 120.83 g//"day"`