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

Power of reactor = 200 MW = `200 xx 10^(6) W = 2 xx 10^(8) J//s`
Total amount of energy drawn from the power reactor per day (E) = `2 xx 10^8 xx 24 xx 3600 = 1,728 xx 10^10` J Now we shall calculate energy produced in fusion of two deuterium atoms. To calculate energy produced per fusion reaction we need to calculate m ass defect first.
Mass defect `(Deltam) = 2 xx m(""_(1)H^(2)) -m(""_(2)He^(4))`
`=2 xx 2.0141 - 4.0026 = 0.0256` amu
We know that 1 am u produces 931 MeV energy, hence energy produced in fusion reaction can be w ritten as follows:
`DeltaE = (0.0256) xx (931) MeV`
`rArr DeltaE = 0.0256 xx 931 xx 10^(6) xx 1.6 xx 10^(-19)` joules.
`rArr DeltaE = 38.13 xx 10^(-13)` Joules.
We know th at the above reaction is being used with 25% efficiency. And hence energy available to us per fusion reaction can be w ritten as follows: `DeltaE.=0.25 xx 38.13 xx 10^(-12)"joules"`. The above calculated am ount of energy `DeltaE.` is energy generated when two deuterium atoms are fused. Hence number of deuterium atoms fused to meet our requirem ent of energy per day can be written as follows:
Number of atoms fused per day.
`=(1,728 xx 10^(10))/((9.53 xx 10^(-13))/2) = (1,728 xx 10^(10) xx 2)/(9.53 xx 10^(-13)) = 362.64 xx 10^(23)`
Mass of `6.02 xx 10^(23)` atoms of deuterium is 2 grams. Hence mass of `362.64 xx 10^23` atoms can be w ritten as follows:
Mass of deuterium consumed per day.
`=(362.64 xx 10^(23))/(6.02 xx 10^(23)) xx 2` grams =120.48 grams.