Question

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 atommic mass units and 4.0026 atomic mass units respectively.)

Answer 1

Correct Answer - A::B
Mass defect in the given nuclear reaction,
`Deltam=2(mass of deuterium)-(mass of helium)`
`=2(2.0141)-(4.0026)=0.0256`
Therefore, energy released
`DeltaE=(Deltam)(931.48)MeV=23.85MeV`
`=23.85xx1.6xx10^-13J=3.82xx10^-12J`
Efficiency is only 25%, therefore,
`25% of DeltaE=(25/100)(3.82xx10^-12)J`
`=9.55xx10^-13J`
i.e. by the fusion of two deuterium nuclei, `9.55xx10^-13J` energy is available to the nuclear reactor.
Total energy required in one day to run the reactor with a given power of 200 MW,
`E_(Total)=200xx10^6xx24xx3600=1.728xx10^13J`
:. Total number of deuterium nuclei required for this purpose,
`n=(E_(Total))/(DeltaE//2)=(2xx1.728xx10^13)/(9.55xx10^-13)`
`=0.362xx10^26`
:. `Mass of deuterium required = (Number of g-moles of deuterium required)xx2g`
`=((0.362xx10^26)/(6.02xx10^23))xx2=120.26g`