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It is proposed to use the nuclear fusion reaction

\mathrm{{ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \longrightarrow{ }_2 \mathrm{He}^4}

in a nuclear reactor with an electrical power rating of 200 MW. If the energy from the above reaction is used with 25 percent efficiency in the reactor, how many grams of deuterium fuel will be needed per day? (The masses of \mathrm{{ }_1 \mathrm{H}^2 \text { and }{ }_2 \mathrm{He}^4} are 2.0141 amu and 4.0026 amu respectively.)

Option: 1

60 gm


Option: 2

80 gm


Option: 3

100 gm


Option: 4

120 gm


Answers (1)

best_answer

Mass defect occurring in one fusion reaction

\mathrm{\begin{aligned} & \Delta \mathrm{m}=(2 \times 2.0141-4.0026) \mathrm{amu}=0.0256 \mathrm{amu} \\ & \text { Energy released }=0.0256 \times 931 \mathrm{MeV} \end{aligned}}

Energy used in reactor per fusion reaction

\mathrm{\begin{aligned} & =\frac{25}{100} \times 0.0256 \times 931 \mathrm{MeV}=5.9584 \mathrm{MeV}=9.5334 \times \\ & 10^{-13} \mathrm{~J} \\ & \text { Total energy required per day }=(200 \mathrm{MW}) \times(24 \times 60 \\ & \times 60 \text { sec.) } \\ & \text { Mass of deuterium fuel needed per reaction }=2 \times \\ & 2.0141 \mathrm{amu} \\ & =\frac{4.0282}{6.02 \times 10^{23}} \mathrm{gm}=0.6691 \times 10^{-23} \mathrm{gm} \\ & \text { Mass of deuterium required = } \\ & \frac{0.6691 \times 10^{-23} \times 200 \times 10^6 \times 24 \times 60 \times 60}{9.5334 \times 10^{-13}}=120 \mathrm{gm} . \\ & \end{aligned}}

 

Posted by

Pankaj

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