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  • It is proposed to use the nuclear fusion reaction <img alt="{ }_1 \mathrm{H}^2+{ }_1 \mathrm{H}^2 \longrightarrow{ }_2 \mathrm{He}^4" src="https://entrancecorner.oncodecogs.com/gif.latex?

It is proposed to use the nuclear fusion reaction

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

in a nuclear reactor with an electrical power rating of200\mathrm{~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 { }_1 \mathrm{H}^2 and { }_2 \mathrm{He}^4 are 2.0141 \mathrm{~amu} and 4.0026 \mathrm{~amu} respectively.)
 

Option: 1

96 \mathrm{~gm}


Option: 2

106 \mathrm{~gm}


Option: 3

68 \mathrm{~gm}


Option: 4

\mathrm{121 \mathrm{~gm}}


Answers (1)

best_answer

Mass defect occurring in one fusion reaction

\Delta \mathrm{m}=(2 \times 2.0141-4.0026) \mathrm{amu}=0.0256 \mathrm{amu}

Energy released = 0.0256 \times 931 \mathrm{MeV}
Energy used in reactor per fusion reaction = \frac{25}{100} \times 0.0256 \times 931 \mathrm{MeV}

=5.9584 \mathrm{MeV}=9.5334 \times 10^{-13} \mathrm{~J}

Total energy required per day = (200 \mathrm{MW}) \times(24 \times 60 \times 60 \mathrm{sec}. )

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}

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}}=121 \mathrm{gm}

Posted by

rishi.raj

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