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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Nuclear Physics

JIPMER 2018 · Physics · Nuclear Physics

In the fusion reaction _1^2 H+ _1^2 H _2^3 He+ _0^1 n , the masses...

In the fusion reaction 12H+12H⟶{ }_1^2 \mathrm{H}+{ }_1^2 \mathrm{H} \longrightarrow 23He+01n{ }_2^3 \mathrm{He}+{ }_0^1 \mathrm{n}, the masses of deuteron, helium and neutron expressed in amu are 2.015,3.0172.015,3.017 and 1.009 , respectively. If 1 kg of deuterium undergoes complete fusion, find the amount of total energy released. (1 amu =931.5MeV)=931.5 \mathrm{MeV}).

Options

  1. A9.0×1013 J9.0 \times 10^{13} \mathrm{~J}
  2. B20×105 J20 \times 10^5 \mathrm{~J}
  3. C5×1016 J5 \times 10^{16} \mathrm{~J}
  4. D8×105 J8 \times 10^5 \mathrm{~J}

Correct answer

A. 9.0×1013 J9.0 \times 10^{13} \mathrm{~J}

Step-by-step solution

Δm=2(2.015)−(3.017+1.009)=0.004amu∴\Delta \mathrm{m}=2(2.015)-(3.017+1.009)=0.004 \mathrm{amu} \therefore Energy released =(0.004×931.5)MeV=3.726MeV=(0.004 \times 931.5) \mathrm{MeV} =3.726 \mathrm{MeV} Energy released per deuteron =3.7262=\frac{3.726}{2}=1.863 MeV\mathrm{MeV}Number of deuteronsin1kg=6.02×10262=3.01×1026∴deuterons in 1 kg =\frac{6.02 \times 10^{26}}{2}=3.01 \times 10^{26} \therefore Energy released per kg of deuterium fusion =(3.01×1026×1.863)=5.6×1026MeV≈9.0×1013 J\begin{aligned} & =\left(3.01 \times 10^{26} \times 1.863\right) \\ & =5.6 \times 10^{26} \mathrm{MeV} \approx 9.0 \times 10^{13} \mathrm{~J}\end{aligned}
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