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JIPMER 2018
Exam
JIPMER
Year
2018
Subject
Physics
Chapter
Nuclear Physics

JIPMER 2018 · Physics · Nuclear Physics

An atomic power nuclear reactor can deliver 300 MW. The energy...

An atomic power nuclear reactor can deliver 300 MW. The energy released due to fission of each nucleus of uranium atoms U238\mathrm{U}^{238} is 170 MeV . The number of uranium atoms fissioned per hour will be

Options

  1. A30×102530 \times 10^{25}
  2. B4×10224 \times 10^{22}
  3. C10×102010 \times 10^{20}
  4. D5×10155 \times 10^{15}

Correct answer

B. 4×10224 \times 10^{22}

Step-by-step solution

Power = Energy  Time =300×106 W=3×108 J/s=\frac{\text{ Energy }}{\text{ Time }}=300 \times 10^6 \mathrm{~W}=3 \times 10^8 \mathrm{~J} / \mathrm{s} Energy released due to fission =170MeV=170×106×1.6×10−19=27.2×10−12 J\begin{gathered}=170 \mathrm{MeV}=170 \times 10^6 \times 1.6 \times 10^{-19} \\ =27.2 \times 10^{-12} \mathrm{~J}\end{gathered} Number of atoms fissioned per second =3×10827.2×10−12=3×102027.2=\frac{3 \times 10^8}{27.2 \times 10^{-12}}=\frac{3 \times 10^{20}}{27.2} Number of atoms fissioned per hour =3×1020×360027.2=3×3627.2×1022=4×1022 m\begin{aligned} & =\frac{3 \times 10^{20} \times 3600}{27.2} \\ & =\frac{3 \times 36}{27.2} \times 10^{22}=4 \times 10^{22} \mathrm{~m}\end{aligned}
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