Q125 / 180 · JIPMER 2018 · Physics

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

  • A
    $30 \times 10^{25}$
  • Correct answer
    B
    $4 \times 10^{22}$
  • C
    $10 \times 10^{20}$
  • D
    $5 \times 10^{15}$

Solution

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

Hence, the correct answer is option: 'B'.

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