Q157 / 160 · TS EAMCET 2024 (09 May Shift 2) · Physics

The energy released by the fission of one uranium nucleus is 200 MeV . The number of fissions per second required to produce 128 W power is

  • A
    $6 \times 10^{12}$
  • B
    $2 \times 10^{12}$
  • C
    $8 \times 10^{12}$
  • Correct answer
    D
    $4 \times 10^{12}$

Solution

Energy per fission $=200 \mathrm{MeV}$ $\begin{aligned} & =.200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J} \\ & \text { Fission rate }=\frac{128}{200 \times 10^6 \times 1.6 \times 10^{-19}}=0.4 \times 10^{13} \\ & =4 \times 10^{12}\end{aligned}$

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

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