Q153 / 160 · TS EAMCET 2023 (14 May Shift 2) · Physics

113. Energy released in the fission of a single uranium nucleus is $200 \mathrm{MeV}$. Then the number of fissions per second to produce $5 \mathrm{~mW}$ power is

  • Correct answer
    A
    $1.56 \times 10^8$
  • B
    $1.56 \times 10^{13}$
  • C
    $3.12 \times 10^8$
  • D
    $3.12 \times 10^{13}$

Solution

Number of fissions per second, $\begin{aligned} & \mathrm{n}=\frac{5 \times 10^{-3}}{200 \times 10^6 \times 1.6 \times 10^{-19}} \\ & =\frac{50}{32} \times 10^8=1.56 \times 10^8 \mathrm{~m} / \mathrm{s}\end{aligned}$

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

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