Q108 / 180 · AIIMS 2013 · Physics

When a slow neutron is captured by a ${ }_{92}^{235} \mathrm{U}$ nucleus, a fission energy releasing $200 \mathrm{MeV}$. If power of nuclear reactor is $100 \mathrm{~W}$ then rate of nuclear fission is

  • A
    $3.6 \times 10^6 \mathrm{~s}^{-1}$
  • Correct answer
    B
    $3.1 \times 10^{12} \mathrm{~s}^{-1}$
  • C
    $1.8 \times 10^4 \mathrm{~s}^{-1}$
  • D
    $4.1 \times 10^6 \mathrm{~s}^{-1}$

Solution

Number of fission per second $=\frac{\text { total power }}{\text { energy/fission }}$. Here, total power $=100 \mathrm{~W}$ energy/fission $=200 \mathrm{MeV}=200 \times 10^6 \times 1.6 \times 10^{-19} \mathrm{~J}$ $=3.2 \times 10^{-11} \mathrm{~J}$ $\therefore \quad$ fission rate $=\frac{100}{3.2 \times 10^{-11}}=3.1 \times 10^{12} \mathrm{~s}^{-1}$

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

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