Q157 / 160 · AP EAMCET 2024 (19 May Shift 2) · Physics
The decrease in each day. in the Uranium mass of the material in a Uranium reactor operating at a power of 12 MW is (Energy released in one ${ }_{92} \mathrm{U}^{235}$ fission is about $200 \mathrm{MeV})$
- A$12.64 \times 10^{-2} \mathrm{~kg}$
- B$11.50 \times 10^{-2} \mathrm{~kg}$
- C12.64 kg
- Correct answerD12.64 g
Solution
Energy/fission = E = 200 meV
$\begin{aligned} & \therefore \text { Power, } \mathrm{P}=\frac{\mathrm{nE}}{\mathrm{t}} \Rightarrow \mathrm{n}=\frac{\mathrm{p.t}}{\mathrm{E}} \\ & \therefore \mathrm{n}=\frac{12 \times 10^6 \times 24 \times 3600}{200 \times 10^6 \times 1.6 \times 10^{-19}}=324 \times 10^{20} \\ & \therefore \text { Number of moles }=\frac{\mathrm{n}}{\mathrm{N}_{\mathrm{A}}}=\frac{\mathrm{m}}{\mathrm{M}_{\mathrm{o}}} \\ & \Rightarrow \frac{324 \times 10^{20}}{6.023 \times 10^{23}}=\frac{\mathrm{m}}{235} \\ & \therefore \mathrm{~m}=\frac{235 \times 324 \times 10^{20}}{6.023 \times 10^{23}} \\ & =12.64 \mathrm{~g}\end{aligned}$
Hence, the correct answer is option: 'D'.