A) 10 kg
B) 100 kg
C) 1 g
D) 10-2 g
Correct Answer: C
Solution :
Key Idea: 1 amu is one twelfth part of the mass of carbon \[{{(}_{6}}{{C}^{12}})\] atom. Firstly total energy released is \[\text{1 MW = 1}{{\text{0}}^{\text{6}}}\text{W}\] Energy released per day \[\therefore \] \[\text{E}={{10}^{6}}\times 24\times 60\times 60\] \[\text{E = 86400 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{6}}}\text{ J}\] Energy released by 235 amu = 200 MeV \[\because \] \[\text{amu}=1.66\times {{10}^{-27}}\,\text{kg}\] \[\therefore \] \[235\,\text{amu = 235}\times \text{1}\text{.6}\times {{10}^{-27}}\,\text{kg}\] Also \[1.6\times {{10}^{-19}}\text{J =1}\,\text{eV}\] Using these relations, we have \[200\times 1.6\times {{10}^{-13}}\,\text{J}\]energy is released by \[235\times 1.66\times {{10}^{-27}}\,\text{kg }\]of \[{{\text{U}}^{\text{235}}}\] So, \[\text{86400}\times {{10}^{6}}\text{J}\] energy would be released by \[\frac{235\times 1.66\times {{10}^{-27}}\times 86400\times {{10}^{6}}}{200\times 1.6\times {{10}^{-13}}}\] \[=1\times {{10}^{-3}}\,\text{kg}\] \[=1\,g\,\]of \[{{\text{U}}^{\text{235}}}\]
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