A) \[20\mu g\]
B) \[40\mu g\]
C) \[1\mu g\]
D) \[10\mu g\]
Correct Answer: B
Solution :
Let assume power P = 1000 W Energy per hour = 1000 x 3600 J Energy per fission = 200 MeV \[=200\times 1.6\times {{10}^{-13}}J\] \[\therefore \]Number of fission per hour \[n=\frac{1000\times 3600}{200\times 1.6\times {{10}^{-13}}}\] Number of mole per hour\[=\frac{n}{N}\times 235\] \[\therefore \] Mass per hour\[=\frac{n}{N}\times 235\] \[=\frac{1000\times 3600\times 235}{200\times 1.6\times {{10}^{-13}}\times 6.02\times {{10}^{23}}}\] \[=43.9\times {{10}^{-6}}g\] This \[43.9\times {{10}^{-6}}g\]is nearest value of 40 micron so option (b) is correct.
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