A) \[1.6\times {{10}^{-5}}W\]
B) \[2.6\times {{10}^{-3}}W\]
C) \[3.3\times {{10}^{-5}}W\]
D) \[5.1\times {{10}^{-3}}W\]
Correct Answer: C
Solution :
[c] The total energy released E= Energy released in fission process + energy released in \[\alpha \]- decay process \[={{N}_{F}}\times 200+{{N}_{\alpha }}\times (0.005517\times 931)\] \[=\left( \frac{8}{100}\times {{10}^{20}} \right)\times 200+\left( \frac{92}{100}\times {{10}^{20}} \right)\] \[\times (0.005517\times 931)\] \[=20.725\times {{10}^{20}}MeV\] Power output \[P=E/t\] \[=\frac{20.725\times {{10}^{20}}\times 1.6\times {{10}^{-13}}}{{{10}^{13}}}\] \[=3.3\times {{10}^{-5}}W.\]You need to login to perform this action.
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