A) \[0.5\times 1014\]
B) \[0.5\times {{10}^{12}}\]
C) \[5\times {{10}^{12}}\]
D) \[5\times {{10}^{14}}\]
Correct Answer: D
Solution :
Given : Energy released per second E = 200 MeV \[=200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\] \[=3.2\times {{10}^{-11}}\]joule Power \[P=16kW=16\times {{10}^{3}}\]watt As the number of nuclei is given by \[n=\frac{P}{E}=\frac{16\times {{10}^{3}}}{3.2\times {{10}^{-11}}}=5\times {{10}^{14}}\]
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