A) \[3.1\times {{10}^{13}}\]
B) \[1.3\times {{10}^{16}}\]
C) \[1.3\times {{10}^{15}}\]
D) \[3.1\times {{10}^{16}}\]
Correct Answer: A
Solution :
\[Acts\text{ }of\text{ }fission=\frac{power}{energy\text{ }released}\] Given, \[P=1\text{ }kW={{10}^{3}}W,\] \[E=200\text{ }MeV=200\times {{10}^{6}}eV\] Also, \[1\,eV=1.6\times {{10}^{-19}}\] \[\therefore \] \[E=200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}J\] \[\therefore \] \[Acts\text{ }of\text{ }fission=\frac{1000}{200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}}\] \[=\frac{{{10}^{14}}}{3.2}=3.1\times {{10}^{13}}\]fission/s
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