A) \[3.125\times {{10}^{9}}\]
B) \[3.125\times {{10}^{12}}\]
C) \[3.125\times {{10}^{13}}\]
D) \[3.125\times {{10}^{11}}\]
Correct Answer: C
Solution :
Energy released by fission of an atom of \[_{\text{92}} & {{\text{U}}^{\text{235}}}\]is \[\text{E = 200}\,\,\,\text{MeV}\text{.}\] We know that \[1\,eV=1.6\times {{10}^{-19}}\,J\] \[\therefore \] \[E=200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}\,\text{J}\] \[=3.2\times {{10}^{-11}}\,J\] Hence, \[3.2\times {{10}^{-11}}\,J\]energy is released in one fission. So, no. of fissions required to produce a power of 1 kW i.e., \[=\frac{1000}{3.2\times {{10}^{-11}}}=3.125\times {{10}^{13}}\]
You need to login to perform this action.
You will be redirected in
3 sec
