A) \[0.5\times {{10}^{14}}\]
B) \[0.5\times {{10}^{12}}\]
C) \[5\times {{10}^{12}}\]
D) \[5\times {{10}^{14}}\]
Correct Answer: D
Solution :
Energy released in the fission of one nucleus = 200 MeV \[=200\times {{10}^{6}}\times 1.6\times {{10}^{-19}}J=3.2\times {{10}^{-11}}J\] \[P=16KW=16\times {{10}^{3}}Watt\] Now, number of nuclei required per second \[n=\frac{P}{E}=\frac{16\times {{10}^{3}}}{3.2\times {{10}^{-11}}}=5\times {{10}^{14}}\].
You need to login to perform this action.
You will be redirected in
3 sec
