JEE Main & Advanced Physics Nuclear Physics And Radioactivity Question Bank Topic Test - Nuclear Physics and Radioactivity

200 MeV of energy may be obtained per fission of \[{{U}^{235}}\]. A reactor is generating 1000 kW of power. The rate of nuclear fission in the reactor is

A) 1000

B) \[2\times {{10}^{8}}\]

C) \[3.125\times {{10}^{16}}\]

D) 931

Correct Answer: C

Solution :

[c] Power \[\Rightarrow {{\left( \frac{1}{2} \right)}^{4}}={{\left( \frac{1}{2} \right)}^{t/48}}\Rightarrow t=192\ hour.\]

Rate of nuclear fission \[=\frac{{{10}^{6}}}{200\times 1.6\times {{10}^{-13}}}\]= 3.125 ´ 1016.