On bombarding U ²³⁵ by slow neutron, 200 MeV energy is release(d)If the power output of atomic reactor is 1.6 MW , then the rate of fission will be

Step-by-step solution

Given, output power, P_ o =1.6 MW =1.6 10⁶ ~W Energy released per fission, aligned E₁ &=200 MeV &=200 10⁶ 1.6 10⁻¹⁹ ~J &=320 10⁻¹³ ~J &=3.2 10⁻¹¹ ~J aligned The rate of fission R is given as aligned &R= P₀ E₁ &= 1.6 10⁶ 3.2 10⁻¹¹ &=0.5 10¹⁷=5 10¹⁶ / s aligned