A radionuclide with half-life 1620 s is produced in a reactor at a constant rate of 1000 nuclei per second. During each decay energy, 200 MeV is released. If the production of radionuclides started at t = 0 , then the rate of release of energy at t = 3240 s is

Step-by-step solution

Let N be the number of nuclei at time t, then net rate of increase of nuclei at instant t is,z d ⁡ N ⁡ d ⁡ t = α - λ N ⁡ (where α = rate of production of nuclei) ∴ ∫ 0 N ⁡ d ⁡ N ⁡ α - λ N ⁡ = ∫ 0 t d ⁡ t ∴ N ⁡ = α λ 1 - e ⁡ - λ t ...(i) Rate of decay at this instant R = λN = α ( 1 – e – λt ) Hence, rate of release of energy at this time = R (energy released in each decay) = α 1 - e ⁡ - λ t 2 0 0 MeV / s Substituting the values, we have rate of release of energy = 1 0 0 0 1 - e ⁡ - 0 . 6 9 3 1 6 2 0 × 3 2 4 0 2 0 0