Atoms & Nuclei

JEE Main level, numerical answer, medium

196 of 239

Question 196 of 239

In a nuclear fission process, a high mass nuclide (A≈236)(A\approx 236) with binding energy 7.6MeV/7.6MeV/Nucleon dissociated into two middle mass nuclides (A≈118)(A\approx 118), having binding energy of 8.6MeV/8.6MeV/Nucleon. The energy released in the process would be _______ MeVMeV.

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Answer: 236

The formula to calculate the energy released in the given process can be written as
Q=BEProduct−BEReactant...(1)Q=B{E}_{\text{Product}}-B{E}_{\text{Reactant}}...\left(1\right)
From equation (1), it follows that
Q=2(118)(8.6)−236(7.6)MeV=236MeV\begin{matrix}Q & = & 2(118)(8.6)-236(7.6)MeV \\ & = & 236MeV\end{matrix}

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