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Find the energy released in the fission reaction

$$

\mathrm{n}+_{92}^{235} \mathrm{U} \quad \rightarrow \quad_{40}^{98} \mathrm{Zr}+\frac{135}{52} \mathrm{Te}+3 \mathrm{n}

$$

The atomic masses of the fission products are 97.9120 u for 98 Zr and 134.9087 u for $_{52}^{135} \mathrm{Te}$ .

192 $\mathrm{MeV}$

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National University of Singapore