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Problem 4 Easy Difficulty

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 $_{52}^{135} \mathrm{Te}$ .

Answer

192 $\mathrm{MeV}$

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Video Transcript

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National University of Singapore
Top Physics 103 Educators
Elyse G.

Cornell University

Andy C.

University of Michigan - Ann Arbor

Farnaz M.

Other Schools

Meghan M.

McMaster University