Question
Calculate the energy released in the fission reaction of Eq. (29-30). The atomic masses of ${ }_{56}^{141} \mathrm{Ba}$ and ${ }_{36}^{92} \mathrm{Kr}$ are $140.914 \mathrm{u}$ and $91.926 \mathrm{u}$, respectively.
Step 1
It is given as: \[{}_{92}^{235}U + {}_{0}^{1}n \rightarrow {}_{56}^{141}Ba + {}_{36}^{92}Kr + 3{}_{0}^{1}n\] Show more…
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Calculate the energy released in the fission reaction of Eq. $(29-30) .$ The atomic masses of 141 56 Ba and 36 Kr are $140.914 \mathrm{u}$ and $91.926 \mathrm{u},$ respectively.
(I) What is the energy released in the fission reaction of Eq. $42-5 ?$ (The masses of ${ }_{56}^{141} \mathrm{Ba}$ and ${ }_{36}^{92} \mathrm{Kr}$ are $140.914411 \mathrm{u}$ and 91.926156 u, respectively.)
Find the energy released in the fission reaction $$ \mathrm{n}+{ }_{92}^{295} \mathrm{U} \rightarrow{ }_{40}^{98} \mathrm{Zr}+{ }_{52}^{135} \mathrm{Te}+3 \mathrm{n} $$ The atomic masses of the fission products are $97.9120 \mathrm{u}$ for ${ }_{40}^{98} \mathrm{Zr}$ and $134.9087 \mathrm{u}$ for ${ }^{135} \mathrm{Te}$.
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