Nucleus: Practice Problem & Solution
For nuclear reaction $_{92}U^{235} + _{0}n^{1} \rightarrow _{56}Ba^{144} + ...... + 3 _{0}n^{1}$: (1998)
Solution Explained:
To solve this problem, we apply the core principles of Nucleus. Understanding the underlying formula is key to arriving at the correct answer below:
Let the unknown product be $^{A}_{Z}X$. Balancing the atomic numbers: $92 + 0 = 56 + Z + 0 \Rightarrow Z = 36$. Balancing the mass numbers: $235 + 1 = 144 + A + 3 \Rightarrow A = 236 - 147 = 89$. The element with $Z=36$ is Krypton (Kr), so the product is $_{36}Kr^{89}$.
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