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Complete the following nuclear fission reaction. ¹₀n+²³⁵₉₂U→²³⁶₉₂U→¹⁴⁴₅₆Ba+ ______ + 3 ¹₀n

Asked in GSEB Board August 2020 · Q-value, decay and conservation laws

Answer: (1) ⁸⁹₃₆Kr

Step-by-step solution

Idea: nucleon number and charge are conserved. Call the missing fragment ^A_ZX and write one equation for each.

Nucleons: 1+235=236=144+A+3, so A=236-147=89.

Charge: 0+92=56+Z+0, so Z=36.

Z=36 is krypton, so the missing fragment is ⁸⁹₃₆Kr.

This is the textbook fission channel of ²³⁵₉₂U: barium, krypton and three neutrons.

Why the other options are wrong

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