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Question
When \(_{92}^{235}U\) captures a bombarding neutron, the intermediate nucleus formed before it splits is:
Options
\(_{56}^{144}Ba\)
\(_{92}^{236}U\)
\(_{92}^{235}U\)
\(_{36}^{89}Kr\)
MCQ
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Solution
A neutron has mass number 1 and atomic number 0, so its capture raises the mass number from 235 to 236 while the atomic number stays at 92, forming \(_{92}^{236}U\). This intermediate nucleus then splits into two lighter nuclei with the release of neutrons.
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