1. ¹⁴⁴₅₆ Ba 2. ⁹¹₄₀ Zr 3. ¹⁰¹₃₆ Kr 4. ¹⁰³₃₆ Kr

Answer: 1. ¹⁴⁴₅₆ Ba Balance mass number and atomic number. Left: ²³⁵₉₂U+¹₀n gives A=236,\ Z=92 . Right: ⁸⁹₃₆Kr+3 ¹₀n+^A_Z X . A=236-89-3=144 Z=92-36=56 , which is barium So the product is ¹⁴⁴₅₆Ba : option 1.

Topic: Nuclear fission and fusion· Atoms and Nuclei

When a uranium isotope ^235_92U is bombarded with a neutron, it generates ^89_36Kr, three neutrons and :

1. \(^{144}_{56}\)Ba

2. \(^{91}_{40}\)Zr

3. \(^{101}_{36}\)Kr

4. \(^{103}_{36}\)Kr

Doubtly Asked 30 Oct 2026 NEET UG 2020 Practice test neet Save WhatsApp Report

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Answer: 1. \(^{144}_{56}\)Ba

Balance mass number and atomic number.

Left: \(^{235}_{92}\mathrm{U}+{}^1_0n\) gives \(A=236,\ Z=92\).

Right: \(^{89}_{36}\mathrm{Kr}+3\,^1_0n+{}^A_Z X\).

\(A=236-89-3=144\)
\(Z=92-36=56\), which is barium

So the product is \(^{144}_{56}\mathrm{Ba}\): option 1.

Doubtly AI · 30 Oct 2026 AI-generated · not yet verified Report

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When a uranium isotope ^235_92U is bombarded with a neutron, it generates ^89_36Kr, three neutrons and :

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