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\( ^{235} _{92}U + ^1 _0n → ^{144} _{56}Ba + ^{90} _{36}Kr +2X ...

\( ^{235} _{92}U + ^1 _0n → ^{144} _{56}Ba + ^{90} _{36}Kr +2X \)
In the reaction above, X is
Take Free Practice Test On 2026 JAMB UTME, Post UTME, WAEC SSCE, GCE, NECO SSCE
  • A Electron
  • B Proton
  • C Neutrino
  • D Neutron
Correct Answer: Option D
Explanation:
235U + 1n → 144Ba + 90Kr + 2x9205636
In the reaction above, balancing the left with the right, show that X is neutron

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