In a nuclear reaction: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + x(¹₀n). The value of x is:

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Question

In a nuclear reaction: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + x(¹₀n). The value of x is:

Accepted Answer

Correct Answer: C — x = 3Step-by-step Solution:In any nuclear reaction, both mass number (A) and atomic number (Z) must be conserved.Conservation of Mass Number (A):235 + 1 = 141 + 92 + x × 1236 = 233 + xx = 3Verification with Atomic Number (Z):92 + 0 = 56 + 36 + x × 092 = 92 ✓ (neutrons have Z = 0)This is the famous nuclear fission of Uranium-235, which releases 3 neutrons and enormous energy. These 3 neutrons can trigger further fissions — the basis of a chain reaction.Why Options A (1), B (2), D (4) are wrong: These do not satisfy conservation of mass number. Only x = 3 gives 141 + 92 + 3 = 236 = 235 + 1. ✓Key Concepts from Nuclei (NCERT Class 12, Chapter 13):Nuclear fission: A heavy nucleus splits into two lighter nuclei (called fission fragments) and neutrons, releasing energy (Q-value). The mass of products is less than reactants — this 'mass defect' appears as energy (E = mc²).Conservation laws in nuclear reactions: Mass number A, atomic number Z, momentum, energy, and lepton number are conserved.The fission of ²³⁵U releases ~200 MeV per fission (about 3.2 × 10⁻¹¹ J). This is the basis of nuclear reactors and atomic bombs.Chain reaction: The 3 neutrons released can each trigger further fissions of U-235, leading to a self-sustaining chain reaction (controlled in reactors, uncontrolled in bombs).Binding energy: BE = Δm × c², where Δm is the mass defect. Greater binding energy per nucleon = more stable nucleus. Fe-56 has the highest binding energy per nucleon.

In a nuclear reaction: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + x(¹₀n). The value of x is:

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In a nuclear reaction: ²³⁵₉₂U + ¹₀n → ¹⁴¹₅₆Ba + ⁹²₃₆Kr + x(¹₀n). The value of x is:

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