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Q.158 Physics ⭐ 360° Option Elimination Active
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NEET 2020 Physics Question 158 Solution

Topic: Nuclei · Official NTA Booklet Code E1
NEET 2020 Physics Question 158: Nuclear Fission Reaction Balancing: 235_92 U + 1_0 n -> 144_56 Ba + 89_36 Kr + 3 1_0 n

Question

When a uranium isotope ²³⁵₉₂U is bombarded with a neutron, it generates ⁸⁹₃₆Kr, three neutrons and:

✓
Verified Against NTA Final Answer Key
Option (1) : 144_56 Ba
📚 Nuclei Easy
📋 Given Parameters & Problem Setup
Target Nucleus 235_92 U
Incident Projectile Thermal neutron (1_0 n)
Known Products 89_36 Kr and 3 neutrons (3 1_0 n)
Question Objective Determine the missing fission fragment (A_Z X)
⚡

The 30-Second Exam Hall Shortcut: ⚡ "Conservation of Mass & Charge: Z = 92 - 36 = 56 (Barium); A = 236 - (89 + 3) = 144 -> 144_56 Ba!"

⏱️ 15 seconds

Conservation of charge (atomic number): 92+0=36+Z+3(0)  ⟹  Z=92−36=5692 + 0 = 36 + Z + 3(0) \implies Z = 92 - 36 = 56 (Element with Z=56Z = 56 is Barium, Ba!). Conservation of mass number: 235+1=89+A+3(1)  ⟹  236=92+A  ⟹  A=236−92=144235 + 1 = 89 + A + 3(1) \implies 236 = 92 + A \implies A = 236 - 92 = 144. The missing fragment is 56144Ba^{144}_{56}\text{Ba}! Mark Option (1) in 5 seconds.

1 Balanced nuclear equation: 235_92 U + 1_0 n -> 89_36 Kr + A_Z X + 3 1_0 n.
2 Balance atomic number Z: 92 + 0 = 36 + Z + 0 => Z = 92 - 36 = 56.
3 Element with Z = 56 is Barium (Ba).
4 Balance mass number A: 235 + 1 = 89 + A + 3(1) => 236 = 92 + A => A = 144.
5 Missing nucleus is 144_56 Ba.
6 Select Option (1).
🛡️
Why You Can 100% Trust This Shortcut: NCERT Physics · Nuclei (Nuclear Energy: Nuclear Fission of Uranium-235)
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Formulate the Unbalanced Nuclear Fission Equation

When a uranium-235 nucleus absorbs a slow thermal neutron, it undergoes nuclear fission according to the reaction: 92235U+01n→3689Kr+ZAX+301n^{235}_{92}\text{U} + ^1_0\text{n} \to ^{89}_{36}\text{Kr} + ^A_Z\text{X} + 3 ^1_0\text{n} where ZAX^A_Z\text{X} is the unknown co-product nucleus.

Step 2

Apply Conservation of Atomic Number (Charge Conservation)

Total charge on reactant side equals total charge on product side: ∑Zreactants=∑Zproducts\sum Z_{\text{reactants}} = \sum Z_{\text{products}} 92+0=36+Z+3(0)92 + 0 = 36 + Z + 3(0) 92=36+Z92 = 36 + Z Z=92−36=56Z = 92 - 36 = 56 The element with atomic number Z=56Z = 56 is Barium (Ba\text{Ba}).

Step 3

Apply Conservation of Mass Number (Nucleon Conservation)

Total mass number on reactant side equals total mass number on product side: ∑Areactants=∑Aproducts\sum A_{\text{reactants}} = \sum A_{\text{products}} 235+1=89+A+3(1)235 + 1 = 89 + A + 3(1) 236=89+A+3236 = 89 + A + 3 236=92+A236 = 92 + A A=236−92=144A = 236 - 92 = 144

Step 4

Identify the Nucleus

The missing nuclide is barium-144, represented as 56144Ba^{144}_{56}\text{Ba}. This matches Option (1).

⭐

360° Option Elimination Engine Key Feature

Diagnose every choice: Know exactly why the right option works and why distractors fail.

4-Option Deep Scan
1 A
✓ 100% Correct
Verdict & Scientific Reason:

Factually and nuclear-mechanically exact; balances both atomic number (Z=56Z = 56) and mass number (A=144A = 144).

💡
Why this is airtight: None; textbook nuclear fission reaction.
2 B
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; atomic number 40+36=76≠9240 + 36 = 76 \neq 92, violating charge conservation.

⚠️
Student Fallacy (Why candidates pick this trap): Failing to balance nuclear charge.
3 C
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; duplicate krypton isotope, charge does not balance (36+36=72≠9236 + 36 = 72 \neq 92).

⚠️
Student Fallacy (Why candidates pick this trap): Assuming symmetric fission into two krypton nuclei.
4 D
✗ Incorrect
Verdict & Scientific Reason:

Incorrect; violates conservation of electric charge.

⚠️
Student Fallacy (Why candidates pick this trap): Random fission fragment option.
⚠️

NTA Trap Alert: Don't Forget the 3 Neutrons in Mass Balance!

The reaction produces THREE neutrons (3×1=33 \times 1 = 3 mass units). If you forget the 3 neutrons, you might calculate A=236−89=147A = 236 - 89 = 147. Always account for 301n3 ^1_0\text{n}!

🛡️ How to avoid the -1 mark: Charge: 92 - 36 = 56 (Barium!). Mass: 236 - 89 - 3 = 144 -> 56144Ba^{144}_{56}\text{Ba}!
📖

Core NEET Syllabus Standard

Physics · Nuclei Standard Benchmark
Core Scientific Principle Tested:

""

Syllabus Focus: Nuclear Fission Reaction Balancing: 235_92 U + 1_0 n -> 144_56 Ba + 89_36 Kr + 3 1_0 n
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

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