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Q.46 Chemistry ⭐ 360° Option Elimination Active
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NEET 2025 Chemistry Question 46 Solution

Topic: Structure of Atom · Official NTA Booklet Code 45
NEET 2025 Chemistry Question 46: Ratio of Wavelengths of Absorbed Light in Rydberg Transitions

Question

The ratio of the wavelengths of the light absorbed by a Hydrogen atom when it undergoes n = 2 → n = 3 and n = 4 → n = 6 transitions, respectively, is

✓
Verified Against NTA Final Answer Key
Option (4) : \frac{1}{4}
📚 Structure of Atom Easy
📋 Given Parameters & Problem Setup
Transition 1 n = 2 → n = 3
Transition 2 n = 4 → n = 6
Target Ratio λ₁ / λ₂ (wavelength of light absorbed in transition 1 to transition 2)
⚡

The 30-Second Exam Hall Shortcut: ⚡ Scaled Orbit Rydberg Shortcut

⏱️ Save ~40 seconds

Notice transition 2 has double the quantum numbers of transition 1: (4 = 2×2, 6 = 2×3). Since 1/λ ∝ 1/n², doubling all n values scales 1/λ by 1/2² = 1/4 ⟹ λ₂ = 4 λ₁ ⟹ λ₁ / λ₂ = 1/4!

1 Notice quantum scaling: n₁ = 2, n₂ = 3 for transition 1; n₁' = 4 = 2(2), n₂' = 6 = 2(3) for transition 2.
2 Wavenumber formula: 1/λ = R_H (1/n₁² - 1/n₂²).
3 When both quantum numbers are multiplied by a factor of 2: 1/λ₂ = R_H [1/(2×2)² - 1/(2×3)²] = (1/4) × R_H (1/2² - 1/3²) = (1/4) (1/λ₁).
4 Therefore: λ₁ / λ₂ = 1/4.
5 Select Option (4) in under 10 seconds without calculating full fractions!
🛡️
Why You Can 100% Trust This Shortcut: Rydberg Formula for Hydrogen Atom Transitions (NCERT Chemistry · Structure of Atom: 1/λ = R_H Z² (1/n₁² - 1/n₂²)).
📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Apply Rydberg Formula to Transition 1 (n = 2 → n = 3)

According to the Rydberg formula, the reciprocal of the absorbed wavelength for hydrogen is:

1λ1=RH(122−132)=RH(14−19)=RH(9−436)=536RH\frac{1}{\lambda_1} = R_H \left(\frac{1}{2^2} - \frac{1}{3^2}\right) = R_H \left(\frac{1}{4} - \frac{1}{9}\right) = R_H \left(\frac{9 - 4}{36}\right) = \frac{5}{36} R_H
  • • Inverting gives: λ₁ = 36 / (5 R_H).
Step 2

Apply Rydberg Formula to Transition 2 (n = 4 → n = 6)

For the transition between n = 4 and n = 6:

1λ2=RH(142−162)=RH(116−136)=RH(9−4144)=5144RH\frac{1}{\lambda_2} = R_H \left(\frac{1}{4^2} - \frac{1}{6^2}\right) = R_H \left(\frac{1}{16} - \frac{1}{36}\right) = R_H \left(\frac{9 - 4}{144}\right) = \frac{5}{144} R_H
  • • Inverting gives: λ₂ = 144 / (5 R_H).
Step 3

Calculate the Wavelength Ratio

Divide the two wavelengths:

λ1λ2=365RH1445RH=36144=14\frac{\lambda_1}{\lambda_2} = \frac{\frac{36}{5 R_H}}{\frac{144}{5 R_H}} = \frac{36}{144} = \frac{1}{4}
  • • Matches Option (4) exactly.
⭐

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
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Forgets common factor 5/R_H or squares denominator.

⚠️
Student Fallacy (Why candidates pick this trap): Squaring the ratio 1/6.
2 B
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Squaring the 1/4 factor.

⚠️
Student Fallacy (Why candidates pick this trap): Squaring the scaling factor (1/4)² = 1/16.
3 C
✗ Incorrect
Verdict & Scientific Reason:

Incorrect. Ratio of 1/n² terms.

⚠️
Student Fallacy (Why candidates pick this trap): Dividing 4 by 36.
4 D
✓ 100% Correct
Verdict & Scientific Reason:

100% correct. λ₁ / λ₂ = 36 / 144 = 1/4.

💡
Why this is airtight: None; flawless Rydberg calculation.
⚠️

NTA Trap Alert: Inverting Wavenumber Ratio vs Wavelength Ratio

Calculating (1/λ₁) / (1/λ₂) = 144/36 = 4, then forgetting that λ₁/λ₂ is the reciprocal 1/4.

🛡️ How to avoid the -1 mark: Remember: 1/λ is wavenumber; wavelength λ is its inverse!
📖

Core NEET Syllabus Standard

NEET Chemistry Core Syllabus · Structure of Atom Standard Benchmark
Core Scientific Principle Tested:

"1/λ = R_H (1/n₁² - 1/n₂²)."

Syllabus Focus: Bohr’s Model & Rydberg Formula for Hydrogen Spectrum
✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗
Doubled levels ⟹ 1/λ drops by 2² = 4 ⟹ λ quadruples ⟹ λ₁ / λ₂ = 1/4!

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