Q.15 Physics · Section A ⭐ 360° Option Elimination Active

NEET 2024 Physics Question 15 Solution

Topic: Atoms · Official NTA Booklet Code T1

Question

Match List I (Spectral Lines of Hydrogen for transitions from) with List II (Wavelengths (nm)). List I: A. n₂ = 3 to n₁ = 2; B. n₂ = 4 to n₁ = 2; C. n₂ = 5 to n₁ = 2; D. n₂ = 6 to n₁ = 2. List II: I. 410.2; II. 434.1; III. 656.3; IV. 486.1. Choose the correct answer from the options given below:

✓

Verified Against NTA Final Answer Key

Option (4) : A-III, B-IV, C-II, D-I

📚 Atoms Easy

📋 Given Parameters & Problem Setup

Spectral Series Balmer series with lower state n1 = 2

Transition A n2 = 3 to n1 = 2 (H-α line)

Transition B n2 = 4 to n1 = 2 (H-β line)

Transition C n2 = 5 to n1 = 2 (H-γ line)

Transition D n2 = 6 to n1 = 2 (H-δ line)

⚡

The 30-Second Exam Hall Shortcut: ⚡ Energy-Wavelength Inverse Ordering Hack

⏱️ Save ~50 seconds

Larger jump = higher photon energy = SHORTER wavelength! n2=3 is longest (656.3 nm) -> A-III -> Instant single-match!

1 In the Balmer series, transitions from higher n2 represent larger energy drops ΔE.

2 Since λ = hc / ΔE, higher n2 yields SHORTER wavelengths.

3 Transition A (n2 = 3 to 2) has the smallest energy gap, hence the LONGEST wavelength: 656.3 nm -> A must match III.

4 Looking at the options grid: Only Option (4) has A-III!

5 Solved in 5 seconds without calculating anything else!

🛡️

Why You Can 100% Trust This Shortcut: Rydberg Formula for Hydrogen Transitions (1/λ = R_H [1/n1² - 1/n2²]).

📐

Step-by-Step Rigorous Workflow & Mathematical Proof

Step 1

Governing Formula (Rydberg Formula)

For transitions in the hydrogen atom ending at n1 = 2 (the Balmer series in the visible spectrum), the wavelength λ is given by the Rydberg formula:

\frac{1}{\lambda} = R_H \left(\frac{1}{2^2} - \frac{1}{n_2^2}\right) = R_H \left(\frac{1}{4} - \frac{1}{n_2^2}\right)

Step 2

Calculation for Transition A (n2 = 3 to n1 = 2, H-α line)

The transition from n2 = 3 to n1 = 2 has the smallest energy difference:

\frac{1}{\lambda_A} = R_H \left(\frac{1}{4} - \frac{1}{9}\right) = \frac{5}{36} R_H \implies \lambda_A = \frac{36}{5 R_H} \approx 656.3\,\text{nm} \implies \text{A-III}

Step 3

Calculation for Transition B (n2 = 4 to n1 = 2, H-β line)

The transition from n2 = 4 to n1 = 2 has intermediate energy:

\frac{1}{\lambda_B} = R_H \left(\frac{1}{4} - \frac{1}{16}\right) = \frac{3}{16} R_H \implies \lambda_B = \frac{16}{3 R_H} \approx 486.1\,\text{nm} \implies \text{B-IV}

Step 4

Calculation for Transition C (n2 = 5 to n1 = 2, H-γ line)

The transition from n2 = 5 to n1 = 2:

\frac{1}{\lambda_C} = R_H \left(\frac{1}{4} - \frac{1}{25}\right) = \frac{21}{100} R_H \implies \lambda_C \approx 434.1\,\text{nm} \implies \text{C-II}

Step 5

Calculation for Transition D (n2 = 6 to n1 = 2, H-δ line)

The highest energy transition in the set from n2 = 6 to n1 = 2 produces the shortest wavelength:

\lambda_D \approx 410.2\,\text{nm} \implies \text{D-I}

⭐

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 (1)

✗ Incorrect

Verdict & Scientific Reason:

Mismatched wavelengths.

⚠️

Student Fallacy (Why candidates pick this trap): Associating A with the second longest wavelength.

2 (2)

✗ Incorrect

Verdict & Scientific Reason:

Inverts the wavelength relationship (assigns shortest wavelength to smallest transition).

⚠️

Student Fallacy (Why candidates pick this trap): Confusing wavelength with frequency/energy.

3 (3)

✗ Incorrect

Verdict & Scientific Reason:

Completely jumbled matching.

⚠️

Student Fallacy (Why candidates pick this trap): Guesswork without checking ordering.

4 (4)

✓ 100% Correct

Verdict & Scientific Reason:

Perfect monotonic match: smallest gap (n=3) has longest λ (656.3 nm), largest gap (n=6) has shortest λ (410.2 nm).

💡

Why this is airtight: None.

⚠️

NTA Trap Alert: Energy vs Wavelength Inverse Proportionality

Students frequently assign the highest wavelength to the highest energy transition (n=6 to 2).

🛡️ How to avoid the -1 mark: Always recite: E = hc / λ. Big energy gap = Tiny wavelength!

📖

Core NEET Syllabus Standard

Physics Core Modern Physics · Atoms Standard Benchmark

Core Scientific Principle Tested:

"The Balmer series corresponds to electron transitions to n1 = 2; wavelengths lie in the visible region from 656.3 nm (red, n2=3) to 364.6 nm (series limit)."

Syllabus Focus: Bohr’s Model of the Hydrogen Atom and Spectral Series

✓ Core NEET UG Syllabus Anchor Syllabus & Academic Standards ↗

Red (n=3) is Longest (656 nm); Violet (n=6) is Shortest (410 nm)!

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NEET 2024 Physics Question 15 Solution

Question

The 30-Second Exam Hall Shortcut: ⚡ Energy-Wavelength Inverse Ordering Hack

Step-by-Step Rigorous Workflow & Mathematical Proof

Governing Formula (Rydberg Formula)

Calculation for Transition A (n2 = 3 to n1 = 2, H-α line)

Calculation for Transition B (n2 = 4 to n1 = 2, H-β line)

Calculation for Transition C (n2 = 5 to n1 = 2, H-γ line)

Calculation for Transition D (n2 = 6 to n1 = 2, H-δ line)

360° Option Elimination Engine Key Feature

NTA Trap Alert: Energy vs Wavelength Inverse Proportionality

Core NEET Syllabus Standard

Ready to attempt under real exam pressure?

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