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Hydrogen spectrum and Rydberg formula: Class 11 practice

1/λ = R(1/n₁² − 1/n₂²) with R = 1.097 × 10⁷ m⁻¹ and n₂ > n₁. Read the method, follow the three worked examples, then try the six questions.

What you should know first

  • Fractions
  • Scientific notation

The idea behind hydrogen spectrum and rydberg formula

1/λ = R(1/n₁² − 1/n₂²) with R = 1.097 × 10⁷ m⁻¹ and n₂ > n₁. n₁ = 1 is the Lyman series (UV), 2 Balmer (visible), 3 Paschen (IR). Energy of level n is −13.6/n² eV.

Step-by-step method

  1. Identify n₁ and n₂.
  2. Compute 1/λ with the Rydberg formula.
  3. Invert to get λ (nm); or ΔE = 13.6(1/n₁² − 1/n₂²) eV.

Worked example 1 (Basic)

Find the wavelength of the line for the transition n = 5 → n = 1 in hydrogen.

  1. 1/λ = 1.097 × 10⁷ (1/1 − 1/25) = 1.053e+7 m⁻¹
  2. λ = 94.96 nm (Lyman, UV)
  3. Answer: 94.96 nm

Worked example 2 (Standard)

Find the wavelength of the line for the transition n = 3 → n = 2 in hydrogen.

  1. 1/λ = 1.097 × 10⁷ (1/4 − 1/9) = 1.524e+6 m⁻¹
  2. λ = 656.34 nm (Balmer)
  3. Answer: 656.34 nm

Worked example 3 (Challenge)

Find the wavelength of the line for the transition n = 5 → n = 3 in hydrogen.

  1. 1/λ = 1.097 × 10⁷ (1/9 − 1/25) = 7.801e+5 m⁻¹
  2. λ = 1281.91 nm (Paschen, IR)
  3. Answer: 1281.91 nm

Common mistakes and misconceptions

  • Swapping n₁ and n₂
  • Forgetting to invert

Practice questions

  1. (Basic) Find the wavelength of the line for the transition n = 3 → n = 1 in hydrogen.
  2. (Basic) Find the wavelength of the line for the transition n = 2 → n = 1 in hydrogen.
  3. (Standard) Find the wavelength of the line for the transition n = 4 → n = 2 in hydrogen.
  4. (Standard) Find the wavelength of the line for the transition n = 5 → n = 2 in hydrogen.
  5. (Challenge) Find the wavelength of the line for the transition n = 6 → n = 3 in hydrogen.
  6. (Challenge) Find the wavelength of the line for the transition n = 4 → n = 3 in hydrogen.

Answer key with full working

  • Q1: 1/λ = 1.097 × 10⁷ (1/1 − 1/9) = 9.751e+6 m⁻¹ → λ = 102.55 nm (Lyman, UV) → Answer: 102.55 nm
  • Q2: 1/λ = 1.097 × 10⁷ (1/1 − 1/4) = 8.228e+6 m⁻¹ → λ = 121.54 nm (Lyman, UV) → Answer: 121.54 nm
  • Q3: 1/λ = 1.097 × 10⁷ (1/4 − 1/16) = 2.057e+6 m⁻¹ → λ = 486.17 nm (Balmer) → Answer: 486.17 nm
  • Q4: 1/λ = 1.097 × 10⁷ (1/4 − 1/25) = 2.304e+6 m⁻¹ → λ = 434.08 nm (Balmer) → Answer: 434.08 nm
  • Q5: 1/λ = 1.097 × 10⁷ (1/9 − 1/36) = 9.142e+5 m⁻¹ → λ = 1093.89 nm (Paschen, IR) → Answer: 1093.89 nm
  • Q6: 1/λ = 1.097 × 10⁷ (1/9 − 1/16) = 5.333e+5 m⁻¹ → λ = 1875.24 nm (Paschen, IR) → Answer: 1875.24 nm

Syllabus note

This skill is listed under "Structure of Atom" for Class 11 in the NCERT/CBSE books. ICSE, IGCSE, IB and state boards may teach it in a different chapter or year.

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

Related practice and notes

Frequently asked questions

Which chapter is hydrogen spectrum and rydberg formula in?

"Structure of Atom", Class 11 Chemistry (NCERT/CBSE). Other boards may place it differently.

How were the answers checked?

Answers on this page are generated and automatically re-checked by a second, independent calculation. They have not been reviewed by a teacher — if you spot a problem, tell us on WhatsApp.

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