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The wave number is given by the formula ν = 1/λ = kZ² [1/n₁² - 1/n₂²]. For the Lyman series, the shortest wavelength (λₘₐₓ) occurs when n₁ = 1, n₂ = ∞, and Z = 1 (for a hydrogen atom). This simplifies to 1/λ = kZ² [1/1² - 1/∞²] ⇒ λ = 1/x. Which of the following statements is correct?

  1. For the shortest wavelength in the Lyman series of Li²⁺, n₁ = 1, n₂ = ∞, and Z = 3. Thus, 1/λ = x × 3² [1/1² - 1/∞²] ⇒ λ = x/9.
  2. For the longest wavelength in the Lyman series, n₁ = 1, n₂ = 2. This gives 1/λ = x × 3² [1/1² - 1/2²] = 27/4x ⇒ λ = 4x/27.
  3. For the shortest wavelength in the Balmer series, n₁ = 2, n₂ = ∞. Therefore, 1/λ = x × 3² [1/2² - 1/∞²] ⇒ λ = 4x/9.
  4. For the longest wavelength in the Balmer series, n₁ = 2, n₂ = 3. Hence, 1/λ = x × 3² [1/2² - 1/3²] ⇒ λ = 4x/5.

Correct answer: For the shortest wavelength in the Lyman series of Li²⁺, n₁ = 1, n₂ = ∞, and Z = 3. Thus, 1/λ = x × 3² [1/1² - 1/∞²] ⇒ λ = x/9.

Solution

For the shortest wavelength in the Lyman series of Li²⁺, the correct equation is 1/λ = x × 3² [1/1² - 1/∞²] ⇒ λ = x/9 because the atomic number Z is 3 for Li²⁺, and the shortest wavelength occurs when n₁ = 1 and n₂ = ∞.

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