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In hydrogen emission spectrum, for any series, the principal quantum number is n. Corresponding maximum wavelength λ is (R = Rydberg’s constant)

Options

  1. AR 2 n + 1 n 2 n + 1
  2. Bn 2 n + 1 2 R 2 n + 1
  3. Cn 2 n + 1 R 2 n + 1
  4. DR 2 n + 1 n 2 n + 1 2

Correct answer

B. n 2 n + 1 2 R 2 n + 1

Step-by-step solution

In hydrogen emission spectrum, wavelength is given by 1 λ = R 1 n 1 2 - 1 n 2 2 For maximum wavelength of any principal quantum number n, n 1 = n a n d n 2 = n + 1 ∴ 1 λ m a x = R 1 n 2 - 1 n + 1 2 = R n + 1 2 - n 2 n 2 n + 1 2 = R n 2 + 2 n + 1 - n 2 n 2 n + 1 2 1 λ m a x = R 2 2 n + 1 n 2 n + 1 2 ∴ λ m a x = n 2 n + 1 2 R 2 n + 1

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