The wavelength of the first spectral line in the Balmer series of the hydrogen atom is 6561 A ∘ . The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is

Question

The wavelength of the first spectral line in the Balmer series of the hydrogen atom is 6561 A ∘ . The wavelength of the second spectral line in the Balmer series of singly ionized helium atom is, 1215 A ∘, 1640 A ∘, 2430 A ∘, 4687 A ∘

MARKS App · Accepted Answer

For hydrogen or hydrogen type atoms 1 λ = R Z 2 1 n f 2 - 1 n i 2 In the transition from n i → n f ∴ λ ∝ 1 Z 2 1 n f 2 - 1 n i 2 ∴ λ 2 λ 1 = Z 1 2 1 n f 2 - 1 n i 2 1 Z 2 2 1 n f 2 - 1 n i 2 2 λ 2 = λ 1 Z 1 2 1 n f 2 - 1 n i 2 1 Z 2 2 1 n f 2 - 1 n i 2 2 Substituting the values, we have = ( 6561 ) 1 2 1 2 2 - 1 3 2 2 2 1 2 2 - 1 4 2 = 1215 A ∘

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