Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength λ 1 : λ 2 emitted in the

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Electron in hydrogen atom first jumps from third excited state to second excited state and then from second excited to the first excited state. The ratio of the wavelength λ 1 ​ : λ 2 ​ emitted in the two cases is

TestHub · Accepted Answer

The wave number ( v ˉ ) of the radiation = λ 1 ​ = R ∞ ​ [ n 1 2 ​ 1 ​ − n 2 2 ​ 1 ​ ] ​ Now for case (I) n 1 ​ = 3 , n 2 ​ = 2 λ 1 ​ 1 ​ = R ∞ ​ [ 9 1 ​ 4 − 1 ​ ] , R ∞ ​ = Rydberg constant λ 1 ​ 1 ​ = R ∞ ​ [ 36 4 − 9 ​ ] = 36 − 5 R ∞ ​ ​ ⇒ λ 1 ​ = 5 R ∞ ​ − 36 ​ λ 2 ​ 1 ​ = R ∞ ​ [ 4 1 ​ − 1 1 ​ ] = 4 − 3 R ∞ ​ ​ ⇒ λ 2 ​ = 3 R ∞ ​ − 4 ​ ⇒ λ 2 ​ λ 1 ​ ​ = 5 R ∞ ​ − 36 ​ × − 4 3 R ∞ ​ ​ ⇒ λ 2 ​ λ 1 ​ ​ = 5 27 ​ ​

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