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JWL 3.R 1.096776 x 107 m) L) 384.6 nm C) 683.8 nm D) 1282 nm E) >1500 nm Ans: D According to the Rydberg equation, the line with the shortest wavelength in the emission spectrum of atomic hydrogen is predicted to lie at a wavelength (in nm) of A) 91.2 nm B) 1.10 x 102 nm C) 1.10 x 102 nm Ans: A D) 1.10 x 1016 nm E) None of these choices is correct.

User Richie Li
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Answer:

91.2 nm

Explanation:

The Rydberg equation is given by the formula

1/ λ = Rh ( 1/ n₁² - 1/ n₂²)

where

λ is the wavelength

Rh is Rydberg constant

and n₁ and n₂ are the energy levels of the transion.

We can see from this equation that the wavelength is inversely proportional to the difference of the squares of the inverse of the quantum numbers n₁ and n₂. It follows then that the smallest wavelength will be given when the the transitions are between the greatest separation between n₁ and n₂ whicg occurs when n1= 1 and n₂= ∞ , that is the greater the separation in energy levels the shorter the wavelength.

Substituting for n₁ and n₂ and solving for λ :

1/λ = 1.0974 x 10⁷ m⁻¹ x ( 1/1² -1/ ∞²) = 1.0974 x 10⁷ m⁻¹ x ( 1/1² - 0) =

λ = 1/1.0974 x 10⁷ m = 9.1 x 10⁻8 m = 91.2 nm

User Dan Hogan
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