With some manipulation, the Rydberg equation can be rewritten in the form

$\mathbf{E}\mathbf{}\mathbf{=}\mathbf{}\mathbf{constant}\mathbf{\times}\mathbf{(}\frac{\mathbf{1}}{{{\mathbf{n}}_{\mathbf{f}}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{{\mathbf{n}}_{\mathbf{i}}}^{\mathbf{2}}}\mathbf{)}$

which allows you to calculate the energy of the emitted light. Express this constant in terms of the constants h, c, and R_{H} using relationships between wavelength and energy as well as the Rydberg equation from the introduction.

The Rydberg equation expresses the wavelength, λ, of emitted light based on the initial and final energy states (n_{i} and n_{f}) of an electron in a hydrogen atom:

$\frac{\mathbf{1}}{\mathbf{\lambda}}\mathbf{=}\mathbf{}{\mathbf{R}}_{\mathbf{H}}\mathbf{\times}\mathbf{(}\frac{\mathbf{1}}{{{\mathbf{n}}_{\mathbf{f}}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{{\mathbf{n}}_{\mathbf{i}}}^{\mathbf{2}}}\mathbf{)}$

where R_{H} = 1.097 × 10^{7} m^{− 1} = 1.097 × 10^{−2} nm^{−1}.

You may also see this equation written as

$\frac{\mathbf{1}}{\mathbf{\lambda}}\mathbf{=}\mathbf{}\mathbf{-}{\mathbf{R}}_{\mathbf{H}}\mathbf{\times}\mathbf{(}\frac{\mathbf{1}}{{{\mathbf{n}}_{\mathbf{i}}}^{\mathbf{2}}}\mathbf{-}\frac{\mathbf{1}}{{{\mathbf{n}}_{\mathbf{f}}}^{\mathbf{2}}}\mathbf{)}$

Since

$(\frac{1}{{{n}_{f}}^{2}}-\frac{1}{{{n}_{i}}^{2}})\mathbf{=}\mathbf{-}(\frac{1}{{{n}_{i}}^{2}}-\frac{1}{{{n}_{f}}^{2}})$

the two formulas are equivalent and sometimes used interchangeably. It can help to remember that when light is emitted, E is negative. When light is absorbed, E is positive.

Frequently Asked Questions

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