What will be the wavelength of a photon that is emitted when an electron of a hydrogen atom undergoes a transition from n=7 to n=4?

The wavelength of the transition from n = 7 to n=4 is 2166nm.

Using the Rydberg formula, we have;

1/λ = R(1/n^2f - 1/n^2i)

λ = wavelength

R = Rydberg constant = 1.097 × 10^7 m-1

nf = 4

ni = 7

Substituting values;

1/λ = 1.097 × 10^7 m-1(1/4^2 - 1/7^2)

1/λ = 2.166 × 10^7 m or 2166nm

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Explanation:

Using Rydberg's Equation:

Where,

= Wavelength of radiation

= Rydberg's Constant  =

= Higher energy level = 7  

= Lower energy level = 4

Z= atomic number = 1 (for hydrogen)

Thus wavelength is

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