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What is the wavelength (in nm) of the line in the spectrum of the hydrogen atom that arises from the transition of the electron from the orbital with n =5 to the orbital with n = 2?

Answer :

maacastrobr

Answer:

433 nm

Explanation:

To solve this problem we can use Rydberg's equation:

1 / λ = R *[tex](\frac{1}{n^{2}_{final}}-\frac{1}{n^{2}_{initial}}  )[/tex]

Where λ is the wavelength (in m), R is a constant (1.0974x10⁷ m⁻¹) and n are the initial and final states.

We put the data given by the problem and solve for λ:

1 / λ = 1.0974x10⁷ m⁻¹ *[tex](\frac{1}{2^2} -\frac{1}{5^2} )[/tex]

1 / λ = 2304540 m⁻¹

λ = 4.34x10⁻⁷ m

Converting m to nm:

4.34x10⁻⁷m * 10⁹nm / 1m = 433 nm

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