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An electron in an unknown energy level of a hydrogen atom transitions to the n=2 level and emits a photon with wavelength 410 nm in the process. What was the initial energy level? Use R[infinity]=2.179×10−18J for the hydrogen atom Rydberg constant. Use h=6.626×10−34 Js for Planck's constant. Use c=2.998×108ms for the speed of light. Your answer should be a whole number.

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Answer:

6

Explanation:

[tex]n_1[/tex] = Final energy level = 2

[tex]R_H[/tex] = Rydberg constant = [tex]10967758.3\ \text{m}^{-1}[/tex]

[tex]\lambda[/tex] = Wavelength = 410 nm

From the Rydberg formula we have

[tex]\dfrac{1}{\lambda}=R_H(\dfrac{1}{n_1^2}-\dfrac{1}{n_2^2})\\\Rightarrow n_2=(\dfrac{1}{n_1^2}-\dfrac{1}{\lambda R_H})^{-\dfrac{1}{2}}\\\Rightarrow n_2=(\dfrac{1}{2^2}-\dfrac{1}{410\times 10^{-9}\times 10967758.3})^{-\dfrac{1}{2}}\\\Rightarrow n_2=\pm6.01\approx 6[/tex]

The initial energy level is 6.

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