# Problem: An electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ = 410 nm. { m ; m nmFind the principal level to which the electron relaxed.

###### FREE Expert Solution

We’re being asked to calculate for the principal level to which the electron relaxed.

We’re going to use the Balmer Equation which relates wavelengths to a photon’s electronic transitions.

$\overline{)\frac{\mathbf{1}}{\mathbf{\lambda }}{\mathbf{=}}{{\mathbf{RZ}}}^{{\mathbf{2}}}\left(\frac{\mathbf{1}}{{{\mathbf{n}}^{\mathbf{2}}}_{\mathbf{final}}}\mathbf{-}\frac{\mathbf{1}}{{{\mathbf{n}}^{\mathbf{2}}}_{\mathbf{initial}}}\right)}$

λ = wavelength, m
R = Rydberg constant = 1.097x107 m-1
Z = atomic number of the element
ninitial = initial energy level
nfinal = final energy level

Calculate the final energy level (nfinal):

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###### Problem Details

An electron in the n = 6 level of the hydrogen atom relaxes to a lower energy level, emitting light of λ = 410 nm. Find the principal level to which the electron relaxed.