Calculate each emission state from n = 4 to n = 3, 2 and 1 using Balmer equation and find which will produce a wavelength at Visible spectrum

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 = 1.097x10^{7} m^{-1}

Z = atomic number of the element

n_{initial }= initial energy level

n_{final} = final energy level

According to the electromagnetic spectrum, visible light has a wavelength at the range 400-700 nm or 4x10^{-7}- 7x10^{-7} m

If a single electron in a hydrogen atom is in the excited *n* = 4 state and eventually relaxes to lower states, there are six possible emission or spectral lines. How many of these lines are in the visible region of the EM spectrum?

a. 1

b. 2

c. 3

d. 4

e. all of them

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