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**Problem**: 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. 1b. 2c. 3d. 4e. all of them

###### FREE Expert Solution

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

###### Problem Details

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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