Chemistry Practice Problems Bohr and Balmer Equations Practice Problems Solution: An atomic emission spectrum of hydrogen shows the ...

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# Solution: An atomic emission spectrum of hydrogen shows the following three wavelengths: 1875 nm, 1282 nm, and 1093 nm. Assign these wavelengths to transitions in the hydrogen atom.For nm = 1282 { m nm}.

###### Problem

An atomic emission spectrum of hydrogen shows the following three wavelengths: 1875 nm, 1282 nm, and 1093 nm. Assign these wavelengths to transitions in the hydrogen atom.

For nm = 1282 .

###### Solution

We’re being asked to assign 1282 nm to transitions in the hydrogen atom.

We can see that the three wavelengths correspond to Paschen/Bohr series with nfinal = 3

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

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 initial energy level (ninitial):

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