# Problem: Calculate the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the n = 6 state and then fall to states with smaller values of  n.

###### FREE Expert Solution

Distance = Energy

lower-numbered shell there’s a bigger distance between them
the distance between shells gets smaller the higher up you go
the smaller the distance then the less energy is required for an electron to travel

$\overline{){\mathbf{E}}{\mathbf{=}}\frac{\mathbf{hc}}{\mathbf{\lambda }}}$

↓ smaller distance, ↓ less energy required, ↑ wavelength

longest wavelength:     n = 6 to n = 5

↑ bigger distance, ↑ higher energy required,  wavelength

shortest wavelength:     n = 6 to n = 1

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 wavelength of light (λ):

n = 6 to n = 5

Z = atomic number of Hydrogen = 1 (refer to the periodic table)
R = 1.097x107 m
-1
ninitial = 1
nfinal = 3

λ = 7458 nm

n = 6 to n = 1

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###### Problem Details

Calculate the longest and shortest wavelengths of light emitted by electrons in the hydrogen atom that begin in the n = 6 state and then fall to states with smaller values of  n.