The end (final) value of n in a hydrogen atom transition is 2.
Using the Rydberg formula ;
1/lamda = R ( 1/[tex]n_{f} ^{2}[/tex] - 1/[tex]n_{i} ^{2}[/tex] )
where , lamda = wavelength = 486 nm = 486 ×[tex]10^{-9} m[/tex]
[tex]n_{f}[/tex] in a hydrogen atom transition = final state = ?
[tex]n_{i}[/tex] in a hydrogen atom transition = initial state = 4
R = Rydberg constant = 1.097 ×[tex]10^{7} m^{-1}[/tex]
1/ 486 ×[tex]10^{-9} m[/tex] = 1.097 ×[tex]10^{7} m^{-1}[/tex] ( 1/[tex]n_{f} ^{2}[/tex] - 1/[tex]4^{2}[/tex] )
[tex]n_{f}[/tex] = 2
Missing parts :
Determine the end (final) value of n in a hydrogen atom transition, if the electron starts in n=4 and the atom emits a photon of light with a wavelength of 486 nm.
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