# Energy vs wavelength relation is given by, E ( e V ) = 1241 n m

Energy vs wavelength relation is given by, $E\left(eV\right)=\frac{1241}{nm}$. Energy of H-atom in the ground state is -13.6eV as form the relation $E\left(eV\right)=-13.6\frac{{Z}_{eff}^{2}}{{n}^{2}}\left(eV\right)$. Find that in which transition on hydrogen atom is the wavelength 486.1 nm produced? To which seriesdoes it belong ?
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Kaylin Barry
Solution:
The energy corresponding to the given wavelength is calculated as;

This energy corresponds to the beta Balmer series. This spectrum occurs when there is a transition from n=4 to n=2. Thus, this wavelength corresponds to the Balmer series.