The Balmer series for the hydrogen atom corresponding to electronic transition that terminate in the state of quantum number `n = 2`. Find the longest-wavelength photons emitted and determine its enegry.
Strategy: The longest-wavelength photon is associated with the smallest enegry difference. Its emission in the Balmer series results from the transition form `n_(i) = 3`to `n_(f) = 2`.

Explanation

Step `1. bar(v) = (1)/(lambda) = R_(H) |((1)/(n_(f)^(2))-(1)/(n_(i)^(2)))|`
`(1)/(lambda_(max)) = R_(H) ((1)/(2^(2))-(1)/(3^(2)))| = (5)/(36).R_(H)`
`(1)/(lambda_(max)) = (36)/(5R_(H)) = (36)/(5(1.097xx10^(7)m^(-1)))`
`= 6.564 xx 10^(-7) m = 656.4 nm`
This wavelength lies in the red region of the visible spectrum.
Step `2.` The energy of this phton is given by
`E_("Photon") = hv = (hc)/(lambda_(max))`
`((6.6xx10^(-34)Js)(3.00xx10^(8)ms^(-1)))/((656.4xx10^(-9)m))`
`= 3.02 xx 10^(-19) J = 1.89 eV`