What is the wavelength of ligth for the least energetic photon emitted in the Lyman series of the hydrogen spectrum. (Take , hc =1240 eV -nm)

Explanation

KEY IDEAS
(1) For any series, the transition that produces the least energetic photon is the transition between the home base level that defines the series and the level immediately above it. (2) For the Lyman series, the home - base level is at n = 1. thus, the transition that produces the least energetic photon is the transition from n = 2 level to the n = 1 level.
Calculations : From Eq. 38-29 the energy difference is `DeltaE=E_(2)-E_(1)=-(13.60eV)((1)/(2^(2))-(1)/(1^(2)))=10.20eV`.
Then from Eq. 38-6 `(DeltaE=hf)`, with `c//lambda` replacing f, we have
`lambda=(hc)/(DeltaE)=((6.63xx10^(-34)J.s)(3.00xx10^(8)m//s))/((10.20eV)(1.60xx10^(-19)J//eV))`
`=1.22xx10^(-7)m=122nm.`
Light with this wavelength is in the ultraviolet range.