Fig. shows energy level diagram of hydrogen atom. Find out the transition which results in the emission of a photon of wavelength 496 nm. Which transition corresponds to emission of radiation of maximum wavelength? Justify your answer.

Answer

`E_(4)to E_(2) : 4 to 3`

Explanation

Energy levels of hydrogen atom are
`E_(1)=-13.6eV, E_(2)=-3.4eV`
`E_(3)=-1.51eV, E_(4)=-0.85eV`
Energy of a photon of wavelength 496nm is
`E=(hc)/lambda=(6.63xx10^(-34)xx3xx10^(8))/(496xx10^(-9)(1.6xx10^(-19)))eV=2.5eV`
As `E_(4)-E_(2)=-0.85+3.4=2.55eV=E`, therefore, the transition form n=4 to n=2 level results in the emission of photon of wavelength 496 nm. Further, wavelength emitted will be maximum, when energy emitted is minimum. The transition n=4 to n=3 level will give maximum wavelength as energy emitted is minimum.