In a hydrogen atom, a transition takes place from n = 3 to n = 2 orbit. Calculate the wavelength of the emitted photon. Will the photon be visible ? To which spectral series will this photon belong? Given `R =1.097 xx 10^(7) m^(-1)`
The wavelength of the emitted photon is given by
`(1)/(lambda)=R((1)/n_(1)^(2)-(1)/(3^(2)))`
When the transition takes place n =3 to n=2, then
`(1)/(lambda)=(1.097xx10^(7))((1)/(2^(2))-(1)/(3^(3)))`
`=1.097xx10^(7)xx(4)/(36)`
`:." " lambda=(36)/(1.097xx10^(7)xx5)=6.563xx10^(-7) m`
`=6563Å`
`lambda` falsl in the visible (red) part of the specturm, hence the photon will be visible. This photon is the first member of the Balmar series.