Energy level diagram of a hydrogen atom is shown below :

(i)If a transition takes place from n=3 to n=2 orbit, calculate the wavelength of the radiation emitted. Identify the spectral series to which it belongs.
(ii)Which transition results in emission of radiation of maximum wavelength ?

Explanation

(i)For n=2 ,
Energy level is , `E_2=E/2^2=(-13.6)/4`=-3.4eV
For n=3,
Energy level is , `E_3=E/3^2=-13.6/9`=-1.51 eV
For transition from n=3 to n=2 , we have
`E_3-E_2=(hc)/lambda`
where `lambda` is the wavelength of the photon emitted
`therefore lambda=(hc)/(E_3-E_2)=(6.63xx10^(-34)xx3xx10^8)/(1.89xx1.6xx10^(-19))`
`=6.577xx10^(-7)` m = 6577Å
This wavelength belongs to Balmer series of hydrogen spectrum.
(ii)For transition from n=4 to n=3 level , energy difference is minimum amongst all the transitions.
As wavelength emitted `prop 1/"energy"`
Maximum wavelength will be emitted for this transition.