(a) A hydrogen atom is excited by radiation of wavelength 97.5 nm. Find the principal quantum number of the excited state.
(b) Show that the total number of lines in emission spectrum is `(n(n-1))/2` and compute the total number of possible lines in emission spectrum.

Explanation

(a) Wave length of radiation `lamda=97.5` (or)
`975Å=975xx10^(-10)m`
Energy associated with it,
`DeltaE=12400/(lamdaA)eV`
`DeltaE=12400/975=12.71eV`
We know enrgy associated `DeltaE=13.6(1/n_(1)^(2)-1/n_(2)^(2))...(1)`
Where `DeltaE=12.7eV`
`n_(1)=1" "[therefore"the "H_(2)" atom is in the ground stae"]`
Sub `n_(1)=1,n_(2)=n&n&DeltaE=12.7eV" in "(1)`
`12.71=13.6(1/1^(2)-1/n^(2))rArr12.71/13.6=1-1/n^(2)`
`rArr12.71/13.6-1=1/n^(2)rArr1/n^(2)=0.89/13.6`
`n^(2)=13.6/0.89rArrn=sqrt15.28`
`n=3.9~~4`
(b) Hence total number of spectral lines
= `(n(n-1))/2=(4(4-1))/2=4xx3/2=6`
6 possible transitions.