If shortest wavelength of H-atom in Balmer series is X then
(i) What is the shortest wave length in Lyman series.
(ii) What is the longest wave length in Paschen series.

Explanation

(i) `(1)/(lambda)=Z^(2)R_(H)[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
`lambda=[(1)/(Z^(2)R_(H))][(n_(1)^(2)n_(1)^(2))/(n_(2)^(2)-n_(1)^(2))]`
Shortest wavelength correspond to maximum energy `(E=(hc)/(lambda))` hence transition must be `n_(2)=oo, n_(1)=2`
`=[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]=((1)/(4))=(n_(1)^(2)n_(2)^(2))/(n_(2)^(2)-n_(1)^(2))=4`
`x=((1)/(Z^(2)R_(H)))4=(4)/(R_(H))implies R_(H)=(4)/(x)`
(a) For shortest wavelength Lyman series
`n_(1)=1,n_(2)=oo`
`lambda=((1)/(R_(H)))implieslambda=(x)/(4)`
(b) For longest wavelength in Paschen series
`n_(1)=3,n_(2)=4`
`implies (n_(2)^(2)n_(1)^(2))/(n_(2)^(2)-n_(1)^(2))=(144)/(16-9)=[(144)/(7)]=lambda=((1)/(R_(H)))xx(144)/(7)=(x x 144)/(4xx7)=(36x)/(7)`