The last member of Lymann series of Hydrogen atom is `912Å` Calculate The wavelength of series limit of blamer series.

Explanation

Given : `lamda=912xx10^(-10)m`
for Lyman series, `n_(1)=1,n_(2)=2,3,4,…..oo`
for the member , the transition is form infinite to `n_(1)=1` state.
`:. 1/lamda=R(1/n_(1)^(2)-1/n_(2)^(2))` may be written as `1/(912xx10^(-10))=((1)/(1^(2)))" where " n_(2)=oo`
`:.R=1.096xx10^(7)m^(-1)`
For Balmer series `n_(1)=2,n_(2)=3,4,5,……oo`
Hence, `1/lamda=1.096xx10^(7)((1)/(2^(2)))=0.274xx10^(7)m^(-1)`
Series limit , `lamda_(L)=1/(0.274xx10^(7))`
i.e., `lamda_(L)=3.6496xx10^(-7)m`
or, `lamda_(L)=3649.6Å`