Calculate the shortest and longest wavelength in Lyman series of hydrogen spectrum.

Explanation

For Lyman series, `n_(1)=1 and n_(2)=2,3,4, . . .oo`
Rydberg equation, `overline(v)=R[(1)/(n_(1)^(2))-(1)/(n_(2)^(2))]`
The difference between `n_(1) and n_(2)` is maximum for the shortest wavelength in Lyman series, hence `n_(2)=oo`
`therefore (1)/(lamda_("min"))=overline(v_(max))=R[(1)/(1^(2))-(1)/(oo^(2))]=R=109678cm^(-1)`
`therefore lamda_("min")=(1)/(109678)=9.117xx10^(-6)cm=911.7` Å
Similarly, the difference betweenn `n_(1) and n_(2)` is minimum for the longest wavelength in Lyman series. hence, `n_(2)=2`
`therefore (1)/(lamda_(max))=overline(v_("min"))=R[(1)/(1^(2))-(1)/(2^(2))]=(3)/(4)R`
`therefore lamda_(max)=(4)/(3R)=(4)/(3xx109678)`
`=1215.7xx10^(-8)cm=1215.7` Å