Using the Rydberg formula, calculate the wavelengths of the first four spectral lines in the Lyman series of the hydrogen spectrum.

Explanation

From Rydberg formula,
`(h_(c))/(lambda_(if))=(me^(4))/(8 epsi_(0)^(2)h^(2))[(1)/(n_(f)^(2))-(1)/(n_(f)^(2))]=21.76xx10^(-19)[(n_(i)^(2)-1)/(m_(f)^(2))]`
`:. lambda_(if)=(h_(c))/(21.76xx10^(-19))[(n_(i)^(2))/(n_(i)^(2)-1)]`
`:. Lambda_(if)=(6.625xx10^(-34)xx3xx10^(8))/(21.76xx10^(-19))xx[(n_(i)^(2))/(n_(i)^(2)-1)]`
`:. lambda_(if)=(n_(i)^(2)xx913.4xx10^(-10)m)/(n_(i)^(2)-1)....(1)`
`rArr` Now putting `n_(i)=2` in equation, (1), we get the wavelenght of `alpha`- line of Lyman series.
`:. lambda_(21)=((2)^(2)xx913.4xx10^(-10))/((2)^(2)-1)`
`=(4xx913.4xx10^(-10))/(3)`
`=1217.8xx10^(-10)m=1218Å`
`rArr` Putting `n_(i)=3` in equation (1), the wavelength of `beta`- line of Lyman series,
`:. lambda_(31)=((3)^(2)xx931.4xx10^(-10))/((3)^(2)-1)`
`:. lambda_(31)=(9xx931.4xx10^(-10))/(8)`
`:. lambda_(31)=1027.5xx10^(-10)m=1028Å`
`rArr` Putting `n_(i)=4` in equation (1), the wavelength of `gamma`- line of Lyman series,
`:. lambda_(41)=((4)^(2)931.4xx10^(-10))/((4)^(2)-1)`
`=(16xx913.4xx10^(-10))/(15)`
`:. lambda_(41)=674.29xx10^(-10)m=974.3Å`
`rArr` Putting `n_(i)=5` in equation (1), the wavelength of `delta`- line of Lyman series
`:. lambda_(51)=((5)^(2)931.4xx10^(-10))/((5)^(2)-1)`
`=(25xx913.4xx10^(-10))/(24)`
`:. lambda_(51)= 951.45xx10^(-10)m=951.4Å`