The wavelength of the first line of Balmer series in hydrogen atom is `6562.8Å`. Calculate ionisation potential of hydrogen and also, the wavelength of first line of Lyman series.

Explanation

Here, `lambda_1=6562.8Å=6562.8xx10^(-10)m`
`:. 1/(lambda_1)=R(1/(2^2)-1/(3^2))=Rxx5/36`
`R=36/(5lambda_1)=36/(5xx6562.8xx10^(-10))`
`R=1.0971xx10^7m^-1`
The energy of electron in nth energy state
`E_n=R(hc)/(n^2)`
`:.` Ionisation energy of Hydrongen atom
`E=E_1-E_(oo)=Rhc-0`
`=1.0971xx10^7xx6.6xx10^(-34)xx3xx10^8`
`=21.788xx10^(-19)J`
`=(21.78xx10^(-19))/(1.6xx10^(-19))eV=13.6eV`
`:.` Ionisation potential of hydrogen atom
`=13.6V`
The wavelength of first line of Lyman series is given by
`1/lambda=R(1/(1^2)-1/(2^2))=3/4R`
`lambda=4/(3R)=4/(3xx1.0971xx10^7)`
`=1.2153xx10^-7m=1215.3Å`