Determine the wavelength of the first Lymanline,the transition from `n=2"to"n=1`.In what region of the electromagnetic spectrum does this line lie ?

Explanation

Strategy : We use , `hf=E_u-E-l`, with the energies obtained from energy level diagram to find the energy and the wavelength of the transition. The region of the electromagnetic spectrum is found using the electromagnetic spectrum.
In this case , `hf=E_2-E_1`
={-3.4 eV -(-13.6 eV)}
=10.2 eV
`=(10.2 eV) (1.60xx10^(-19) J//eV)=1.63xx10^(-18)` J
Since `lambda`=c/f, we have
`lambda=c/f="hc"/(E_2-E_1)`
`=((6.63xx10^(-34) J.s)(3.00xx10^8 m//s))/(1.63xx10^(-18)J)`
`=1.22xx10^(-7)` m
or 122 nm , which is in the UV region of the electromagnetic spectrum
`lambda=1/((2.44xx10^6 m^(-1))=4.10xx10^(-7)`m or 410 nm. This is the fourth line in the Balmer series and is violet in color.