A doubly ionized lithium atom is hydrogen like with atomic number 3. Find the wavelength of the radiation required to excite the electron in `Li^(++)` from to the third Bohr orbit (ionization energy of the hydrogen atom equals 13.6 eV).

Explanation

We know, `E_(n)=(Z^(2))/(n^(2))(13.6 eV)`
By putting Z=3 (given), we have
`E_(n)=-(122.4)/(n^(2)) eV`
As, excitation is from Ist to 3rd Bohr orbit
`i.e.,n_(2)=3`.
`rArr" " E_(1)=-(122.4)/((1)^(2))=-122.4 eV`
and `" " E_(3)(122.4)/((3)^(2))=-13.6 eV`
`:. " " DeltaE=E_(3)-E_(1)=108.8 eV`
The corresponding wavelength is ,
`lambda=(12375)/(DeltaE("in"eV))Å=(12375)/(108.8)Å=113.74 Å`