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

The energy of `n^(th)` orbit of a hydrogen -like atom is given as` E_(n) =-(13.6Z^(2))/n^(2)` Thus for `Li^(2+)` atom , as `Z= 3`, the electron energies of the first and third Bohr orbits are For `n =1,E_(1) =- 122.4eV`, for `n=3`,
an elecron from `E_(1)` level to `E_(3)` level to `E-(3)` level is `E=E_(3) -E_(1)`
` =-13.6-(-122.4)=108.8e`. Therefore, the radiation needed to cause this transition should have photons of this energy. `hv = 108.8 eV`. The wavelengtth of this radiation is or `=lambda=(hc)/(108.8eV)= 114.25Å`