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 the first 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 , `E_(3)`= - 13.6eV. Thus the level is, E = `E_(3) - E_(1) `= - 13.6 - (-122.4) = 108.8eV
Therefore , the radiation needed to cause this transition should have photons of this energy. hv = 108.8 eV
The wavelength of this radiation is or
`lambda = (hc)/(108.8eV) = 114.25`Å