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 energy required to transfer an electron from `E_(1)` level to `E_(3)` 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)=114.25Å`