A doubly ionised lithium atom is hydrogen-like with atomic number 3:
(i) Find the wavelength of the radiation required to excite the electron in Li⁺⁺ from the first to the third Bohr orbit. (Ionisation energy of the hydrogen atom equals 13.6 eV.)
(ii) How many spectral lines are observed in the emission spectrum of the above excited system?

Solution

Given, a doubly ionised lithium atom with atomic no. Z=3.

(i) The energy difference of electron in Li⁺⁺ between the first and the third orbit is,

= E_{3} - E_{1}

∴

E_{3} - E_{1} = 13.6 \times Z^{2} (\frac{1}{n_{1}^{2}} - \frac{1}{n_{2}^{2}})

= 13.6 \times (3)^{2} (\frac{1}{1^{2}} - \frac{1}{3^{2}}) = 13.6 \times 9 \times \frac{8}{9} \times 1.6 \times 10^{- 19} J

Therefore, the equivalent wavelength λ is given by equating difference in energy with energy of a photon.

E_{3} - E_{1} = \frac{hc}{λ}

\Rightarrow λ = \frac{hc}{E_{3} - E_{1}} = \frac{6.63 \times 10^{- 34} \times 3 \times 10^{8}}{13.6 \times 8 \times 1.6 \times 10^{- 19}} = 1.137 \times 10^{- 8} m = 113.7 Å .

(ii) The following three spectral lines are observed due to the following transitions:
3rd to 1st orbit
3rd to 2nd orbit
2nd to 1st orbit

Some More Questions From Wave Optics Chapter

For Daily Free Study Material Join wiredfaculty