Question

A doubly ionised lithium atom is hydrogen-like with atomic number 3.
(a) Find the wavelength of the radiation required to excite the electron in $L{i}^{2+}$ from the first to the third Bohr orbit. (Ionisation energy of the hydrogen atom equals 13.6 eV.)
(b) How many spectral lines are observed in the emission spectrum of the above excited system?

Answer 1

Given $Z=3:E_n\propto \frac{Z^2}{n^2}$
(a) To excite the atom from n=1 to n=3, energy of photon required is,
$E_{1-3}=E_3-E_1=\frac{(-13.6)(3{)}^2}{(3{)}^2}-[\frac{(-13.6)(3{)}^2}{(1{)}^1}$
=108.8eV
Corresponding wavelength will be,
$\lambda (in\mathring{A})=\frac{12375}{E(ineV)}=\frac{12375}{108.8}=113.74\mathring{A}$ (b) From $n^{th}$ orbit total number of emission lines can be
\frac{n(n-1)}{2}
$\therefore$ Number of emission lines $=\frac{3(3-1)}{2}=3$