Question

A doubly ionized lithium ion is hydrogen atom-like. Find wavelength of the radiation required to excite the electron in $Li ^{2+}$ from first orbit to third orbit. How many spectral lines are observed when the electron in the third orbit deexcites? Ionization energy of H-atom $=13 \cdot 6 eV$.

Answer 1

SOLUTION : For hydrogen-like ions, energy $E_{n}=-13 \cdot 6 \times \frac{Z^{2}}{n^{2}} eV$

When $\quad n=1, E_{1}=-9 \times 13 \cdot 6 eV$, here $Z=3$

When $n=3, E_{3}=-13 \cdot 6 \times \frac{9}{9} \quad eV =-13 \cdot 6 eV$

$$\begin{aligned}\Delta E &=E_{3}-E_{1}=+13 \cdot 6(9-1) eV =13 \cdot 6 \times 8 eV \\&=13 \cdot 6 \times 1 \cdot 60 \times 10^{-19} \times 8 J\end{aligned}$$

By Planck's equation

$$\begin{aligned}\Delta E &=\frac{h c}{i} \\\therefore \quad \lambda &=\frac{h c}{\Delta E}=\frac{6 \cdot 626 \times 10^{-34} \times 3 \times 10^{8}}{13 \cdot 6 \times 8 \times 1 \cdot 60 \times 10^{-19}} m =0 \cdot 114 \times 10^{-7} m =11 \cdot 4 n m\end{aligned}$$

The no. of spectral lines obtained by deexcitation from $n^{\text {th }}$ orbit $=n(n-1) / 2$ Here $n=3$, no. of spectral lines $=3 \times 2 / 2=3$.