Question

Find the ionization energy of the hydrogen-like ion when wavelength of second spectral line in the Balmer series is equal to $1218 Å\left( R _{H}=1 \cdot 097 \times 10^{7} m ^{-1}\right)$.

Answer 1

SOLUTION — For $H$-like species

$\begin{array}{l}\frac{1}{\lambda}=R_{H} Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right) ; \quad \frac{1}{1218 \times 10^{-10}}=1.097 \times 10^{7} \times Z^{2}\left(\frac{1}{2^{2}}-\frac{1}{4^{2}}\right) \\\text { or, } \quad Z^{2}=\frac{16}{3 \times 1218 \times 1.097 \times 10^{7} \times 10^{-10}}=3.99 \approx 4 \quad \therefore \quad Z=2\end{array}$

The species is $He ^{+}$.

Ionization energy is equal to the energy of the first orbit with opposite sign.

$\therefore \quad I . E =13 \cdot 6 \times \frac{Z^{2}}{n^{2}} eV =13 \cdot 6 \times 2^{2} eV = 5 4 \cdot 4 eV \text {. }$