The difference between wavelengths of first lines of Balmer series and Lyman series is 59.3 nm for a hydrogen-like ion.

Question

The difference between wavelengths of first lines of Balmer series and Lyman series is $59 \cdot 3 nm$ for a hydrogen-like ion. Identify it. $\left(R_{H}=109678 cm ^{-1}\right)$

Answer 1

SOLUTION : We know that $\frac{1}{\lambda}=R_{H} Z^{2}\left(\frac{1}{n_{1}^{2}}-\frac{1}{n_{2}^{2}}\right)$

For first line of Lyman series $\frac{1}{\lambda_{1}}=R_{H} Z^{2}\left(1-\frac{1}{4}\right)$;

$\quad \therefore \lambda_{1}=\frac{4}{3 R_{H} Z^{2}} cm$ For first line of Balmer series $\frac{1}{\lambda_{2}}=R_{H} Z^{2}\left(\frac{1}{4}-\frac{1}{9}\right) ;$

$\therefore \lambda_{2}=\frac{36}{5 R_{H} Z^{2}} cm$ Given $\lambda_{2}-\lambda_{1}=59 \cdot 3 \times 10^{-7} cm , \quad \therefore \frac{1}{R_{H} Z^{2}}\left(\frac{36}{5}-\frac{4}{3}\right)=59 \cdot 3 \times 10^{-7}$ or, $\quad \frac{1}{R_{H} Z^{2}} \times \frac{88}{15}=59 \cdot 3 \times 10^{-7}$

$\therefore \quad Z^{2}=\frac{88}{15 \times 59 \cdot 3 \times 10^{-7} \times 109678}=9 \cdot 0, \quad \therefore \quad Z=3 \cdot 0$

The $H$-like ion is $L i ^{2+}$.