Question

The wave number of the first line of Balmer series of H-atom is 15200 ${cm}^{-1}$. The wave number of first Balmer line of ${Li}^{2+}$ ion is:

• A. ${15200cm}^{-1}$
• B. ${60800cm}^{-1}$
• C. ${76000cm}^{-1}$
• D. ${136800cm}^{-1}$

Answer 1

The correct option is D ${136800cm}^{-1}$

We know that,
$\frac{1}{\lambda }\propto z^2$

$\nu \propto z^2$

$\frac{{\nu }_1}{{\nu }_2}=\frac{{Z_1}^2}{{Z_2}^2}$

$\therefore$ Wavenumber of the first line of Balmer series of $L{i}^{2+}=9\times$ wave number of the first line of Balmer series of $H$.

$=9\times 15200=136800c{m}^{-1}$

Hence, option D is correct.