Question

The wave number of first line of Balmer series of hydrogen atom is 15200 $c{m}^{-1}$. What is the wave number of first line of Balmer series of $L{i}^{2+}$ ion?

• A. $15200$ $c{m}^{-1}$
• B. $13680000$ $m^{-1}$
• C. $76000$ $c{m}^{-1}$
• D. $13680$ $c{m}^{-1}$

Answer 1

The correct option is B $13680000$ $m^{-1}$

The wave number can be given as follow:

$\overline{v}=\frac{1}{\lambda }=R{Z}^2$ $[\frac{1}{n_2^2}-\frac{1}{n_1^2}]$

$\overline{v}\propto Z^2$

Thus wave number is directly proportional to the square of the atomic number.
Thus, $\frac{\overline{v_1}}{\overline{v_2}}=\frac{Z_1^2}{Z_2^2}=\frac{1}{9}$

$\frac{15200}{x}=\frac{1}{9}$
$x=9x15200$

$x=136800$ $c{m}^{-1}$
Option B is correct.