Question

The wave number of the first line of Balmer series of hydrogen is $15200{cm}^{-1}$. The wave number of the first line of Balmer series for $L{i}^{2+}$ ion is:

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

Answer 1

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

As we know,
$\overline{v}=\frac{1}{\lambda }=R_H\times Z^2[\frac{1}{n_1^2}-\frac{1}{n_2^2}]$

For 1st line of Balmer series, $n_1=2;n_2=3$

$\overline{v_H}=\frac{1}{{\lambda }_H}=R_H\times 1^2[\frac{1}{2^2}-\frac{1}{3^2}]$

For $Li$ ion,

$\overline{v_{L{i}^{2+}}}=\frac{1}{{\lambda }_{L{i}^{2+}}}=R_H\times 3^2[\frac{1}{2^2}-\frac{1}{3^2}]$

$\therefore \overline{v_{L{i}^{2+}}}=9\times \overline{v_H}=9\times 15200=136800c{m}^{-1}$