The wave number of the first line of the Balm | vedclass

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Question

(C) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^

Vedclass · Accepted Answer

$136800$

Vedclass · Answer

(C) The wave number $\bar{
u}$ for a hydrogen-like species is given by the Rydberg formula: $\bar{
u} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$.For the Balmer series,$n_1 = 2$ and for the first line,$n_2 = 3$.For hydrogen $(H)$,$Z = 1$,so $\bar{
u}_H = R(1)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight) = 15200 \, cm^{-1}$.For $Li^{2+}$ ion,$Z = 3$. The wave number for the first line of the Balmer series is $\bar{
u}_{Li^{2+}} = R(3)^2 \left( \frac{1}{2^2} - \frac{1}{3^2} ight)$.Substituting the value of $R \left( \frac{1}{2^2} - \frac{1}{3^2} ight)$ from the hydrogen equation:$\bar{
u}_{Li^{2+}} = 3^2 	imes \bar{
u}_H = 9 	imes 15200 = 136800 \, cm^{-1}$.

The wave number of the first line of the Balmer series in the hydrogen spectrum is $15200 \, cm^{-1}$. What will be the wave number of the first line of the Balmer series for the $Li^{2+}$ ion in $cm^{-1}$?

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