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

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Question

(C) The Rydberg formula for wave number is $\bar{
u} = R_H Z^2 [\frac{1}{n_1^2} - \frac{1}{n_2^2}]$.For the first line of the Balmer series of $H$ at

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$738720$

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(C) The Rydberg formula for wave number is $\bar{
u} = R_H Z^2 [\frac{1}{n_1^2} - \frac{1}{n_2^2}]$.For the first line of the Balmer series of $H$ atom $(Z=1, n_1=2, n_2=3)$: $\bar{
u}_1 = R_H (1)^2 [\frac{1}{2^2} - \frac{1}{3^2}] = R_H [\frac{1}{4} - \frac{1}{9}] = R_H (\frac{5}{36}) = 15200 \ cm^{-1}$.For the first line of the Lyman series of $Li^{2+}$ $(Z=3, n_1=1, n_2=2)$: $\bar{
u}_2 = R_H (3)^2 [\frac{1}{1^2} - \frac{1}{2^2}] = R_H (9) [1 - \frac{1}{4}] = R_H (9) (\frac{3}{4}) = R_H (\frac{27}{4})$.Taking the ratio: $\frac{\bar{
u}_2}{15200} = \frac{R_H (27/4)}{R_H (5/36)} = \frac{27}{4} 	imes \frac{36}{5} = \frac{27 	imes 9}{5} = \frac{243}{5} = 48.6$.Therefore,$\bar{
u}_2 = 15200 	imes 48.6 = 738720 \ cm^{-1}$.

The wave number of the first line of the Balmer series of $H$ atom is $15200 \ cm^{-1}$. What is the wave number of the first line of the Lyman series of $Li^{2+}$? (in $cm^{-1}$)

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