The wavenumber of the first spectral line of | vedclass

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Question

(A) The Rydberg formula for wavenumber is $\bar{
u} = R_H Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the first spectral line of the Ly

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$\frac{9 x}{16}$

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(A) The Rydberg formula for wavenumber is $\bar{
u} = R_H Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For the first spectral line of the Lyman series of $He^{+}$ $(Z=2)$: $n_1=1, n_2=2$.$\bar{
u}_1 = R_H 	imes 2^2 	imes \left[ \frac{1}{1^2} - \frac{1}{2^2} ight] = R_H 	imes 4 	imes \frac{3}{4} = 3 R_H = x$.Thus, $R_H = \frac{x}{3}$.For the second spectral line of the Balmer series of $Li^{2+}$ $(Z=3)$: $n_1=2, n_2=4$.$\bar{
u}_2 = R_H 	imes 3^2 	imes \left[ \frac{1}{2^2} - \frac{1}{4^2} ight] = R_H 	imes 9 	imes \left[ \frac{1}{4} - \frac{1}{16} ight] = R_H 	imes 9 	imes \frac{3}{16} = \frac{27 R_H}{16}$.Substituting $R_H = \frac{x}{3}$:$\bar{
u}_2 = \frac{27}{16} 	imes \frac{x}{3} = \frac{9 x}{16}$.

The wavenumber of the first spectral line of the Lyman series of $He^{+}$ ion is $x \ m^{-1}$. What is the wavenumber (in $m^{-1}$) of the second spectral line of the Balmer series of $Li^{2+}$ ion?

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