The wavelength of the spectral line obtained | vedclass

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(D) Let $n_1$ be the lower energy level and $n_2$ be the higher energy level.Given: $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$.Adding the two equations: $2n_

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$1.14 	imes 10^{-6} \ cm$

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(D) Let $n_1$ be the lower energy level and $n_2$ be the higher energy level.Given: $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$.Adding the two equations: $2n_2 = 6 \implies n_2 = 3$.Subtracting the two equations: $2n_1 = 2 \implies n_1 = 1$.Using Rydberg's formula: $\frac{1}{\lambda} = R_H Z^2 \left[ \frac{1}{n_1^2} - \frac{1}{n_2^2} ight]$.For $Li^{2+}$, $Z = 3$. Substituting the values: $\frac{1}{\lambda} = R_H (3)^2 \left[ \frac{1}{1^2} - \frac{1}{3^2} ight]$.$\frac{1}{\lambda} = 9 R_H \left[ 1 - \frac{1}{9} ight] = 9 R_H \left( \frac{8}{9} ight) = 8 R_H$.$\lambda = \frac{1}{8 R_H}$.Taking $R_H \approx 1.097 	imes 10^5 \ cm^{-1} \approx 1.1 	imes 10^5 \ cm^{-1}$.$\lambda = \frac{1}{8 	imes 1.1 	imes 10^5} \ cm = \frac{1}{8.8 	imes 10^5} \ cm \approx 0.1136 	imes 10^{-5} \ cm = 1.136 	imes 10^{-6} \ cm$.Thus, $\lambda \approx 1.14 	imes 10^{-6} \ cm$.

The wavelength of the spectral line obtained in the spectrum of $Li^{2+}$ ion, when the transition takes place between two levels whose sum is $4$ and difference is $2$, is

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