In the spectrum of Li^{2+},the difference bet | vedclass

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(A) Let the energy levels be $n_1$ and $n_2$ where $n_2 > n_1$. Given that $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$. Solving these equations,we get $n_2 =

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

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(A) Let the energy levels be $n_1$ and $n_2$ where $n_2 > n_1$. Given that $n_1 + n_2 = 4$ and $n_2 - n_1 = 2$. Solving these equations,we get $n_2 = 3$ and $n_1 = 1$. For $Li^{2+}$,the atomic number $Z = 3$. The Rydberg formula for the wavelength $\lambda$ is given by: $\frac{1}{\lambda} = R_H 	imes Z^2 	imes \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} ight)$. Substituting the values: $\frac{1}{\lambda} = 109678 	imes 3^2 	imes \left( \frac{1}{1^2} - \frac{1}{3^2} ight) = 109678 	imes 9 	imes \left( 1 - \frac{1}{9} ight) = 109678 	imes 9 	imes \frac{8}{9} = 109678 	imes 8 \ cm^{-1}$. $\lambda = \frac{1}{109678 	imes 8} \approx 1.14 	imes 10^{-6} \ cm$.

In the spectrum of $Li^{2+}$,the difference between two energy levels is $2$ and their sum is $4$. Find the wavelength of the photon emitted during the transition between these two energy states.

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