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(A) For the Lyman series of the hydrogen atom $(Z=1)$,the shortest wavelength occurs for the transition from $n_2 = \infty$ to $n_1 = 1$. Using the Ry

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$\frac{9x}{5}$

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(A) For the Lyman series of the hydrogen atom $(Z=1)$,the shortest wavelength occurs for the transition from $n_2 = \infty$ to $n_1 = 1$. Using the Rydberg formula: $1/\lambda = R Z^2 (1/n_1^2 - 1/n_2^2)$. $1/x = R(1)^2 (1/1^2 - 1/\infty^2) = R$. Therefore,$x = 1/R$. For the Balmer series of $He^+$ $(Z=2)$,the longest wavelength occurs for the transition from $n_2 = 3$ to $n_1 = 2$. $1/\lambda' = R Z^2 (1/n_1^2 - 1/n_2^2) = R(2)^2 (1/2^2 - 1/3^2)$. $1/\lambda' = 4R (1/4 - 1/9) = 4R (5/36) = 5R/9$. Substituting $R = 1/x$: $1/\lambda' = 5/(9x)$. Thus,$\lambda' = 9x/5$.

If the shortest wavelength of the hydrogen atom in the Lyman series is $x$,then the longest wavelength in the Balmer series of $He^+$ is:

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