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(D) For the Lyman series,the shortest wavelength occurs at $n_1 = 1$ and $n_2 = \infty$. Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left(

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$\frac{36A}{7}$

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(D) For the Lyman series,the shortest wavelength occurs at $n_1 = 1$ and $n_2 = \infty$. Using the Rydberg formula: $\frac{1}{\lambda} = R Z^2 \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$. For hydrogen $(Z=1)$: $\frac{1}{A} = R(1)^2 \left( \frac{1}{1^2} - \frac{1}{\infty^2} \right) = R$,so $R = \frac{1}{A}$. For the Paschen series,the longest wavelength occurs at the first line,where $n_1 = 3$ and $n_2 = 4$. For $He^{+}$ $(Z=2)$: $\frac{1}{\lambda} = R(2)^2 \left( \frac{1}{3^2} - \frac{1}{4^2} \right)$. $\frac{1}{\lambda} = 4R \left( \frac{1}{9} - \frac{1}{16} \right) = 4R \left( \frac{16-9}{144} \right) = 4R \left( \frac{7}{144} \right) = \frac{7R}{36}$. Substituting $R = \frac{1}{A}$: $\frac{1}{\lambda} = \frac{7}{36A}$. Therefore,$\lambda = \frac{36A}{7}$.

If the shortest wavelength in the Lyman series of a hydrogen atom is $A$,then the longest wavelength in the Paschen series of $He^{+}$ is:

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