Wavelength Ranges and Overlap in Hydrogen Spectrum

To determine the correct statement(s) regarding the emission spectrum of a hydrogen atom, we use the Rydberg formula for the wavelength $\lambda$ of a spectral line corresponding to a transition from principal quantum number $n_2$ to $n_1$ ($n_2 > n_1$):

$\frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right)$

where $R$ is the Rydberg constant.

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Analysis of Option A:

For the Balmer series ($n_1 = 2$):

1. Shortest wavelength ($\lambda_{\text{min}}$): Occurs for the transition from $n_2 = \infty$: $\frac{1}{\lambda_{\text{min}}} = R \left( \frac{1}{2^2} - \frac{1}{\infty^2} \right) = \frac{R}{4} \implies \lambda_{\text{min}} = \frac{4}{R}$

2. Longest wavelength ($\lambda_{\text{max}}$): Occurs for the transition from $n_2 = 3$: $\frac{1}{\lambda_{\text{max}}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = \frac{5R}{36} \implies \lambda_{\text{max}} = \frac{36}{5R}$

3. Ratio of longest to shortest wavelength: $\frac{\lambda_{\text{max}}}{\lambda_{\text{min}}} = \frac{\frac{36}{5R}}{\frac{4}{R}} = \frac{36}{20} = \frac{9}{5}$

Thus, Option A is correct.

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Analysis of Option B:

To check if the Balmer and Paschen series overlap, we compare their wavelength ranges:

• Balmer Series ($n_1 = 2$): $\lambda_{\text{min, Balmer}} = \frac{4}{R}$ $\lambda_{\text{max, Balmer}} = \frac{36}{5R} = \frac{7.2}{R}$

• Paschen Series ($n_1 = 3$): $\frac{1}{\lambda_{\text{min, Paschen}}} = R \left( \frac{1}{3^2} - \frac{1}{\infty^2} \right) = \frac{R}{9} \implies \lambda_{\text{min, Paschen}} = \frac{9}{R}$ $\frac{1}{\lambda_{\text{max, Paschen}}} = R \left( \frac{1}{3^2} - \frac{1}{4^2} \right) = R \left( \frac{1}{9} - \frac{1}{16} \right) = \frac{7R}{144} \implies \lambda_{\text{max, Paschen}} = \frac{144}{7R}$

Since $\lambda_{\text{max, Balmer}} = \frac{7.2}{R} < \lambda_{\text{min, Paschen}} = \frac{9}{R}$, the wavelength range of the Balmer series ends before the Paschen series begins. Therefore, there is no overlap.

Thus, Option B is incorrect.

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Analysis of Option C:

For the Lyman series ($n_1 = 1$), the shortest wavelength ($\lambda_0$) corresponds to $n_2 = \infty$: $\frac{1}{\lambda_0} = R \left( \frac{1}{1^2} - 0 \right) = R \implies \lambda_0 = \frac{1}{R}$

The general expression for the wavelength $\lambda$ of a line in the Lyman series ($n_2 = n \ge 2$) is: $\frac{1}{\lambda} = R \left( 1 - \frac{1}{n^2} \right) = \frac{1}{\lambda_0} \left( \frac{n^2 - 1}{n^2} \right)$ $\implies \lambda = \lambda_0 \left( \frac{n^2}{n^2 - 1} \right) = \lambda_0 \left( 1 + \frac{1}{n^2 - 1} \right)$

For $\lambda$ to be expressed in the form $\left( 1 + \frac{1}{m^2} \right)\lambda_0$ where $m$ is an integer, $n^2 - 1$ must be a perfect square $m^2$. However, $n^2 - 1$ is never a perfect square for any integer $n \ge 2$ (since the difference between consecutive squares $n^2$ and $(n-1)^2$ is greater than $1$).

Thus, Option C is incorrect.

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Analysis of Option D:

To check if the Lyman and Balmer series overlap:

• Lyman Series ($n_1 = 1$): $\lambda_{\text{min, Lyman}} = \lambda_0 = \frac{1}{R}$ $\frac{1}{\lambda_{\text{max, Lyman}}} = R \left( 1 - \frac{1}{2^2} \right) = \frac{3R}{4} \implies \lambda_{\text{max, Lyman}} = \frac{4}{3R}$

• Balmer Series ($n_1 = 2$): $\lambda_{\text{min, Balmer}} = \frac{4}{R}$

Since $\lambda_{\text{max, Lyman}} = \frac{4}{3R} < \lambda_{\text{min, Balmer}} = \frac{4}{R}$, the wavelength ranges of the Lyman and Balmer series do not overlap.

Thus, Option D is correct.

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Conclusion:

The correct statements are A and D.