Hydrogen Spectrum: Wavelength Calculations

This solution addresses the calculation of the difference between the largest wavelengths of the Paschen and Balmer series for hydrogen, referencing the provided Lyman series data.

Understanding Spectral Series

The hydrogen spectrum is categorized into series based on the final energy level ($n_1$) of electron transitions:

• Lyman series: Transitions ending at the ground state ($n_1 = 1$).
• Balmer series: Transitions ending at the first excited state ($n_1 = 2$).
• Paschen series: Transitions ending at the second excited state ($n_1 = 3$).

Rydberg Formula Application

The wavelength ($\lambda$) of emitted light is determined by the Rydberg formula:

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

For hydrogen, the atomic number $Z=1$. $R$ is the Rydberg constant. $n_2$ is the principal quantum number of the initial energy level, and $n_1$ is the principal quantum number of the final energy level ($n_2 > n_1$).

Largest Wavelength Calculation

The largest wavelength ($\lambda_{\text{max}}$) in any series corresponds to the smallest energy difference between levels. This occurs for the transition where the electron drops from the level immediately above the final level, i.e., $n_2 = n_1 + 1$.

• Balmer series ($n_1=2$): The largest wavelength is for the transition $n_2=3 \to n_1=2$. $ \frac{1}{\lambda_{\text{max, Balmer}}} = R \left( \frac{1}{2^2} - \frac{1}{3^2} \right) = R \left( \frac{1}{4} - \frac{1}{9} \right) = R \left( \frac{5}{36} \right) $ $ \lambda_{\text{max, Balmer}} = \frac{36}{5R} $
• Paschen series ($n_1=3$): The largest wavelength is for the transition $n_2=4 \to n_1=3$. $ \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) = R \left( \frac{7}{144} \right) $ $ \lambda_{\text{max, Paschen}} = \frac{144}{7R} $

Wavelength Difference Derivation

The difference between the largest wavelengths of the Paschen and Balmer series is:

$ \Delta \lambda = \lambda_{\text{max, Paschen}} - \lambda_{\text{max, Balmer}} = \frac{144}{7R} - \frac{36}{5R} $

To simplify, find a common denominator:

$ \Delta \lambda = \frac{1}{R} \left( \frac{144 \times 5 - 36 \times 7}{35} \right) = \frac{1}{R} \left( \frac{720 - 252}{35} \right) = \frac{468}{35R} $

Data Interpretation and Result Analysis

The question provides the smallest wavelength of the Lyman series ($\lambda_{\text{min, Lyman}}$) as 91 nm. This corresponds to the transition $n_2=2 \to n_1=1$.

$ \frac{1}{\lambda_{\text{min, Lyman}}} = R \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = R \left( 1 - \frac{1}{4} \right) = R \left( \frac{3}{4} \right) $

Using the given value $\lambda_{\text{min, Lyman}} = 91 \, \text{nm}$:

$ \frac{1}{91 \, \text{nm}} = R \left( \frac{3}{4} \right) $

This implies the value of $1/R$ is:

$ \frac{1}{R} = 91 \, \text{nm} \times \frac{4}{3} = \frac{364}{3} \, \text{nm} \approx 121.33 \, \text{nm} $

Substituting this into the difference formula:

$ \Delta \lambda = \frac{468}{35} \times \left( \frac{364}{3} \right) \, \text{nm} \approx 1621 \, \text{nm} $

Note on Discrepancy: The calculation derived directly from the provided 91 nm value yields approximately 1621 nm. Standard calculations using the widely accepted Rydberg constant value result in a difference of approximately 1219 nm (closest to Option 2). The provided correct answer is Option D (1550 nm). Due to these significant inconsistencies between the given data, standard physical constants, the options, and the designated answer, a definitive step-by-step derivation reaching 1550 nm is not possible under standard interpretation.