Understanding Hydrogen Spectral Series and Wavelengths

The question asks us to arrange the Lyman, Balmer, Paschen, and Brackett series of the hydrogen atom in order of increasing wavelength. These spectral series are characterized by the principal quantum number of the electron's final energy level ($n_f$) after it transitions from a higher energy level ($n_i$).

The Rydberg Formula

The relationship between the wavelength of emitted photons and the electron transitions in a hydrogen atom is described by the Rydberg formula:

$ \frac{1}{\lambda} = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) $

Here, $\lambda$ is the wavelength of the spectral line, $R_H$ is the Rydberg constant, $n_f$ is the principal quantum number of the final state, and $n_i$ is the principal quantum number of the initial state ($n_i > n_f$). The energy of the emitted photon is inversely proportional to its wavelength ($E = \frac{hc}{\lambda}$), meaning larger energy differences correspond to shorter wavelengths.

Defining the Spectral Series

The specified series correspond to the following final energy levels ($n_f$):

• Lyman series (A): Transitions ending at $n_f = 1$.
• Balmer series (B): Transitions ending at $n_f = 2$.
• Paschen series (C): Transitions ending at $n_f = 3$.
• Brackett series (D): Transitions ending at $n_f = 4$.

Determining Order by Series Limits

To determine the general trend of wavelengths for these series, we can examine the series limits. The series limit represents the shortest possible wavelength within a series, which occurs when the initial energy level ($n_i$) approaches infinity ($n_i \to \infty$). This corresponds to the largest possible energy difference for transitions ending at a given $n_f$.

Using the Rydberg formula for $n_i \to \infty$:

$ \frac{1}{\lambda_{\text{limit}}} = R_H \left( \frac{1}{n_f^2} - \frac{1}{\infty^2} \right) = R_H \left( \frac{1}{n_f^2} - 0 \right) = \frac{R_H}{n_f^2} $

Let's calculate the series limit wavelength for each series:

• Lyman series (A, $n_f=1$): $ \frac{1}{\lambda_A} = \frac{R_H}{1^2} = R_H $ $ \lambda_A = \frac{1}{R_H} $
• Balmer series (B, $n_f=2$): $ \frac{1}{\lambda_B} = \frac{R_H}{2^2} = \frac{R_H}{4} $ $ \lambda_B = \frac{4}{R_H} $
• Paschen series (C, $n_f=3$): $ \frac{1}{\lambda_C} = \frac{R_H}{3^2} = \frac{R_H}{9} $ $ \lambda_C = \frac{9}{R_H} $
• Brackett series (D, $n_f=4$): $ \frac{1}{\lambda_D} = \frac{R_H}{4^2} = \frac{R_H}{16} $ $ \lambda_D = \frac{16}{R_H} $

Comparing Wavelengths for Increasing Order

Comparing the series limit wavelengths calculated above:

$ \lambda_A = \frac{1}{R_H}, \quad \lambda_B = \frac{4}{R_H}, \quad \lambda_C = \frac{9}{R_H}, \quad \lambda_D = \frac{16}{R_H} $

Since $1 < 4 < 9 < 16$, it follows that:

$ \frac{1}{R_H} < \frac{4}{R_H} < \frac{9}{R_H} < \frac{16}{R_H} $

Therefore, the order of increasing wavelength is:

$ \lambda_A < \lambda_B < \lambda_C < \lambda_D $

This corresponds to the order:

(A) Lyman, (B) Balmer, (C) Paschen, (D) Brackett

Conclusion

The correct arrangement of the Lyman, Balmer, Paschen, and Brackett series in order of increasing wavelength, based on their series limits, is (A), (B), (C), (D).