Problem 15 - 43 Years JEE ADVANCED (1978-2020) + JEE MAIN Chapterwise & Topicwise Solved Papers Chemistry

Question

If the shortest wavelength in Lyman series of hydrogen atom is \(\mathrm{A}\), then the longest wavelength in Paschen series of \(\mathrm{He}^{+}\)is : (a) \(\frac{5 \mathrm{~A}}{9}\) (b) \(\frac{9 \mathrm{~A}}{5}\) (c) \(\frac{36 \mathrm{~A}}{5}\) (d) \(\frac{36 \mathrm{~A}}{7}\)

Step-by-Step Solution

Answer

(d) \( \frac{36 \mathrm{~A}}{7} \)

Step 1: Understanding the Lyman Series

The Lyman series corresponds to electron transitions in a hydrogen atom where the final energy level is n=1. The shortest wavelength occurs when the transition is from the highest possible energy level, which theoretically is from n=∞ to n=1. The formula for the shortest wavelength \( \lambda \) in terms of Rydberg's constant \( R_H \) is given by: \( \frac{1}{\lambda} = R_H (1^2 - \frac{1}{\infty^2}) = R_H \). Therefore, \( \lambda = \frac{1}{R_H} \).

Step 2: Applying to the Paschen Series

For the Paschen series of \( \mathrm{He}^{+} \), electron transitions end at n=3. The longest wavelength occurs for the transition from n=4 to n=3. This can be calculated using the modified Rydberg formula: \( \frac{1}{\lambda_{P}} = R_{He} ( \frac{1}{3^2} - \frac{1}{4^2} ) \), where \( R_{He} = Z^2 R_H \) and \( Z \) is the atomic number of helium, which is 2, therefore \( R_{He} = 4R_H \).

Step 3: Calculate the Wavelength

Substitute the values into the formula for \( \lambda_{P} \): \[ \frac{1}{\lambda_P} = 4R_H ( \frac{1}{9} - \frac{1}{16} ) \] This simplifies to: \[ \frac{1}{\lambda_P} = 4R_H \left( \frac{16 - 9}{144} \right) = 4R_H \left( \frac{7}{144} \right) = \frac{28}{144} R_H \]Thus, \[ \lambda_P = \frac{144}{28} \times \frac{1}{R_H} \] Substituting \( \lambda = \frac{1}{R_H} \) from the Lyman series gives: \[ \lambda_P = \frac{144}{28} A \].

Step 4: Simplifying the Expression

Simplify \( \frac{144}{28} \):\[ \frac{144}{28} = \frac{36}{7} \]Thus, the longest wavelength in the Paschen series of \( \mathrm{He}^{+} \) is \( \frac{36}{7} A \).

Key Concepts

Lyman Series

The Lyman series is a set of spectral lines corresponding to electron transitions in a hydrogen atom. Specifically, all transitions end at the energy level known as the ground state, or n=1.
This series is important in the context of atomic spectra because it represents the UV light emitted by these transitions, which is why it's typically observed in the ultraviolet spectrum.
When an electron falls back to the n=1 level from a higher energy level, light is emitted. This can be visualized as the electron losing energy and releasing a photon (a particle of light) in the process.

• Shortest wavelength: Occurs when the electron falls from an infinitely high energy level to n=1.
• Formula: The shortest wavelength in the Lyman series is given by the formula: \[ \frac{1}{\lambda} = R_H \left( 1^2 - \frac{1}{\infty^2} \right) = R_H \]which simplifies to \[ \lambda = \frac{1}{R_H} \]

Paschen Series

The Paschen series is another set of spectral lines for the hydrogen atom, with transitions ending at n=3. The energy transitions observed in this series emit infrared light, since the energy differences here are lesser than those in the Lyman series, yet significant enough to still cause an electron transition.
In the context of this exercise, we're looking at the ionized helium atom, represented as \( \mathrm{He}^{+} \). Here, the transitions that end at n=3 form the Paschen series.

• Longest wavelength: This occurs when the electron transitions from the n=4 level to the n=3 level.
• These wavelengths can be calculated using a modified version of the Rydberg formula: \[ \frac{1}{\lambda_{P}} = R_{He} \left( \frac{1}{3^2} - \frac{1}{4^2} \right) \]where \( R_{He} = 4R_H \) due to helium's atomic number being 2.

Rydberg Formula

The Rydberg Formula is a fundamental equation used to predict the wavelengths of spectral lines of many chemical elements. It plays a crucial role in calculating the wavelengths associated with electron transitions in atoms.
In essence, the formula relates the principal quantum numbers of electron orbits before and after a transition, along with the Rydberg constant, \( R \).

• General Formula: The general format of the Rydberg formula is \[ \frac{1}{\lambda} = R \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]Here, \( n_1 \) and \( n_2 \) are the principal quantum numbers.
• Modified for Helium ion (\( \mathrm{He}^{+}) \): Since \( Z = 2 \) for helium, the formula becomes: \[ \frac{1}{\lambda} = Z^2 R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]
• This adaptability makes the Rydberg formula widely applicable to different atom types and series like the Balmer, Lyman, and Paschen.

By utilizing the Rydberg formula, one can effectively determine electron transition wavelengths necessary for analyzing atomic spectra.