Problem 105 - A Complete Resource Book in Chemistry for JEE Main

Question

What is the wavelength of the radiation emitted produced in a line in the Lyman series when an electron falls from fourth stationary state in hydrogen atom? \(\left(\mathrm{R}_{\mathrm{H}}=1.1 \times 10^{7} \mathrm{~m}^{-1}\right)\) (a) \(96.97 \mathrm{~nm}\) (b) \(969.7 \mathrm{~nm}\) (c) \(9.697 \mathrm{~nm}\) (d) none

Step-by-Step Solution

Answer

The wavelength is \(9.697 \mathrm{~nm}\) (option c).

Step 1: Understand the Lyman Series

The Lyman series involves electron transitions from higher energy levels to n = 1 in a hydrogen atom. The wavelength of the radiation can be calculated using the Rydberg formula for hydrogen.

Step 2: Recall the Rydberg Formula

The formula to find the wavelength of emitted radiation is: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \(n_1 = 1\) and \(n_2 = 4\) for this problem, and \(R_H = 1.1 \times 10^7 \mathrm{~m}^{-1}\).

Step 3: Substitute Values into the Formula

Substitute the values into the Rydberg formula: \[ \frac{1}{\lambda} = 1.1 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \] Calculate \(\frac{1}{1^2} - \frac{1}{4^2} = 1 - \frac{1}{16} = \frac{15}{16}\).

Step 4: Calculate the Wavelength (\(\lambda\))

Substitute \(\frac{15}{16}\) into the equation: \[ \frac{1}{\lambda} = 1.1 \times 10^7 \times \frac{15}{16} \] Simplify to find \(\frac{1}{\lambda} = \frac{16.5 \times 10^7}{16} = 1.03125 \times 10^7 \). Therefore, \( \lambda = \frac{1}{1.03125 \times 10^7} \approx 9.697 \times 10^{-9} \mathrm{~m}\), which is \(9.697 \mathrm{~nm}\).

Step 5: Compare Against Options

Compare the calculated wavelength \(9.697 \mathrm{~nm}\) against the provided options. The correct option is (c) \(9.697 \mathrm{~nm}\).

Key Concepts

Rydberg Formula

The Rydberg formula is crucial for calculating the wavelengths of light resulting from electron transitions within hydrogen atoms. This formula helps us understand the spectrum of light emitted by hydrogen under various conditions. It is expressed as:

• \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]

In this formula:

• \( \lambda \) is the wavelength of the emitted light.
• \( R_H \) is the Rydberg constant, approximately \( 1.1 \times 10^7 \ \text{m}^{-1} \).
• \( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron's initial and final energy states.

The significance of the Rydberg formula lies in its ability to predict the wavelength of light emitted when an electron moves between different energy levels. By knowing these transitions, we can gather valuable information about atomic structure and spectral lines.

Hydrogen Atom Transitions

Hydrogen atom transitions occur when an electron changes its energy level within an atom. These transitions are fundamental in understanding atomic spectra. When electrons move to lower energy levels, they emit electromagnetic radiation, which can be described using specific series.

In the case of hydrogen, these series arise from electron transitions to or from specific energy levels:

• Lyman series: Electron transitions to the \( n=1 \) level.
• Balmer series: Transitions to the \( n=2 \) level.
• Paschen series: Transitions to the \( n=3 \) level.

For the Lyman series, electrons falling to the \( n=1 \) level from higher levels are involved. This results in the emission of ultraviolet light. Understanding these transitions is key to examining the spectral lines of hydrogen, giving us insight into its structural properties.

Wavelength Calculation

The wavelength calculation is a significant task when dealing with electronic transitions in hydrogen atoms. This can be done by applying the Rydberg formula. Let's explore the required steps using our specific exercise:

First, identify the initial \( n_2 \) and final \( n_1 \) energy levels. For the Lyman series, the final level \( n_1 \) is 1, and let's say our initial level \( n_2 \) is 4:

• \[ \frac{1}{\lambda} = R_H \left( \frac{1}{1^2} - \frac{1}{4^2} \right) \]

By substituting the values:

• \( R_H = 1.1 \times 10^7 \ \text{m}^{-1} \)
• \( \frac{1}{1} - \frac{1}{16} = \frac{15}{16} \)

Calculate \( \frac{1}{\lambda} = 1.1 \times 10^7 \times \frac{15}{16} \), simplifying to find \( \lambda \). This approach helps determine how light's wavelength changes based on the electron's transitions, a process essential for understanding atomic spectra.

Electron Energy Levels

Electron energy levels, also known as energy states, represent the fixed or quantized levels where electrons can exist in an atom. Electrons in hydrogen atoms can jump between these distinct energy levels, and when they do, they either absorb or release energy in the form of electromagnetic radiation.

This behavior of electrons is mostly governed by quantum mechanics. Each energy level is associated with a principal quantum number \( n \), which can be any positive integer. As \( n \) increases, the corresponding energy level becomes less negative and higher, suggesting the electron is further from the nucleus:

• \( n=1 \): Lowest energy level, where electrons are closest to the nucleus.
• \( n=2, 3, ... \): Higher energy levels with electrons further out.

Changes between these levels, when electrons absorb or emit a photon, result in the creation of a spectral line. These transitions lead to different series of lines, such as the Lyman series, providing a wealth of information about atomic structure and behavior.