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 \(9.697 \, \text{nm}\), corresponding to option (c).

Step 1: Understanding the Lyman Series

The Lyman series involves transitions where the electron falls to the first energy level (n = 1) from higher levels. In this problem, the electron falls from the fourth stationary state (n = 4) to n = 1.

Step 2: Using the Rydberg Formula

The wavelength for transitions in the hydrogen atom can be calculated using the Rydberg formula: \[\frac{1}{\lambda} = R_H\left(\frac{1}{n_1^2} - \frac{1}{n_2^2}\right)\]where \(R_H = 1.1 \times 10^7 \, \text{m}^{-1}\), \(n_1 = 1\), and \(n_2 = 4\).

Step 3: Plug in the Values

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 the difference in inverse squares: \[\frac{1}{1} - \frac{1}{16} = \frac{16}{16} - \frac{1}{16} = \frac{15}{16}\]Now, substitute back into the formula: \[\frac{1}{\lambda} = 1.1 \times 10^7 \times \frac{15}{16}\]

Step 4: Calculate the Value

Perform the multiplication to find \(\frac{1}{\lambda}\): \[\frac{1}{\lambda} = 1.1 \times 10^7 \times \frac{15}{16} = 1.03125 \times 10^7\, \text{m}^{-1}\]Now, take the reciprocal to find \(\lambda\): \[\lambda = \frac{1}{1.03125 \times 10^7} \approx 9.697 \times 10^{-8}\, \text{m}\]Convert to nanometers (1 m = 1,000,000,000 nm): \[9.697 \times 10^{-8}\, \text{m} = 9.697 \, \text{nm}\]

Step 5: Select the Correct Answer

The calculated wavelength is \(9.697 \, \text{nm}\), which matches option (c).

Key Concepts

Rydberg Formula

The Rydberg formula is an essential equation in atomic physics. It is used to predict the wavelengths of spectral lines of hydrogen and other types of atoms, based on quantum mechanics. Specifically, this formula calculates the wavelengths of the photons emitted when an electron transitions between energy levels in a hydrogen atom.

The general form of the Rydberg formula used for calculating the wavelength \( \lambda \) of light emitted is:

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

Here, \( R_H \) is the Rydberg constant for hydrogen, approximately \( 1.1 \times 10^7 \text{ m}^{-1} \). This constant provides a foundational value for calculating wavelengths.
\( n_1 \) and \( n_2 \) are the principal quantum numbers of the electron's initial and final energy levels. For a transition with an electron falling from a higher energy level to a lower level, \( n_1 \) is always less than \( n_2 \).

Using this formula, we can determine the exact wavelength of light emitted or absorbed during electron transitions, which align with the observed spectral lines in the hydrogen spectrum.

Hydrogen Atom Spectral Lines

Spectral lines in the hydrogen atom are a result of electronic transitions between different energy levels. When electrons in a hydrogen atom fall from higher energy states to lower ones, they release energy in the form of light. This emitted light appears as distinct lines at specific wavelengths which are part of the atomic spectrum.

Each series of lines in the hydrogen spectrum corresponds to electrons falling to a specific lower energy level. For instance:

• Lyman series: Transitions where electrons fall to the first energy level (\(n = 1\))
• Balmer series: Transitions to the second energy level (\(n = 2\))
• Paschen series: Transitions to the third energy level (\(n = 3\))

For the Lyman series, which we are mainly concerned with here, these transitions occur in the ultraviolet region since they involve a significant amount of energy change.

Understanding spectral lines helps us closely observe and interpret the chemical and physical properties of atoms, especially in distant stars and galaxies by analyzing their light spectrum.

Wavelength Calculation

Calculating the wavelength of light emitted by a hydrogen atom involves using the Rydberg formula, which provides the necessary framework for these computations. The critical step involves plugging in known values for the Rydberg constant and principal quantum numbers associated with the electron's energy levels.

For the problem with the Lyman series, where an electron transitions from \( n_2 = 4 \) to \( n_1 = 1 \), you would substitute these values into the formula:

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

This results in a calculation where you find the difference in the squares of the quantum numbers, leading to:

• \( \frac{1}{1} - \frac{1}{16} = \frac{15}{16} \)

Substitute this into the formula and perform the multiplication to compute \( \frac{1}{\lambda} \), and finally take the reciprocal to calculate \( \lambda \).

The precision of this process allows us not just to solve theoretical problems but also to explore practical applications such as in spectroscopy, aiding technological advancements in fields ranging from astrophysics to laser engineering.