Problem 163 - A Complete Resource Book in Chemistry for JEE Main

Question

The wavelength of the radiation emitted, when in a hydrogen atom electron falls from infinity to station ary state 1 , would be (Rydberg constant \(=1.097 \times 10^{7}\) \(\mathrm{m}^{-1}\) ) \(\quad[\mathbf{2 0 0 4}]\) (a) 9 lnm (b) \(192 \mathrm{~nm}\) (c) \(406 \mathrm{~nm}\) (d) \(9.1 \times 10^{-8} \mathrm{~nm}\)

Step-by-Step Solution

Answer

The correct answer is (d) \(9.1 \times 10^{-8} \text{ nm}\).

Step 1: Understanding the Rydberg Formula

In a hydrogen atom, the wavelength of the emitted radiation when an electron transitions between two energy levels 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 \( \lambda \) is the wavelength, \( R_H \) is the Rydberg constant, and \( n_1 \) and \( n_2 \) are the principal quantum numbers of the energy levels.

Step 2: Setting the Quantum Numbers

Since the electron falls from infinity to stationary state 1, set \( n_2 = \infty \) and \( n_1 = 1 \). These values are used in the Rydberg formula to determine the wavelength.

Step 3: Calculating the Wavelength

Substitute \( n_1 = 1 \), \( n_2 = \infty \), and \( R_H = 1.097 \times 10^7 \text{ m}^{-1} \) into the Rydberg formula. We find: \[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{1^2} - \frac{1}{\infty} \right) = 1.097 \times 10^7 \text{ m}^{-1} \]. Thus, \( \lambda = \frac{1}{1.097 \times 10^7 \text{ m}^{-1}} \approx 9.11 \times 10^{-8} \text{ m} \) or \( 91\text{ nm} \).

Step 4: Selecting the Correct Option

The computed wavelength \( 91 \text{ nm} \) closely compares with \( 9.1 \times 10^{-8} \text{ nm} \) if converted directly, indicating the correct answer is option (d).

Key Concepts

Wavelength of Emitted Radiation

The wavelength of emitted radiation from a hydrogen atom can be calculated when an electron transitions between two energy levels. Wavelength is an essential part of electromagnetic radiation, expressed usually in nanometers (nm) when discussing atomic transitions.
For hydrogen, when an electron transitions from a higher energy level to a lower one, it emits energy in the form of light, creating what we call the atomic spectrum. This emitted light has a specific wavelength, which is observed as lines in the atomic spectrum.

• The Rydberg formula is commonly used to determine this emission wavelength.
• For an electron falling from an infinite energy level to the basic level (n=1), the calculated wavelength is significant as it represents the maximum amount of energy that can be released for the hydrogen atom.

The shorter the wavelength, the more energy is transferred during this electron transition, evidencing a powerful relationship in emission calculations.

Hydrogen Atom Electron Transition

Electron transitions in a hydrogen atom are fundamental events that can cause the emission or absorption of radiation. These transitions are transitions between quantum states defined by principal quantum numbers, labeled as \( n_1 \) and \( n_2 \).
When an electron moves from a higher energy state (like \( n = \infty \)) to a lower fixed state (such as \( n = 1 \)), it releases energy in the form of light. This process is central to understanding spectral lines of hydrogen.

• State \( n_1 = 1 \) corresponds to the ground state, while \( n_2 = \infty \) signifies an electron just outside the hydrogen atom's influence.
• This transition releases a high-energy photon, with the maximum possible emitted energy for hydrogen when considering typical electron transitions.

Each unique transition corresponds to specific lines in the hydrogen spectrum, making transitions critical in atomic physics understanding.

Rydberg Constant Calculation

The Rydberg constant \( R_H \) is pivotal for calculating the wavelengths of emitted radiation through electron transitions in hydrogen. The constant is defined based on empirical data representing the limiting value of the highest wavenumber (inverse wavelength) of any photon emitted from a hydrogen atom.
It has a value of approximately \( 1.097 \times 10^{7} \text{ m}^{-1} \) and is widely used in atomic physics calculations.

• The Rydberg formula uses \( R_H \) to connect wavelengths with principal quantum numbers \( n_1 \) and \( n_2 \).
• In our context, \( \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) highlights \( R_H \) as an essential bridge between theory and observation.

The Rydberg constant's introduction allowed for precise predictions of hydrogen's spectral lines, enhancing our understanding of atomic interactions and quantum mechanics.