Problem 24 - College Physics

Question

. An electron in an excited state of hydrogen makes a transition from the \(n=5\) level to the \(n=2\) level. (a) Does the atom emit or absorb a photon during this process? How do you know? (b) Calculate the wavelength of the photon involved in the transition.

Step-by-Step Solution

Answer

(a) The atom emits a photon. (b) Wavelength is approximately 434 nm.

Step 1: Understanding Energy Levels

The energy levels in a hydrogen atom are quantized, and transitions between these levels involve either the emission or absorption of a photon. When an electron drops from a higher energy level to a lower one, it emits a photon. When it moves to a higher level, it absorbs a photon. In this case, an electron moves from the \( n=5 \) level to the \( n=2 \) level. Since it moves to a lower energy level, the atom emits a photon.

Step 2: Using the Rydberg Formula

To find the wavelength of the photon emitted during the electron transition, we use the Rydberg formula: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \] where \( R_H = 1.097 \times 10^7 \text{ m}^{-1} \), \( n_1 = 2 \), and \( n_2 = 5 \). Substitute these values into the formula.

Step 3: Calculating the Photon Wavelength

Plug the values for \( n_1 \) and \( n_2 \) into the Rydberg formula:\[ \frac{1}{\lambda} = 1.097 \times 10^7 \left( \frac{1}{2^2} - \frac{1}{5^2} \right) \]Calculate each term: - \( \frac{1}{2^2} = \frac{1}{4} = 0.25 \)- \( \frac{1}{5^2} = \frac{1}{25} = 0.04 \)Thus, \[ \frac{1}{\lambda} = 1.097 \times 10^7 \cdot (0.25 - 0.04) = 1.097 \times 10^7 \cdot 0.21 \]

Step 4: Solving for Wavelength

Complete the calculation to find the reciprocal of the wavelength:\[ \frac{1}{\lambda} = 1.097 \times 10^7 \times 0.21 = 2.3037 \times 10^6 \]Take the reciprocal to find \( \lambda \):\[ \lambda = \frac{1}{2.3037 \times 10^6} \approx 4.34 \times 10^{-7} \text{ m} \]This wavelength can also be expressed as \( 434 \text{ nm} \).

Key Concepts

Rydberg Formula

The Rydberg Formula is a powerful tool used to predict the wavelengths of photons emitted or absorbed during electron transitions in hydrogen atoms. This formula is grounded in the quantum mechanics governing hydrogen energy levels. It allows us to calculate the wavelength of light associated with any electron transition between these levels.

The formula itself is: \[ \frac{1}{\lambda} = R_H \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \]where:

• \( \lambda \) is the wavelength of the photon emitted or absorbed,
• \( R_H \) is the Rydberg constant, approximately \( 1.097 \times 10^7 \text{ m}^{-1} \),
• \( n_1 \) and \( n_2 \) are the principal quantum numbers of the initial and final energy levels.

The simplicity of the Rydberg Formula belies its power in accurately predicting photon wavelengths, making it a staple in understanding atomic spectra.

Photon Emission

Photon emission occurs when an electron in an atom transitions from a higher energy level to a lower one. This process releases energy in the form of a photon, which is a packet of light energy. The energy of this photon corresponds to the difference in energy between the two levels, which can be calculated from the Rydberg Formula.

When considering photon emission:

• The initial energy level is higher than the final energy level (i.e., \( n_2 > n_1 \)).
• The energy of the emitted photon, \( E = h \cdot f \), where \( h \) is Planck's constant and \( f \) is the frequency of the emitted photon.
• The wavelength \( \lambda \) of the photon can be found using the relation \( c = \lambda \cdot f \), where \( c \) is the speed of light.

Understanding photon emission is crucial, as it accounts for the characteristic spectral lines seen in hydrogen spectra, each line representing a specific electron transition.

Energy Levels in Atoms

Atoms consist of a nucleus surrounded by electrons arranged in specific energy levels, also known as orbitals. The energy levels in a hydrogen atom play a vital role in photon absorption and emission. These levels are quantized, meaning electrons can only exist in specific, discrete energy states.

Here are key points about atomic energy levels:

• Each level is indexed by a principal quantum number \( n \), where \( n = 1, 2, 3, \ldots \).
• The energy of each level decreases as \( n \) increases, making lower levels more stable.
• An electron transition from a higher \( n \, (\text{e.g., } n = 5) \) to a lower \( n \, (\text{e.g., } n = 2) \) results in photon emission.

This discrete nature leads to spectral lines, which help in identifying elements and their properties in light sources. Mastery of this concept aids in deepening the understanding of atomic behavior and quantum mechanics.