Problem 58 - Chemistry and Chemical Reactivity

Question

Suppose hydrogen atoms absorb energy so that electrons are excited to the \(n=7\) energy level. Electrons then undergo these transitions, among others: (a) \(n=7 \longrightarrow n=1 ;\) (b) \(n=7 \longrightarrow n=6 ;\) and \((c) n=\) \(2 \longrightarrow n=1 .\) Which transition produces a photon with (i) the smallest energy, (ii) the highest frequency, and (iii) the shortest wavelength?

Step-by-Step Solution

Answer

(i) Transition b; (ii) Transition a; (iii) Transition a.

Step 1: Understand the relationship between energy levels

Electrons in an atom occupy energy levels, which are quantized. When an electron moves from a higher energy level to a lower one, it emits a photon whose energy is equal to the energy difference between the two levels.

Step 2: Apply the Rydberg Formula

To find the energy (E) of the photon emitted during a transition from n_ito n_f, use the Rydberg formula:\[ E = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]where R_H is the Rydberg constant (2.18 \times 10^{-18} J).

Step 3: Calculate energies for each transition

**(a) Transition: \(n=7 \to n=1\):**\[ E_a = R_H \left( \frac{1}{1^2} - \frac{1}{7^2} \right) = 2.18 \times 10^{-18} \left( 1 - \frac{1}{49} \right) \approx 2.18 \times 10^{-18} \times 0.9804 \approx 2.13 \times 10^{-18}\, \text{J} \]**(b) Transition: \(n=7 \to n=6\):**\[ E_b = R_H \left( \frac{1}{6^2} - \frac{1}{7^2} \right) = 2.18 \times 10^{-18} \left( \frac{1}{36} - \frac{1}{49} \right) \approx 2.18 \times 10^{-18} \times 0.0102 \approx 2.22 \times 10^{-20}\, \text{J} \]**(c) Transition: \(n=2 \to n=1\):**\[ E_c = R_H \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = 2.18 \times 10^{-18} \left( 1 - \frac{1}{4} \right) \approx 2.18 \times 10^{-18} \times 0.75 \approx 1.64 \times 10^{-18}\, \text{J} \]

Step 4: Determine the smallest energy transition

Compare the energies:- \(E_a = 2.13 \times 10^{-18}\, \text{J}\)- \(E_b = 2.22 \times 10^{-20}\, \text{J}\)- \(E_c = 1.64 \times 10^{-18}\, \text{J}\)The smallest energy is for transition \(b: n=7 \to n=6\).

Step 5: Determine the highest frequency

The frequency of a photon is directly proportional to its energy. The highest frequency corresponds to the transition with the highest energy, which is transition \(a: n=7 \to n=1\).

Step 6: Determine the shortest wavelength

Wavelength is inversely proportional to both the energy and frequency of the photon. Therefore, the shortest wavelength corresponds to the transition with the highest energy and frequency, which is transition \(a: n=7 \to n=1\).

Key Concepts

Energy Levels

Electrons within a hydrogen atom are not free to have any energy. Instead, they occupy specific energy levels, which are quantized. This means electrons can only exist at distinct energy states, each denoted by a principal quantum number, \( n \). The principal quantum number is integral and positive, starting from \( n = 1 \) (the lowest energy level, closest to the nucleus) and increasing as the energy level expands outward.

Think of these energy levels like rungs on a ladder. Electrons can "jump" from one rung to another, but they can't be in between. When an electron absorbs or releases energy, it moves to a different energy level. The energy of this movement is directly tied to the difference between the initial and final levels.

• Higher energy levels are farther away from the nucleus.
• Lower energy levels are closer and have lesser energy.

Understanding electron transitions among these levels is crucial for explaining photon emissions, as the excited electron releases energy in the form of light.

Rydberg Formula

The Rydberg formula is a mathematical expression that helps us calculate the energy of the photon emitted when an electron transitions between two energy levels in a hydrogen atom. Using this formula, students can understand how the change in the electron's position relates to the energy of the emitted photon.

The formula is expressed as:\[ E = R_H \left( \frac{1}{n_f^2} - \frac{1}{n_i^2} \right) \]Here, \( R_H \) stands as the Rydberg constant (approximately \( 2.18 \times 10^{-18} \) Joules), \( n_i \) is the initial energy level, and \( n_f \) is the final energy level.

When you plug the energy levels into the formula, you calculate the photon's energy that results from the transition:

• As \( n_f \) becomes smaller than \( n_i \), the energy change \( E \) results in photon emission.
• The larger the difference between \( n_i \) and \( n_f \), the greater the energy of the emitted photon.

This formula is a cornerstone of quantum mechanics, helping us predict photon emissions and understand why energy levels are structured as they are.

Photon Emission

Photon emission occurs when an excited electron returns to a lower energy state. It's like an electron catching its breath after a high jump and sliding back to a lower point. When it moves to a lower energy level, it gives off the energy difference as a photon, a particle of light.

The characteristics of this photon, such as its energy, frequency, and wavelength, are determined by the energy transition. For instance:

• The higher the energy of the transition, the higher the frequency of the photon emitted.
• The shorter the wavelength, the more "energetic" the photon is.

In the example transitions:- The smallest energy transition (\( n=7 \rightarrow n=6 \)) results in the least energetic photon.- The transition from \( n=7 \rightarrow n=1 \) results in the highest energy photon. This frequency is the highest, and thus its wavelength is the shortest.

Understanding photon emission provides a clear picture of how electrons interact with light. It exemplifies the concept that light is not just a wave, but also a stream of energetic particles.