Problem 60 - 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) \(n=7 \rightarrow n=1 ;\) (b) \(n=7 \rightarrow n=6 ;\) and (c) \(n=2 \rightarrow n=1 .\) Which of these transitions 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: Understanding Energy Level Transitions

Electron transitions in hydrogen atoms occur between energy levels, each transition corresponding to the emission or absorption of photons. The energy difference between levels determines the photon energy.

Step 2: Calculating Transition Energies

Use the formula for the energy difference for a hydrogen atom transition: E = -13.6 \( \left( \frac{1}{n_1^2} - \frac{1}{n_2^2} \right) \) eV, where \( n_1 \) and \( n_2 \) (\( n_2 > n_1 \)) are the principal quantum numbers of the levels.

Step 3: Calculating Energy for Transition (a)

For \( n=7 \rightarrow n=1 \), substitute into the formula:\[ E = -13.6 \left( \frac{1}{1^2} - \frac{1}{7^2} \right) = -13.6 \left( 1 - \frac{1}{49} \right) \approx 13.6 \times 0.9796 \approx 13.32 \text{ eV} \]

Step 4: Calculating Energy for Transition (b)

For \( n=7 \rightarrow n=6 \), \[ E = -13.6 \left( \frac{1}{6^2} - \frac{1}{7^2} \right) = -13.6 \left( \frac{1}{36} - \frac{1}{49} \right) \approx 0.19 \text{ eV} \]

Step 5: Calculating Energy for Transition (c)

For \( n=2 \rightarrow n=1 \), \[ E = -13.6 \left( \frac{1}{1^2} - \frac{1}{2^2} \right) = -13.6 \left( 1 - \frac{1}{4} \right) = 13.6 \times 0.75 \approx 10.2 \text{ eV} \]

Step 6: Comparing Energies to Answer Questions (i), (ii), (iii)

(i) The smallest energy is from transition (b) \( 0.19 \text{ eV} \). (ii) The highest frequency corresponds to the largest energy transition, which is transition (a) with \( 13.32 \text{ eV} \). (iii) Shortest wavelength also correlates with the highest energy, so this is also transition (a).

Step 7: Conclusion

(i) Transition (b) produces the photon with the smallest energy. (ii) Transition (a) produces the photon with the highest frequency. (iii) Transition (a) produces the photon with the shortest wavelength.

Key Concepts

Energy Levels

In hydrogen atoms, electrons occupy specific energy levels, which are determined by the electron's distance from the nucleus. These levels are often compared to the steps of a ladder.
More energy is required to climb higher up the ladder because each energy level is quantized, or fixed.
This means electrons can only exist at specific energy levels and cannot exist between them.

• The lowest energy level (ground state) is when the electron is closest to the nucleus, represented as \( n=1 \).
• Higher energy levels involve electrons being further from the nucleus, such as \( n=2, n=3, \) and so on.

Each time an electron transitions from a higher level to a lower one, energy is released. This concept is crucial in understanding photon emission within hydrogen atoms.

Photon Emission

When electrons transition between energy levels in hydrogen atoms, photons are either emitted or absorbed. Specifically, photon emission occurs when an electron moves to a lower energy level, releasing energy in the form of light.

The energy of the emitted photon corresponds to the difference between the initial and final energy levels of the electron transition.
This leads to measurable and distinct properties in the photon:

• **Energy**: Higher energy transitions emit photons with more energy.
• **Frequency**: The emitted photon's frequency is directly proportional to its energy. Higher energy transitions result in higher frequency photons.
• **Wavelength**: There is an inverse relationship between wavelength and frequency. Therefore, higher frequency photons have shorter wavelengths.

Understanding these relationships helps in identifying the characteristics of photons resulting from specific electron transitions in hydrogen atoms.

Quantum Numbers

Principal quantum numbers \( n \) are crucial in describing the energy levels of hydrogen atoms.
These numbers denote the electron's energy state and shell, defining the size and energy of the orbitals where electrons reside.

• **n = 1** refers to the ground state, the lowest energy level.
• As \( n \) increases, the energy level and potential energy of the electron also increase.

In the context of the exercise, the transitions mentioned involve different quantum numbers:

• For example, transition \( n=7 \rightarrow n=1 \) involves an electron shedding a significant amount of energy, resulting in high-energy photon emission.
• Similarly, understanding the differences in \( n \) values helps us determine the energy and characteristics of the emitted photons.

The concept of principal quantum numbers provides a basis for predicting the behavior and characteristics of electrons in their atomic paths.