Question

Order the following transitions in the hydrogen atom from smallest to largest frequency of light absorbed: $n=3$ to $n=6,n=4$ to $n=9,n=2$ to $n=3,$ and $n=1$ to $n=2.$

Short answer

The order of transitions from smallest to largest frequency of light absorbed is: $$n=4\to n=9,n=3\to n=6,n=2\to n=3,\text{ and }n=1\to n=2$$

Step-by-step solution

Recall the Rydberg formula

For a hydrogen atom, the Rydberg formula relates the frequency of the absorbed light, $v$, to the energy levels involved in the transition: $$v=R_H(\frac{1}{n_i^2}-\frac{1}{n_f^2})$$ where $R_H$ is the Rydberg constant for hydrogen ($R_H\approx 3.29\times {10}^{15}{s}^{-1}$), $n_i$ is the initial energy level ("lower" energy state), and $n_f$ is the final energy level ("higher" energy state).

Calculate the frequency for each transition

We will now apply the Rydberg formula to determine the frequency of the light absorbed for each of the given transitions: 1. $n=3$ to $n=6$: $v_1=R_H(\frac{1}{3^2}-\frac{1}{6^2})$ 2. $n=4$ to $n=9$: $v_2=R_H(\frac{1}{4^2}-\frac{1}{9^2})$ 3. $n=2$ to $n=3$: $v_3=R_H(\frac{1}{2^2}-\frac{1}{3^2})$ 4. $n=1$ to $n=2$: $v_4=R_H(\frac{1}{1^2}-\frac{1}{2^2})$ Now, we will compute the frequencies for each transition: 1. $v_1=3.29\times {10}^{15}(\frac{1}{9}-\frac{1}{36})\approx 2.47\times {10}^{15}{s}^{-1}$ 2. $v_2=3.29\times {10}^{15}(\frac{1}{16}-\frac{1}{81})\approx 1.91\times {10}^{15}{s}^{-1}$ 3. $v_3=3.29\times {10}^{15}(\frac{1}{4}-\frac{1}{9})\approx 4.95\times {10}^{15}{s}^{-1}$ 4. $v_4=3.29\times {10}^{15}(\frac{1}{1}-\frac{1}{4})\approx 8.23\times {10}^{15}{s}^{-1}$

Order the transitions based on frequency

Finally, we can order the transitions from smallest to the largest based on the frequency of light absorbed: - Smallest frequency: $n=4$ to $n=9$, with $v_2\approx 1.91\times {10}^{15}{s}^{-1}$ - Next smallest frequency: $n=3$ to $n=6$, with $v_1\approx 2.47\times {10}^{15}{s}^{-1}$ - Next largest frequency: $n=2$ to $n=3$, with $v_3\approx 4.95\times {10}^{15}{s}^{-1}$ - Largest frequency: $n=1$ to $n=2$, with $v_4\approx 8.23\times {10}^{15}{s}^{-1}$ So, the order of transitions from smallest to largest frequency of light absorbed is: $$n=4\to n=9,n=3\to n=6,n=2\to n=3,\text{ and }n=1\to n=2$$

Key concepts

Hydrogen Atom Transitions

In the hydrogen atom, electrons can jump between different energy levels or shells. These are known as atomic transitions. When an electron transitions from a lower energy level to a higher one, energy needs to be absorbed, usually in the form of light. Understanding these transitions is crucial for studying atomic spectra because they determine the pattern of lines observed in spectroscopic studies.

Each energy level is represented by a principal quantum number, denoted as $n$. A transition has two energy levels: the initial level $n_i$ and the final level $n_f$. If the final energy level $n_f$ is higher than the initial level $n_i$, the electron absorbs energy; this often means the transition is within the ultraviolet or visible spectrum.

Key points to remember about hydrogen atom transitions include:

• Energy is absorbed when moving from a lower to a higher energy level.
• The energy difference between levels determines the frequency of light absorbed.
• Larger differences between energy levels result in higher frequencies.

This concept is pivotal when calculating the frequency of light absorbed during transitions, as explained by the Rydberg Formula.

Absorbed Light Frequency

The absorbed light frequency during a transition between energy levels in a hydrogen atom can be precisely calculated using the Rydberg formula. The Rydberg formula is given by

$$v=R_H(\frac{1}{n_i^2}-\frac{1}{n_f^2})$$

where $R_H$ is the Rydberg constant, approximately $3.29\times {10}^{15}\text{Hz}$, signifying the strength of the electron's interaction with the hydrogen nucleus.

The frequency, $v$, directly correlates with the energy difference between the initial and final energy levels. Calculating this frequency helps in understanding the characteristics of light absorbed during these transitions. Higher frequency indicates more energy is absorbed due to larger differences between $n_i$ and $n_f$.

Key aspects of absorbed light frequency include:

• The formula involves the inverse square of both initial and final quantum numbers.
• Greater differences in energy levels mean higher frequencies and more energetic photons.
• Lower frequency translates to longer wavelength light, and vice versa.

For example, in the transition from $n=1$ to $n=2$, the frequency is highest among the given transitions, indicating significant energy absorption and resulting in the absorption of photons from the ultraviolet spectrum.

Energy Levels

The concept of energy levels in an atom mostly concerns the regions around the nucleus where electrons are found. These levels are quantized, meaning electrons can exist only at specific energy levels rather than in between two levels. In the hydrogen atom, these levels are associated with different amounts of energy.

Each level is defined by a principal quantum number $n$, where a higher $n$ means a higher energy level. An electron higher up has more energy but is less tightly bound to the nucleus. Calculations involving these energy levels use the concept of potential and kinetic energy to explain how electrons transitions occur and how they absorb or emit energy.

When it comes to energy levels:

• The lowest energy level ($n=1$) is the ground state.
• Higher levels ($n=2,3,...$) are excited states.
• The difference in energy between levels becomes smaller as $n$ increases.

This quantization explains why only certain photon energies (or frequencies) are absorbed or emitted by a hydrogen atom, as transitions between these specific levels require precise amounts of energy. Such discretely separated levels lead to the formation of line spectra, a key observable in atomic spectroscopy.