Problem 32
Question
Without doing detailed calculations, indicate which of the following electron transitions in the hydrogen atom results in the emission of light of the longest wavelength. (a) \(n=4\) to \(n=3 ;\) (b) \(n=1\) to \(n=2\) (c) \(n=1\) to \(n=6 ;\) (d) \(n=3\) to \(n=2\).
Step-by-Step Solution
Verified Answer
The transition that produces the light of the longest wavelength is \(n=3\) to \(n=2\).
1Step 1: Understand the Relationship between Energy, Frequency, and Wavelength
The energy of a photon is given by the equation \(E=hf\), where \(h\) is Planck's constant and \(f\) is the frequency of the light. We also know that the frequency of light is inversely proportional to the wavelength, and so this can also be written as \(E=hc/\lambda\), where \(c\) is the speed of light and \(\lambda\) is the wavelength.
2Step 2: Compare the Energy Changes for Each Transition
The energy levels of the hydrogen atom are given by \(E=-13.6/n^{2} eV\), where \(n\) is the principal quantum number. The energy difference between two levels is given by \(ΔE=E_{f}-E_{i}\), where \(E_{f}\) and \(E_{i}\) represent the final and initial energy states respectively. For (a), the energy difference is \(E_3 - E_4\), for (b) it’s \(E_2 - E_1\), for (c) it’s \(E_6 - E_1\), and for (d) it’s \(E_2 - E_3\). We're looking for the smallest energy difference.
3Step 3: Determine the Transition with the Smallest Energy Difference
Without detailed calculations, observe the energy levels involved in each of the transitions. (a) \(n=4\) to \(n=3\), (b) \(n=1\) to \(n=2\), (c) \(n=1\) to \(n=6\), and (d) \(n=3\) to \(n=2\). Notice that the smallest difference in the quantum numbers is for transition (b) \(n=1\) to \(n=2\), and (d) \(n=3\) to \(n=2\). Both have a change of one energy level. However, energy levels in an atom are not equally spaced; they get closer together as the energy level number increases. Therefore, the transition from \(n=3\) to \(n=2\) will have a smaller energy difference than the transition from \(n=1\) to \(n=2\).
Key Concepts
Energy LevelsPhoton EmissionWavelength and Frequency Relations
Energy Levels
Atoms, particularly hydrogen atoms, have specific locations, called energy levels, that electrons can occupy. These levels are associated with a quantum number, commonly denoted as \(n\).
In a hydrogen atom, the energy of a level is determined by the formula: \(E = -\frac{13.6}{n^2}\) eV. This means that energy levels are quantized, and the energy becomes less negative as \(n\) increases. As such, the very first level, \(n = 1\), is the most tightly bound energy state, while higher values of \(n\) represent higher energy levels that an electron can occupy.
Whenever an electron transitions between two levels, it either absorbs or emits energy. Calculating these energy differences helps us understand the nature of light produced in such processes.
In a hydrogen atom, the energy of a level is determined by the formula: \(E = -\frac{13.6}{n^2}\) eV. This means that energy levels are quantized, and the energy becomes less negative as \(n\) increases. As such, the very first level, \(n = 1\), is the most tightly bound energy state, while higher values of \(n\) represent higher energy levels that an electron can occupy.
Whenever an electron transitions between two levels, it either absorbs or emits energy. Calculating these energy differences helps us understand the nature of light produced in such processes.
Photon Emission
When an electron in a hydrogen atom transitions from a higher energy level to a lower one, it emits energy in the form of a photon. This happens because the electron has excess energy at the upper energy level, which it must release to move to a lower state.
The energy difference between the initial and final states determines the energy (and thus the frequency and wavelength) of the emitted photon. If the energy difference is large, high-energy photons like ultraviolet light are emitted. Conversely, a small energy difference results in low-energy photons like infrared light.
The energy difference between the initial and final states determines the energy (and thus the frequency and wavelength) of the emitted photon. If the energy difference is large, high-energy photons like ultraviolet light are emitted. Conversely, a small energy difference results in low-energy photons like infrared light.
- Photon emission is central to how hydrogen atoms produce light in phenomena such as spectral lines.
- Lesser energy difference means larger wavelength (less energy density per photon).
Wavelength and Frequency Relations
Wavelength and frequency are two key characteristics of light that are inseparably linked.
The relationship between them is given by the formula: \(c = \lambda f\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(f\) is the frequency. This shows that wavelength is inversely proportional to frequency; when the frequency increases, the wavelength decreases and vice versa.
Consequently, in the context of electron transitions, a small energy difference leads to a low frequency and therefore a longer wavelength. This is pivotal when assessing which electron transition produces light of the longest wavelength.
The relationship between them is given by the formula: \(c = \lambda f\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(f\) is the frequency. This shows that wavelength is inversely proportional to frequency; when the frequency increases, the wavelength decreases and vice versa.
Consequently, in the context of electron transitions, a small energy difference leads to a low frequency and therefore a longer wavelength. This is pivotal when assessing which electron transition produces light of the longest wavelength.
- Long wavelength equates to lower energy photons.
- Visible light frequencies are mid-range, while radio waves have the longest wavelengths.
Other exercises in this chapter
Problem 30
Without doing detailed calculations, indicate which of the following electron transitions requires the greatest amount of energy to be absorbed by a hydrogen at
View solution Problem 31
For the Bohr hydrogen atom determine (a) the radius of the orbit \(n=4\) (b) whether there is an orbit having a radius of \(4.00 \AA\) (c) the energy level corr
View solution Problem 33
What electron transition in a hydrogen atom, starting from the orbit \(n=7,\) will produce light of wavelength \(410 \mathrm{nm} ?\)
View solution Problem 34
What electron transition in a hydrogen atom, ending in the orbit \(n=3,\) will produce light of wavelength \(1090 \mathrm{nm} ?\)
View solution