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Chapter 43: Problem 10

Molecules have low-energy vibration modes, and they can make transitions from one such state to another that result in the emission of infrared radiant energy. Suppose two such states are separated by \(0.015 \mathrm{eV}\), and the molecule descends in energy from the higher to the lower. Determine the wavelength of the photon that would be emitted.

Short Answer

Expert verified
The wavelength of the emitted photon is approximately 82.8 渭m.

Step by step solution

01

Understand the Energy Transition

The molecule transitions between two vibration states separated by an energy difference of \(0.015\, \text{eV}\). During this transition, a photon is emitted with energy equal to this difference.
02

Convert Energy Difference to Joules

Energy in electronvolts \(\text{eV}\) can be converted to Joules \(\text{J}\) using the conversion factor \(1\, \text{eV} = 1.602 \times 10^{-19}\, \text{J}\). Thus, \(0.015\, \text{eV} = 0.015 \times 1.602 \times 10^{-19}\, \text{J} = 2.403 \times 10^{-21}\, \text{J}\).
03

Use the Energy-Wavelength Relationship

The relationship between the energy \(E\) of a photon and its wavelength \(\lambda\) is given by \(E = \frac{hc}{\lambda}\), where \(h = 6.626 \times 10^{-34}\, \text{m}^2 \cdot \text{kg} / \text{s}\) is Planck's constant and \(c = 3.00 \times 10^8\, \text{m/s}\) is the speed of light.
04

Solve for Wavelength

Rearrange the equation to solve for wavelength: \(\lambda = \frac{hc}{E}\). Substituting the values, we get \(\lambda = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{2.403 \times 10^{-21}}\), which yields \(\lambda \approx 8.28 \times 10^{-5}\, \text{m}\) or \(82.8 \, \text{micrometers (渭m)}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Energy Transition
Molecules, much like tiny bundles of energy, have distinct states characterized by different energy levels. In the realm of infrared spectroscopy, these are often related to vibration modes. When a molecule undergoes a transition from a higher-energy state to a lower-energy state, it is said to have an energy transition. This process involves the molecule losing energy. Consequently, this energy is emitted in the form of a photon. In simple terms, an energy transition represents the movement of a molecule from a more energetic state to a less energetic state. The energy lost during this transition is precisely equal to the energy difference between the two states.
Understanding these transitions is crucial in fields like chemistry and physics because they tell us about molecular behavior and interactions.
Photon Emission
When a molecule transitions between states, it emits energy in the form of a photon, a tiny particle of light. This concept is crucial in infrared spectroscopy, where the study focuses on the infrared light emitted or absorbed by molecules. Emission occurs because molecules prefer to be in the lowest possible energy state. As they drop in energy levels, they release excess energy.
  • This energy appears as light, specifically infrared light for low-energy transitions.
  • The emitted photon's energy corresponds exactly to the energy gap between the two states.
The exact nature of the emitted photon provides valuable information about the molecule鈥檚 structure and the energies of its possible states.
Planck's Constant
At the heart of the energy-wavelength relationship lies a fundamental constant of nature known as Planck's constant, denoted by the symbol \( h \). Planck's constant is pivotal in quantum mechanics, and it defines the scale at which quantum effects become significant. Its approximate value is \( 6.626 \times 10^{-34} \) joule-seconds. This constant is critical in the formula \( E = \frac{hc}{\lambda} \), where \( E \) is the photon's energy, \( c \) is the speed of light, and \( \lambda \) is the wavelength.
Planck's constant serves as a bridge connecting the energy of photons to their respective wavelengths. This connection is vital for understanding how energy transitions impact photon emission in infrared spectroscopy.
Wavelength Calculation
The process of finding the wavelength of an emitted photon during an energy transition is an insightful application of several intertwined concepts in physics. We start with the energy \( E \) of the transition, which is given or measured. Using the formula \( \lambda = \frac{hc}{E} \), where \( h \) is Planck鈥檚 constant and \( c \) is the speed of light, we can calculate the wavelength \( \lambda \). Substituting the known values, the exact wavelength can be found.
  • For example, if \( E = 2.403 \times 10^{-21} \) Joules, it directs us to compute \( \lambda \) as approximately \( 8.28 \times 10^{-5} \) meters or \( 82.8 \) micrometers.
This calculation shows us not just the size of the photon's wavelength, but also how such transitions can be translated into useful data about molecular structure and dynamics in systems.

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Most popular questions from this chapter

A bright yellow sodium emission line has a wavelength of \(587.561\) \(8 \mathrm{~nm} .\) Determine the difference between the atom's two energy levels defining the transition. Give your answer in \(\mathrm{eV}\) to four significant figures.

One spectral line in the hydrogen spectrum has a wavelength of \(821 \mathrm{~nm}\). What is the energy difference between the two states that gives rise to this line?

Unexcited hydrogen gas is an electrical insulator because it contains no free electrons. What maximum-wavelength photon beam incident on the gas can cause the gas to conduct electricity? The photons in the beam must ionize the atom so as to produce free electrons. (This is called the atomic photoelectric effect.) To do this, the photon energy must be at least \(13.6 \mathrm{eV}\), and so the maximum wavelength is $$ \lambda=(1240 \mathrm{~nm})\left(\frac{1.00 \mathrm{eV}}{13.6 \mathrm{eV}}\right)=91.2 \mathrm{~nm} $$ which is the series limit for the Lyman series.

When a hydrogen atom is bombarded, the atom may be raised into a higher energy state. As the excited electron falls back to the lower energy levels, light is emitted. What are the three longestwavelength spectral lines emitted by the hydrogen atom as it returns to the \(n=1\) state from higher energy states? Give your answers to three significant figures. We are interested in the following transitions (see Fig. \(43-1\) ): $$ \begin{array}{ll} n=2 \rightarrow n=1: & \Delta \mathrm{E}_{2,1}=-3.4-(-13.6)=10.2 \mathrm{eV} \\\ n=3 \rightarrow n=1: & \Delta \mathrm{E}_{3,1}=-1.5-(-13.6)=12.1 \mathrm{eV} \\\ n=4 \rightarrow n=1: & \Delta \mathrm{E}_{4,1}=-0.85-(-13.6)=12.8 \mathrm{eV} \end{array} $$ To find the corresponding wavelengths, proceed as in Problem 43.1, or use \(\Delta \mathrm{E}=h f=h c / \lambda\). For example, for the \(n=2\) to \(n=1\) transition, $$ \lambda=\frac{h c}{\Delta E_{2,1}}=\frac{\left(6.63 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\right)\left(2.998 \times 10^{8} \mathrm{~m} / \mathrm{s}\right)}{(10.2 \mathrm{eV})\left(1.60 \times 10^{-19} \mathrm{~J} / \mathrm{eV}\right)}=1.22 \mathrm{~nm} $$ The other lines are found in the same way to be \(102 \mathrm{~nm}\) and \(96.9\) \(\mathrm{nm}\). These are the first three lines of the Lyman series.

What wavelength does a hydrogen atom emit as its excited electron descends from the \(n=5\) state to the \(n=2\) state? Give your answer to three significant figures. From the Bohr model we know that the energy levels of the hydrogen atom are given by \(\mathrm{E}_{n}=-13.6 / \mathrm{n}^{2} \mathrm{eV}\), and therefore $$ \mathrm{E} 5=-0.54 \mathrm{eV} \quad \text { and } \quad \mathrm{E} 2=-3.40 \mathrm{eV} $$ The energy difference between these states is \(3.40-0.54=2.86\) \(\mathrm{eV}\). Because \(1240 \mathrm{~nm}\) corresponds to \(1.00 \mathrm{eV}\) in an inverse proportion (i.e., the more energetic the photon, the shorter the wavelength), we have, for the wavelength of the emitted photon, $$ \lambda=\left(\frac{1.00 \mathrm{eV}}{2.86 \mathrm{eV}}\right)(1240 \mathrm{~nm})=434 \mathrm{~nm} $$
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