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Show that the wavenumber of any electromagnetic light is proportional to its frequency.

Short Answer

Expert verified
Wavenumber is directly proportional to frequency.

Step by step solution

01

Define Wavenumber and Frequency

The wavenumber is defined as the reciprocal of the wavelength, represented as \( k = \frac{1}{\lambda} \), where \( \lambda \) is the wavelength. Frequency, denoted by \( f \), is the number of cycles of a wave passing a point per unit time. The speed of light \( c \), relates these by the equation \( c = \lambda f \).
02

Express Wavenumber in Terms of Frequency

Substitute the expression for wavelength from the speed of light equation, \( \lambda = \frac{c}{f} \), into the wavenumber formula. Thus, \( k = \frac{1}{\lambda} = \frac{f}{c} \). This derives the wavenumber in terms of frequency and a constant, the speed of light \( c \).
03

Analyze the Proportional Relationship

From \( k = \frac{f}{c} \), observe that \( k \) varies directly with \( f \) since \( \frac{1}{c} \) is a constant. When frequency \( f \) increases, wavenumber \( k \) increases proportionally.

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Key Concepts

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

Wavenumber
The wavenumber is a fundamental property of waves that tells us how many wave cycles exist over a certain distance. It is an inverse relationship with the wavelength, which refers to the actual physical length of one cycle of the wave. Mathematically, this is expressed as:
  • \( k = \frac{1}{\lambda} \)
Here, \( k \) represents the wavenumber, and \( \lambda \) is the wavelength of the wave. The smaller the wavelength, the higher the wavenumber, indicating that more wave cycles are packed into a given distance. In practical terms, this means that as waves become more frequent over a space, the value of the wavenumber increases.
A deep understanding of wavenumber is crucial in fields like spectroscopy and the study of wave behaviors, where knowing the concentration of wave cycles provides significant insights.
Frequency
Frequency is a measure of how often a wave oscillates or cycles in a given amount of time. It tells us how many complete wave cycles pass through a point in one second and is measured in hertz (Hz). In formula terms:
  • Frequency \( f \) of a wave = number of cycles per second
Frequency plays a key role in determining the energy and behavior of waves. High-frequency waves carry more energy than low-frequency waves, a principle that is relevant in technologies like radio, microwaves, and even beyond our visible spectrum into gamma rays.
The relationship between frequency and the speed of the wave helps us understand various wave phenomena, which is fundamental in studying both light and sound waves.
Electromagnetic Waves
Electromagnetic waves are waves of energy that propagate through space at the speed of light. Unlike sound waves, which need a medium to travel through, electromagnetic waves can move through a vacuum, which is why we can see light from distant galaxies.
Some important features of electromagnetic waves include:
  • These waves encompass a wide spectrum that includes radio waves, microwaves, infrared, visible light, ultraviolet, X-rays, and gamma rays.
  • They exhibit both wave-like and particle-like properties, which is fundamental to quantum mechanics.
Understanding electromagnetic waves is critical for the development of various technologies like televisions, radios, and medical imaging techniques. They play a crucial role in communication systems and in understanding the broader universe.
Speed of Light
The speed of light, denoted as \( c \), is a constant that represents the speed at which light waves propagate through a vacuum. It is approximately \( 3.0 \times 10^8 \) meters per second. This constant is fundamental in physics, allowing us to calculate other properties of waves such as the relationship between wavelength and frequency:
  • \( c = \lambda f \), where \( \lambda \) is the wavelength and \( f \) is the frequency.
Given that the speed of light is constant, the equation allows us to find one property of light when the others are known. For instance, if we know the frequency of a wave, we can easily determine its wavelength, and vice versa.
This concept is pivotal in understanding the properties of not just visible light, but all electromagnetic radiation, influencing both theoretical and practical applications in science and engineering.

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Most popular questions from this chapter

Each of the following pairs of molecules has the same number of atoms. In each pair, which one would you expect to have fewer different vibrational frequencies? You may have to determine the symmetry of each molecule before you can make a determination. (a) \(\mathrm{HCl}\) and \(\mathrm{Cl}_{2}\) (b) \(\mathrm{H}_{2} \mathrm{O}_{2}\) and \(\mathrm{C}_{2} \mathrm{H}_{2}\) (c) \(\mathrm{CH}_{4}\) and \(\mathrm{XeF}_{4}\) (d) \(\mathrm{PF}_{5}\) and \(\mathrm{CH}_{3} \mathrm{CN}\) (e) \(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) and \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\)

The moments of inertia for phosphine, \(\mathrm{PH}_{3}\), are \(5.478 \times\) \(10^{-47} \mathrm{~kg} \cdot \mathrm{m}^{2}, 5.478 \times 10^{-47} \mathrm{~kg}-\mathrm{m}^{2}\), and \(6.645 \times 10^{-47} \mathrm{~kg}-\mathrm{m}^{2}\). Calculate the rotational constants \(A, B\), and \(C\) for phosphine.

An acquaintance remarks that a rotational spectrum of \(\mathrm{I}_{2}\) showing the \(J=200 \rightarrow J=201\) transition is predicted very closely by the rigid rotor equations from quantum mechanics. Give two reasons why you should question the validity of that statement.

Which of the following molecules should have pure rotational spectra? (a) Dimethyltriacetylene, \(\mathrm{CH}_{3}-\mathrm{C}=\mathrm{C}-\mathrm{C}=\mathrm{C}-\mathrm{C}=\mathrm{C}-\mathrm{CH}_{3}\) (b) Cyanotetraacetylene, \(\mathrm{H}-\mathrm{C}=\mathrm{C}-\mathrm{C}=\mathrm{C}-\mathrm{C}=\mathrm{C}-\mathrm{C}=\mathrm{C}-\mathrm{C}=\mathrm{N}\) (Such molecules have been detected in interstellar space.) (c) Nitric oxide, NO (d) Nitrogen dioxide, \(\mathrm{NO}_{2}\) (e) Sulfur tetrafluoride, \(\mathrm{SF}_{4}\) (f) Sulfur hexafluoride, \(S F_{6}\)

What is the frequency of light having the following wavelengths? (a) \(1.00 \mathrm{~m}\) (b) \(4.77 \times 10^{-5} \mathrm{~m}\) (c) \(7894 \mathrm{~A}\) (d) \(1.903 \times 10^{3} \mathrm{~m}\)

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