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Carbon dioxide in the atmosphere absorbs energy in the \(4.0-4.5 \mu \mathrm{m}\) range of the spectrum. (a) Calculate the frequency of the \(4.0 \mu \mathrm{m}\) radiation. \((\mathbf{b})\) In what region of the electromagnetic spectrum does this radiation occur?

Short Answer

Expert verified
The frequency of the 4.0 μm radiation is \(7.5 \times 10^{13} Hz\), and this radiation falls within the infrared region of the electromagnetic spectrum.

Step by step solution

01

Convert Wavelength to Meters

To use the formula to calculate frequency, we first need to convert the given wavelength from micrometers to meters. There are 10^6 micrometers in 1 meter, so we can convert 4.0 μm to meters: \(4.0\, \mu m = 4.0 \times 10^{-6}\, m\)
02

Calculate Frequency

Now we can use the formula to calculate the frequency of the 4.0 μm radiation. The speed of light is approximately 3.00 x 10^8 m/s. Plug in the values into the formula: Frequency (ν) = Speed of light (c) / Wavelength (λ) \(ν = \frac{3.00 \times 10^8\, m/s}{4.0 \times 10^{-6}\, m}\)
03

Simplify the Expression

Now, simplify the expression: \(ν = \frac{3.00}{4.0} \times \frac{10^8}{10^{-6}}\) \(ν = 0.75 \times 10^{14}\, Hz\) The frequency of the 4.0 μm radiation is 7.5 x 10^13 Hz.
04

Determine the Electromagnetic Spectrum Region

To determine the region of the electromagnetic spectrum, compare the given wavelength (4.0 μm) to the known ranges of the spectrum. The electromagnetic spectrum can be divided as follows: - Radio waves: > 1 mm - Microwaves: 1 mm - 700 nm - Infrared: 700 nm - 400 nm - Visible light: 400 nm - 10 nm - Ultraviolet: 10 nm - 0.01 nm - X-rays: 0.01 nm - 1 pm - Gamma rays: < 1 pm As we can see, 4.0 μm falls within the infrared part of the spectrum. So the 4.0 μm radiation is in the infrared region of the electromagnetic spectrum.

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Key Concepts

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

Wavelength
Wavelength is a key property of electromagnetic waves, describing the distance between one peak of a wave to the next. In the electromagnetic spectrum, wavelengths vary greatly, from long radio waves to short gamma rays. Understanding wavelength is crucial when analyzing the properties of light and other forms of electromagnetic radiation. Typically, we measure wavelength in meters (m), but smaller units like nanometers (nm) and micrometers (μm) are often used for convenience in specific contexts.

When dealing with conversions, it's helpful to remember:
  • 1 meter = 1,000,000 micrometers (μm)
  • 1 meter = 1,000,000,000 nanometers (nm)
Converting units is often the first step when solving wavelength-related problems, as it simplifies calculations and ensures consistency in our equations.
Frequency Calculation
Frequency refers to how often the crest of a wave passes a fixed point in one second. It's measured in Hertz (Hz) and is an essential concept for understanding the behavior of waves in the electromagnetic spectrum. The relationship between frequency and wavelength is inverse; as one increases, the other decreases.

This relationship is expressed mathematically by the formula:\[u = \frac{c}{\lambda}\]where:
  • \( u \) is the frequency
  • \( c \) is the speed of light (approximately \( 3.00 \times 10^8 \ \text{m/s}\))
  • \( \lambda \) is the wavelength in meters
Using this formula allows you to calculate the frequency if the wavelength is known, and vice versa, highlighting a fundamental property of waves.
Infrared Radiation
Infrared radiation is a type of electromagnetic radiation that lies between visible light and microwaves in the electromagnetic spectrum. It is not visible to the naked eye but can be felt as heat. The wavelength range of infrared radiation is typically from about 700 nanometers (nm) to 1 millimeter (mm).

Infrared radiation plays a significant role in various applications:
  • Remote controls, using infrared light to send signals to devices.
  • Infrared cameras, which detect heat emitted by objects.
  • For medical and military uses, like in thermal imaging.
Understanding infrared radiation is essential, as it has vast implications and uses across different industries, from healthcare to consumer electronics.
Micrometers to Meters Conversion
Converting from micrometers to meters is a straightforward but significant process when working with electromagnetic waves. It involves changing the measurement unit to fit standard formulas and understandings in physics.

The conversion is simple:
  • There are 1,000,000 micrometers in a single meter
  • To convert micrometers to meters, multiply the micrometer value by \( 10^{-6} \).
For example, converting 4.0 micrometers (μm) to meters results in:\[ 4.0 \, \mu\text{m} = 4.0 \times 10^{-6} \text{ m} \]Such conversions help maintain consistency in scientific calculations, ensuring accurate results. This is especially important when calculating properties like the frequency of electromagnetic radiation, which requires measurement consistency for precise computation.

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Most popular questions from this chapter

Einstein's 1905 paper on the photoelectric effect was the first important application of Planck's quantum hypothesis. Describe Planck's original hypothesis, and explain how Einstein made use of it in his theory of the photoelectric effect.

The discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71\) ) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 .\) Write a balanced chemical equation for the reaction. Starting with a \(55.4-\mathrm{g}\) sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and Hf form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

Consider a transition in which the electron of a hydrogen atom is excited from \(n=1\) to \(n=\infty\). (a) What is the end result of this transition? (b) What is the wavelength of light that must be absorbed to accomplish this process? (c) What will occur if light with a shorter wavelength than that in part (b) is used to excite the hydrogen atom? (d) How are the results of parts \((\mathrm{b})\) and \((\mathrm{c})\) related to the plot shown in Exercise \(6.88 ?\)

Sketch the shape and orientation of the following types of orbitals: \((\mathbf{a}) s,(\mathbf{b}) p_{z},(\mathbf{c}) d_{x y}\).

A certain orbital of the hydrogen atom has \(n=4\) and \(l=3\). (a) What are the possible values of \(m_{l}\) for this orbital? (b) What are the possible values of \(m_{s}\) for the orbital?

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