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NASA operates a jetliner fitted with a special telescope that looks into outer space. The telescope has a camera that detects wavelengths between 5 and \(40 \mu \mathrm{m}\). During eight-hour flights at \(12,000 \mathrm{~m}\) above Earth, the plane will gather data of the nearby universe. (a) In what region of the electromagnetic spectrum does the camera operate? (b) Calculate the range of frequencies the camera detects.

Short Answer

Expert verified
(a) Infrared; (b) Frequency range is 7.5 × 10¹² Hz to 6 × 10¹³ Hz.

Step by step solution

01

Identify the Wavelength Range

The problem states that the telescope's camera detects wavelengths between 5 µm and 40 µm. These wavelengths need to be understood in terms of their spectral region.
02

Determine the Spectral Region

The wavelengths from 5 µm to 40 µm fall in the infrared region of the electromagnetic spectrum. Specifically, they are part of the mid-infrared range, which typically spans from roughly 3 µm to 50 µm.
03

Convert the Wavelengths to Meters

To calculate frequencies, first convert the given wavelengths from micrometers to meters: \[ 5\, \mu m = 5 \times 10^{-6}\, m \] \[ 40\, \mu m = 40 \times 10^{-6}\, m \]
04

Use the Speed of Light Equation

The relationship between wavelength \(\lambda\), frequency \(f\), and the speed of light \(c\) is given by:\[ c = \lambda f \]where \( c = 3 \times 10^8 \, m/s \).
05

Calculate the Frequency Range

Using the equation \( f = \frac{c}{\lambda} \): For 5 µm:\[ f = \frac{3 \times 10^8}{5 \times 10^{-6}} = 6 \times 10^{13} \, Hz \] For 40 µm:\[ f = \frac{3 \times 10^8}{40 \times 10^{-6}} = 7.5 \times 10^{12} \, Hz \]Thus, the frequency range is between \(7.5 \times 10^{12}\) Hz and \(6 \times 10^{13}\) Hz.

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Key Concepts

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

Electromagnetic Spectrum
The electromagnetic spectrum is a range of all types of electromagnetic radiation. Radiation is energy that travels and spreads out as it moves. When you hear about "light" in everyday language, it usually refers to visible light, which is just one part of the electromagnetic spectrum. The spectrum itself includes several regions:
  • Radio waves: These have the longest wavelengths and are used for communication, like radio and TV broadcasts.
  • Microwaves: Used in cooking and some forms of communication, like WiFi.
  • Infrared: Detects heat and is used in everything from night-vision technology to remote controls.
  • Visible light: The tiny portion of the spectrum that we can see with our eyes, ranging from violet to red.
  • Ultraviolet: Causes sunburn and is used in black lights and in sterilization processes.
  • X-rays: Penetrate flesh to show bones and are used in medical imaging.
  • Gamma rays: High energy radiation used in cancer treatment and produced by radioactive materials.
Understanding the electromagnetic spectrum is crucial for identifying how different technologies, like NASA's infrared telescope, work by exploiting these diverse kinds of energy. The spectrum serves as a foundation for exploring types of radiation that are invisible to the naked eye.
Wavelength to Frequency Conversion
When discussing electromagnetic waves, two important characteristics are wavelength and frequency. While they seem different, they are interconnected. The wavelength is the distance between consecutive peaks of a wave. Frequency, meanwhile, tells us how many peaks pass a specific point per second. They are inversely related: as one increases, the other decreases. To understand the relationship, we use the equation:\[ c = \lambda f \]Here, \(c\) is the speed of light, approximately \(3 \times 10^8\, m/s\), \( \lambda \) is the wavelength in meters, and \( f \) is the frequency in hertz (Hz). For conversions:
  • Wavelengths in micrometers (µm) should be converted to meters first by multiplying by \(10^{-6}\).
  • Once in meters, apply the equation \( f = \frac{c}{\lambda} \) to find the frequency.
Understanding this conversion allows scientists to interpret data from different parts of the electromagnetic spectrum, translating wavelengths they detect into frequencies, an essential step in utilizing technologies like infrared telescopes effectively.
Infrared Region of Electromagnetic Spectrum
The infrared region is a part of the electromagnetic spectrum that is not visible to the human eye. It lies between the visible spectrum and microwaves. This region is subdivided further:
  • Near-infrared: Closest to visible light, ranging from about 0.7 µm to 1.5 µm, often used in fiber optic communications.
  • Mid-infrared: Includes 1.5 µm to 5 µm, used in detecting heat signatures, making it valuable in environmental monitoring.
  • Far-infrared: Ranges from about 5 µm to 1000 µm, this is where most of the heat radiation emitted by objects is found.
The infrared part of the spectrum is especially important in astronomy.
Space telescopes that "see" in infrared can peer through dense cosmic dust clouds that block visible light.
For NASA's telescope, which detects wavelengths between 5 µm and 40 µm, information gathered in the infrared region helps map and study star formations and cosmic phenomena hidden from telescopes that rely on visible light. By understanding these invisible areas of the spectrum, scientists can make incredible discoveries about the universe.

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Most popular questions from this chapter

The energy of a photon needed to cause ejection of an electron from a photoemissive metal is expressed as the sum of the binding energy of the photon plus the kinetic energy of the emitted electron. Calculate the kinetic energy of an electron that is emitted from a strontium metal surface irradiated with photons of \(4.20 \times 10^{-7} \mathrm{~m}\) light. The binding energy of strontium is \(4.39 \times 10^{-19} \mathrm{~J}\).

A certain minimum energy, \(E_{\text {min }}\) is required to eject an electron from a photosensitive surface. Any energy absorbed beyond this minimum gives kinctic cnergy to the cjected clectron. When \(540 .-\mathrm{nm}\) light falls on a cesium surface, an electron is ejected with a kinctic energy of \(6.69 \times 10^{-20} \mathrm{~J}\). When the wavelength is \(400 \mathrm{nm}\), the kinctic cnergy is \(1.96 \times 10^{-19} \mathrm{~J}\). (a) Calculate \(E_{\text {min }}\) for cesium, in joules. (b) Calculate the longest wavelength, in nanometers, that will eject an electron from cesium.

Consider a \(2+\) ion that has six \(3 d\) electrons; which ion is it? Which \(2+\) ion would have only three \(3 d\) electrons?

Spectroscopists have observed \(\mathrm{He}^{+}\) in outer space. This ion is a one-electron species like a neutral hydrogen atom. Calculate the energy of the photon emitted for the transition from the \(n=5\) to the \(n=3\) state in this ion using the equation: \(E_{n}=-Z^{2} / n^{2}\left(2.179 \times 10^{-18} \mathrm{~J}\right) . \mathrm{Z}\) is the positive charge of the nucleus and \(n\) is the principal quantum number. In what part of the electromagnetic spectrum does this radiation lie?

Use electron configurations to explain why (a) sulfur has a lower electron affinity than chlorine. (b) boron has a lower first ionization energy than beryllium. (c) chlorine has a lower first ionization energy than fluorine. (d) oxygen has a lower tirst ionization energy than nitrogen. (e) iodine has a lower electron affinity than bromine.

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