Ch.6 - Electronic Structure of Atoms

Brown - Chemistry: The Central Science 14th Edition

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Brown 14th Edition

Ch.6 - Electronic Structure of Atoms

Problem 2c

Chapter 6, Problem 2c

A popular kitchen appliance produces electromagnetic radiation with a frequency of 2450 MHz. With reference to Figure 6.4, answer the following: (c) If the radiation is not visible, do photons of this radiation have more or less energy than photons of visible light?

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Identify the frequency of the electromagnetic radiation produced by the kitchen appliance, which is 2450 MHz.

Convert the frequency from MHz to Hz by multiplying by \$10^6$ (since 1 MHz = $10^6$ Hz).

Locate the frequency of 2450 MHz (or $2.45 \times 10^9$ Hz) on the electromagnetic spectrum diagram provided.

Compare the position of 2450 MHz on the spectrum with the visible light range (approximately $4 \times 10^{14}$ Hz to $7.5 \times 10^{14}$ Hz).

Conclude whether the energy of the photons of this radiation is more or less than that of visible light photons by using the relationship $E = h \nu$, where $E$ is energy, $h$ is Planck's constant, and $\nu$ is frequency.

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Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Electromagnetic Spectrum

The electromagnetic spectrum encompasses all types of electromagnetic radiation, arranged by frequency or wavelength. It ranges from low-frequency radio waves to high-frequency gamma rays. Visible light, which is the portion of the spectrum detectable by the human eye, occupies a small range between approximately 400 nm (violet) and 700 nm (red). Understanding where different types of radiation fall within this spectrum is crucial for comparing their properties.

Photon Energy

Photons are particles of light that carry energy, which is directly proportional to their frequency and inversely proportional to their wavelength. The energy of a photon can be calculated using the equation E = hν, where E is energy, h is Planck's constant, and ν is frequency. Therefore, higher frequency radiation, such as ultraviolet light, has more energy per photon than lower frequency radiation, such as microwaves.

Frequency and Energy Relationship

The relationship between frequency and energy is fundamental in understanding electromagnetic radiation. As frequency increases, the energy of the photons also increases. For example, the frequency of 2450 MHz (microwave radiation) is lower than that of visible light, which ranges from about 430 THz to 750 THz. Consequently, photons of microwave radiation have less energy than those of visible light, despite being invisible to the human eye.

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Frequency-Wavelength Relationship

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Textbook Question

Stars do not all have the same temperature. The color of light emitted by stars is characteristic of the light emitted by hot objects. Telescopic photos of three stars are shown below: (i) the Sun, which is classified as a yellow star, (ii) Rigel, in the constellation Orion, which is classified as a blue-white star, and (iii) Betelgeuse, also in Orion, which is classified as a red star. (a) Place these three stars in order of increasing temperature. (i) sun (ii) Rigel (iii) Betelguese

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Textbook Question

A popular kitchen appliance produces electromagnetic radiation with a frequency of 2450 MHz. With reference to Figure 6.4, answer the following: (d) Which of the following is the appliance likely to be? (i) A toaster oven, (ii) A microwave oven, or (iii) An electric hotplate.

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Textbook Question

The speed of sound in dry air at 20 °C is 343 m/s and the lowest frequency sound wave that the human ear can detect is approximately 20 Hz. (a) What is the wavelength of such a sound wave?

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Textbook Question

A popular kitchen appliance produces electromagnetic radiation with a frequency of 2450 MHz. With reference to Figure 6.4, answer the following: (a) Estimate the wavelength of this radiation.

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