Textbook Question
(b) Calculate the energy of a photon of radiation whose wavelength is 413 nm.
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Ch.6 - Electronic Structure of Atoms
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970
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Table of contents
- Ch.1 - Introduction: Matter, Energy, and Measurement111
- Ch.2 - Atoms, Molecules, and Ions166
- Ch.3 - Chemical Reactions and Reaction Stoichiometry138
- Ch.4 - Reactions in Aqueous Solution127
- Ch.5 - Thermochemistry104
- Ch.6 - Electronic Structure of Atoms109
- Ch.7 - Periodic Properties of the Elements107
- Ch.8 - Basic Concepts of Chemical Bonding102
- Ch.9 - Molecular Geometry and Bonding Theories120
- Ch.10 - Gases122
- Ch.11 - Liquids and Intermolecular Forces79
- Ch.12 - Solids and Modern Materials102
- Ch.13 - Properties of Solutions99
- Ch.14 - Chemical Kinetics153
- Ch.15 - Chemical Equilibrium83
- Ch.16 - Acid-Base Equilibria109
- Ch.17 - Additional Aspects of Aqueous Equilibria117
- Ch.18 - Chemistry of the Environment60
- Ch.19 - Chemical Thermodynamics109
- Ch.20 - Electrochemistry113
- Ch.21 - Nuclear Chemistry79
- Ch.22 - Chemistry of the Nonmetals52
- Ch.23 - Transition Metals and Coordination Chemistry57
- Ch.24 - The Chemistry of Life: Organic and Biological Chemistry7
Brown15th EditionChemistry: The Central ScienceISBN: 9780137542970Not the one you use?Change textbook
Chapter 6, Problem 25a
(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is 2.94 × 1014 s-1.
Verified step by step guidance1
Identify the given frequency of the electromagnetic radiation, which is \(2.94 \times 10^{14} \, \text{s}^{-1}\).
Recall the formula for the energy of a photon, which is given by Planck's equation: \(E = h \nu\), where \(E\) is the energy of the photon, \(h\) is Planck's constant (\(6.626 \times 10^{-34} \, \text{J} \cdot \text{s}\)), and \(\nu\) is the frequency of the radiation.
Substitute the given frequency and the value of Planck's constant into the equation.
Perform the multiplication to calculate the energy in joules (J).
Interpret the result, understanding that the energy calculated represents the energy carried by a single photon of the given electromagnetic radiation.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Photon Energy
The energy of a photon is directly proportional to its frequency and can be calculated using the equation E = hν, where E is energy, h is Planck's constant (6.626 x 10^-34 J·s), and ν is the frequency in hertz. This relationship shows that higher frequency photons carry more energy.
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01:40
Photon Energy Formulas
Planck's Constant
Planck's constant is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. It is a key component in the equation E = hν, and its value is approximately 6.626 x 10^-34 J·s. Understanding this constant is essential for calculating photon energy.
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00:50Photons and Planck's Constant
Frequency of Electromagnetic Radiation
Frequency refers to the number of cycles of a wave that pass a point in one second, measured in hertz (Hz). In the context of electromagnetic radiation, frequency is inversely related to wavelength, and it plays a crucial role in determining the energy of the photon, as described by the equation E = hν.
Related Practice
Textbook Question
If human height were quantized in 1-foot increments, what would happen to the height of a child as she grew up?
a. The child’s height would never change.
b. The child’s height would continuously get greater.
c. The child’s height would increase in “jumps” of 1 foot at a time.
d. The child’s height would increase in jumps of 6 inches.
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Textbook Question
A laser pointer used in a lecture hall emits light at 650 nm. What is the frequency of this radiation?
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Textbook Question
A laser pointer used in a lecture hall emits light at 650 nm. Using Figure 6.4, predict the color associated with this wavelength.
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Textbook Question
(c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of 532-nm photons. What is the energy gap between the ground state and excited state in the laser material?
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