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Ch.6 - Electronic Structure of Atoms
Brown - Chemistry: The Central Science 14th Edition
Brown14th EditionChemistry: The Central ScienceISBN: 9780134414232Not the one you use?Change textbook
All textbooksBrown 14th EditionCh.6 - Electronic Structure of AtomsProblem 25a
Chapter 6, Problem 25a

(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is 2.94 × 1014 s-1.

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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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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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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-cm increments, what would happen to the height of a child as she grows up: (i) the child's height would never change, (ii) the child's height would continuously increase, (iii) the child's height would increase in jumps of 6 cm, or (iv) the child's height would increase in 'jumps' of 1 cm at a time?

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Textbook Question

(b) Calculate the energy of a photon of radiation whose wavelength is 413 nm.

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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) What wavelength of radiation has photons of energy 6.06 × 10-19 J?

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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