Ch.6 - Electronic Structure of Atoms

Brown - Chemistry: The Central Science 15th Edition

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Brown 15th Edition

Ch.6 - Electronic Structure of Atoms

Problem 25b

Chapter 6, Problem 25b

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

Verified step by step guidance

Convert the wavelength from nanometers to meters by using the conversion factor: 1 nm = 1 x 10^{-9} m.

Use the speed of light equation: c = \(\lambda\) \(\nu\), where c is the speed of light (3.00 x 10^8 m/s), \(\lambda\) is the wavelength in meters, and \(\nu\) is the frequency in s^{-1}. Rearrange the equation to solve for frequency: \(\nu\) = \(\frac{c}{\lambda}\).

Substitute the wavelength in meters into the equation to calculate the frequency.

Use Planck's equation to find the energy of the photon: E = h\(\nu\), where E is the energy in joules, h is Planck's constant (6.626 x 10^{-34} J·s), and \(\nu\) is the frequency calculated in the previous step.

Substitute the frequency and Planck's constant into the equation to calculate the energy of the photon.

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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 related to its frequency and inversely related to its wavelength. It 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 (nu) is the frequency of the radiation. The frequency can be derived from the wavelength using the speed of light equation, c = λν.

Wavelength and Frequency Relationship

Wavelength (λ) and frequency (ν) are inversely related through the speed of light (c), which is approximately 3.00 x 10^8 m/s. The relationship is expressed as c = λν, meaning that as the wavelength increases, the frequency decreases, and vice versa. This relationship is crucial for converting wavelength measurements into frequency for energy calculations.

Planck's Constant

Planck's constant is a fundamental constant in quantum mechanics that relates the energy of a photon to its frequency. It has a value of approximately 6.626 x 10^-34 J·s. This constant is essential for calculating photon energy and highlights the quantized nature of electromagnetic radiation, where energy is emitted or absorbed in discrete packets called quanta.

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Related Practice

Textbook Question

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

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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. Using Figure 6.4, predict the color associated with this wavelength.

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Textbook Question

An AM radio station broadcasts at 1010 kHz, and its FM partner broadcasts at 98.3 MHz. Calculate and compare the energy of the photons emitted by these two radio stations.

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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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