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 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

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

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

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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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