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Ch 38: Photons: Light Waves Behaving as Particles
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Non è quello che usi tu?Cambia libro di testo
Capitolo 38, Problema 5b

A photon has momentum of magnitude 8.24×10−288.24\(\times\)10^{-28} kg-m/s. What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

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Step 1: Recall the relationship between the momentum of a photon and its wavelength. The equation is given by \( p = \frac{h}{\lambda} \), where \( p \) is the momentum, \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \)), and \( \lambda \) is the wavelength.
Step 2: Rearrange the equation to solve for the wavelength \( \lambda \). This gives \( \lambda = \frac{h}{p} \).
Step 3: Substitute the given values into the equation. The momentum \( p \) is \( 8.24 \times 10^{-28} \, \text{kg} \cdot \text{m/s} \), and \( h \) is \( 6.626 \times 10^{-34} \, \text{J} \cdot \text{s} \). Ensure the units are consistent.
Step 4: Perform the division \( \lambda = \frac{6.626 \times 10^{-34}}{8.24 \times 10^{-28}} \) to calculate the wavelength. This will yield the wavelength in meters.
Step 5: Determine the region of the electromagnetic spectrum by comparing the calculated wavelength to known ranges of wavelengths for different regions (e.g., gamma rays, X-rays, ultraviolet, visible light, infrared, microwaves, and radio waves).

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Wavelength and Frequency Relationship

The wavelength (λ) of a photon is inversely related to its frequency (ν) through the equation c = λν, where c is the speed of light. This means that as the wavelength increases, the frequency decreases, and vice versa, which is crucial for determining the characteristics of electromagnetic radiation.
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