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Confirm the statement in the text that the range of photon energies for visible light is 1.63 to 3.26 eV, given that the range of visible wavelengths is 380 to 760 nm.
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$\lambda_{min} = 380 \textrm{ nm}$
$\lambda_{max} = 761 \textrm{ nm}$
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OpenStax College Physics Solution, Chapter 29, Problem 25 (Problems & Exercises) (1:51)

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Video Transcript
This is College Physics Answers with Shaun Dychko. We are told that the visible spectrum ranges from a wavelength of 380 nanometers at the minimum up to 760 nanometers at a maximum wavelength. And we are going to confirm that these wavelengths correspond to energies of 1.63 electron volts to 3.26 electron volts. So here's our formula for energy of a photon; it's Planck's constant times the frequency of the light and we'll need to replace f with an expression in terms of λ because that's what the question gives us. So we know from the wave equation that the speed of a wave is its frequency times its wavelength and we'll solve for we'll solve for f here actually, not λ. We'll solve for f by dividing both sides by λ. So we have f is c over λ after we switch the sides around too. So f gets replaced by c over λ in this equation here. And we can solve for λ by multiplying both sides by λ over energy. And then we have an expression for the wavelength of a photon, in terms of its energy. So the minimum wavelength is Planck's constant times c, which is 1240 electron volt nanometers divided by the maximum energy which is 3.26 electron volts and we get 380 nanometers which is what we expected and so we have confirmed that is the minimum wavelength possible. And then to find the maximum wavelength possible, we'll substitute in the minimum energy in the denominator and we end up with 761 nanometers which is close enough to 760 and so we have confirmed that as well.