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Chapter 12: Problem 17

The electromagnetic spectrum that lies between \(0.40\) and \(0.76 \mu \mathrm{m}\) is what we call visible light. Within this spectrum, the color violet has the shortest wavelength while the color red has the longest wavelength. Determine which of these colors, violet \((\lambda=0.40 \mu \mathrm{m})\) or red \((\lambda=0.76 \mu \mathrm{m})\), propagates more photon energy.

Short Answer

Expert verified
Answer: Violet light carries more photon energy than red light.

Step by step solution

01

Write down the given values and constants

We are given the following values: - Wavelength of violet light: \(\lambda_v=0.40 \mu \mathrm{m}\) - Wavelength of red light: \(\lambda_r=0.76 \mu \mathrm{m}\) And the following constants: - Planck's constant (h): \(6.63\times 10^{-34} \mathrm{Js}\) - Speed of light (c): \(3\times 10^{8} \mathrm{m/s}\)
02

Convert wavelengths to meters

We need to convert the given wavelengths from micrometers to meters: - Wavelength of violet light: \(\lambda_v = 0.40 \times 10^{-6} \mathrm{m}\) - Wavelength of red light: \(\lambda_r = 0.76 \times 10^{-6} \mathrm{m}\)
03

Calculate the photon energy of violet light

Using the formula \(E = \frac{hc}{\lambda}\), we can calculate the energy of violet light: \(E_v = \frac{6.63\times 10^{-34} \mathrm{Js} \times 3\times 10^{8} \mathrm{m/s}}{0.40 \times 10^{-6} \mathrm{m}} = 4.97\times 10^{-19} \mathrm{J}\)
04

Calculate the photon energy of red light

Using the same formula, we can calculate the energy of red light: \(E_r = \frac{6.63\times 10^{-34} \mathrm{Js} \times 3\times 10^{8} \mathrm{m/s}}{0.76 \times 10^{-6} \mathrm{m}} = 2.61\times 10^{-19} \mathrm{J}\)
05

Compare the photon energies

Now we can compare the photon energies of violet and red light: - Violet light energy: \(4.97\times 10^{-19} \mathrm{J}\) - Red light energy: \(2.61\times 10^{-19} \mathrm{J}\) Since the energy of violet light is higher than the energy of red light, violet light propagates more photon energy.

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Most popular questions from this chapter

Explain why the sky is blue and the sunset is yellow-orange.

A small circular surface of area \(A_{1}=2 \mathrm{~cm}^{2}\) located at the center of a 2-m-diameter sphere emits radiation as a blackbody at \(T_{1}=1000 \mathrm{~K}\). Determine the rate at which radiation energy is streaming through a \(D_{2}=1\)-cm-diameter hole located \((a)\) on top of the sphere directly above \(A_{1}\) and \((b)\) on the side of sphere such that the line that connects the centers of \(A_{1}\) and \(A_{2}\) makes \(45^{\circ}\) with surface \(A_{1}\).

A circular ceramic plate that can be modeled as a blackbody is being heated by an electrical heater. The plate is \(30 \mathrm{~cm}\) in diameter and is situated in a surrounding ambient temperature of \(15^{\circ} \mathrm{C}\) where the natural convection heat transfer coefficient is \(12 \mathrm{~W} / \mathrm{m}^{2} \cdot \mathrm{K}\). If the efficiency of the electrical heater to transfer heat to the plate is \(80 \%\), determine the electric power that the heater needs to keep the surface temperature of the plate at \(200^{\circ} \mathrm{C}\).

It is desired that the radiation energy emitted by a light source reach a maximum in the blue range \((\lambda=0.47 \mu \mathrm{m})\). Determine the temperature of this light source and the fraction of radiation it emits in the visible range \((\lambda=0.40-0.76 \mu \mathrm{m})\).

The sun can be treated as a blackbody at an effective surface temperature of \(10,400 \mathrm{R}\). Determine the rate at which infrared radiation energy \((\lambda=0.76-100 \mu \mathrm{m})\) is emitted by the sun, in Btu/h. \(\mathrm{ft}^{2}\).
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