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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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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photon Energy
Photon energy is a fundamental concept in understanding how light behaves. It refers to the energy carried by a single photon, the smallest unit of light. This energy is crucial because it relates directly to the frequency and wavelength of light. It can be determined using the formula:
\[ E = \frac{hc}{\lambda} \]
where:
  • \( E \) is the photon energy in joules (J)
  • \( h \) is Planck's constant \((6.63 \times 10^{-34} \text{ Js})\)
  • \( c \) is the speed of light \((3 \times 10^8 \text{ m/s})\)
  • \( \lambda \) is the wavelength of the light in meters (m)
This equation shows that the energy of a photon is inversely proportional to the wavelength. Hence, shorter wavelengths mean higher photon energy. This is why violet light, with its shorter wavelength, carries more energy than red light in the visible spectrum.
Visible Light
Visible light is a small portion of the electromagnetic spectrum that can be perceived by the human eye. It ranges from about \(0.40 \mu m\) to \(0.76 \mu m\). Within this range, different wavelengths correspond to different colors, each with unique properties.
  • Violet light has the shortest wavelength around \(0.40 \mu m\) and appears at one end of the visible spectrum.
  • Red light has the longest wavelength around \(0.76 \mu m\) and appears at the other end of the visible spectrum.
The sequence of colors visible to the human eye, often remembered as the acronym ROYGBIV (Red, Orange, Yellow, Green, Blue, Indigo, Violet), represents the order of light from longest to shortest wavelength. Each color corresponds to a specific frequency and therefore a specific energy level, affecting how we perceive color and intensity.
Wavelength
Wavelength is a key parameter in the study of light and electromagnetic waves. It defines the distance between consecutive peaks of a wave and plays an integral role in determining the characteristics of light, such as color and energy.
Wavelength is inversely related to frequency, as expressed in the formula:
\[ c = \lambda u \]
where:
  • \( c \) is the speed of light in a vacuum
  • \( \lambda \) is the wavelength
  • \( u \) is the frequency
Shorter wavelengths correlate with higher frequencies and vice versa. Consequently, shorter-wavelength light (like violet) contributes to higher photon energy, while longer-wavelength light (like red) contributes to lower photon energy. Understanding wavelength helps in determining how different colors of light behave and interact with matter.

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Most popular questions from this chapter

Describe the solar radiation properties of a window that is ideally suited for minimizing the air-conditioning load.

The reflectivity of aluminum coated with lead sulfate is \(0.35\) for radiation at wavelengths less than \(3 \mu \mathrm{m}\) and \(0.95\) for radiation greater than \(3 \mu \mathrm{m}\). Determine the average reflectivity of this surface for solar radiation \((T \approx 5800 \mathrm{~K})\) and radiation coming from surfaces at room temperature \((T \approx 300 \mathrm{~K})\). Also, determine the emissivity and absorptivity of this surface at both temperatures. Do you think this material is suitable for use in solar collectors?

What is the greenhouse effect? Why is it a matter of great concern among atmospheric scientists?

A radiometer can be used to determine the position of an approaching hot object by measuring the amount of irradiation it detects. Consider a radiometer placed at a distance \(H=0.5 \mathrm{~m}\) from the \(x\)-axis is used to measure the position of an approaching small blackbody object. If the radiometer is measuring \(80 \%\) of the irradiation corresponding to the position of the object directly under the radiometer \((x=0)\), determine the position of the object \(L\).

A semi-transparent plate \(\left(A_{1}=2 \mathrm{~cm}^{2}\right)\) has an irradiation of \(500 \mathrm{~W} / \mathrm{m}^{2}\), where \(30 \%\) of the irradiation is reflected away from the plate and \(50 \%\) of the irradiation is transmitted through the plate. A radiometer is placed \(0.5 \mathrm{~m}\) above the plate normal to the direction of viewing from the plate. If the temperature of the plate is uniform at \(350 \mathrm{~K}\), determine the irradiation that the radiometer would detect.
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