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A 75 -W light source consumes 75 W of electrical power. Assume all this energy goes into emitted light of wavelength 600 \(\mathrm{nm}\) . (a) Calculate the frequency of the emitted light. (b) How many photons per second does the source emit? (c) Are the answers to parts (a) and (b) the same? Is the frequency of the light the same thing as the number of photons emitted per second? Explain.

Short Answer

Expert verified
(a) Frequency: \(5 \times 10^{14}\) Hz. (b) \(2.27 \times 10^{20}\) photons/second. (c) No, frequency is not the same as photon count.

Step by step solution

01

Calculate the Frequency of the Emitted Light

To calculate the frequency, we use the formula \( c = \lambda u \), where \( c \) is the speed of light, \( \lambda \) is the wavelength, and \( u \) is the frequency. Given that \( \lambda = 600 \ nm = 600 \times 10^{-9} \ m \) and \( c = 3 \times 10^8 \ m/s \), we rearrange to find \( u = \frac{c}{\lambda} \). \[ u = \frac{3 \times 10^8}{600 \times 10^{-9}} \]Upon calculating, \( u = 5 \times 10^{14} \ Hz \).
02

Calculate the Energy of a Single Photon

To find the energy of a single photon, use the formula \( E = h u \), where \( h = 6.626 \times 10^{-34} \ Js \) is Planck's constant and \( u = 5 \times 10^{14} \ Hz \) is the frequency calculated in Step 1.\[ E = 6.626 \times 10^{-34} \times 5 \times 10^{14} \]Thus, the energy \( E \approx 3.31 \times 10^{-19} \ Joules \).
03

Calculate Number of Photons Emitted Per Second

The total energy consumed per second by the light source is 75 Joules (since 75 Watts = 75 Joules/second). To find the number of photons emitted per second, divide the total energy by the energy of one photon.\[ N = \frac{75}{3.31 \times 10^{-19}} \]Calculating this gives \( N \approx 2.27 \times 10^{20} \) photons per second.
04

Explanation of the Relationship between Frequency and Photon Emission Rate

The answers to (a) and (b) are not the same because they address different concepts. Frequency refers to the number of wave cycles per second (in Hertz), while the number of photons emitted per second is simply a count of photons. The frequency is a property of individual light waves, independent of the total number of photons.

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Key Concepts

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

Frequency of Light
The frequency of light refers to the number of oscillations or cycles a light wave completes per second. It's measured in Hertz (Hz). You can think of it as how fast the light wave is "wiggling" as it travels. This concept is crucial when studying photon emission.

To find the frequency, we often use the equation:
  • \( c = \lambda u \), where:
    • \( c \) = Speed of light (approximately \( 3 \times 10^8 \ m/s \))
    • \( \lambda \) = Wavelength
    • \( u \) = Frequency
An example of this is calculating the frequency of a light with a wavelength of 600 nm, which gives us a frequency of \( 5 \times 10^{14} \ Hz \).

The frequency tells us how "color" or energy-laden a light wave is. Higher frequencies correspond to bluer light, while lower frequencies correspond to redder light.
Photon Energy
Photon energy is the energy carried by a single photon, the fundamental particle of light. This concept is pivotal in understanding how energy is absorbed, emitted, or transferred in light interactions. The energy of a photon is given by the formula:
  • \( E = h u \), where:
    • \( E \) = Energy of the photon
    • \( h \) = Planck's constant (\( 6.626 \times 10^{-34} \ Js \))
    • \( u \) = Frequency of the photon
By using the frequency calculated from our previous step, \( 5 \times 10^{14} \ Hz \), we find the energy of a single photon to be approximately \( 3.31 \times 10^{-19} \) Joules.

Photon energy is crucial in applications like photovoltaics, where light energy is converted into electrical energy, or in analyzing the efficiency of different light sources.
Planck's Constant
Planck's constant is a fundamental quantity in quantum mechanics, represented by \( h \). It relates the energy of photons to their frequency. This tiny constant, approximately \( 6.626 \times 10^{-34} \ Js \), bridges the often distinct worlds of wave-like and particle-like behaviors of light.

When calculating photon energy, Planck's constant is key. Without it, linking frequency and energy in the electromagnetic realm would be impossible.

This constant is named after Max Planck, who was a pioneering physicist in the development of quantum theory. Understanding and utilizing Planck's constant allows scientists to predict the energy outcomes of photons based on frequency and to delve deeper into the mysteries of atomic and subatomic processes.
Wavelength and Frequency Relationship
The relationship between wavelength and frequency is foundational in the study of light and photons. Essentially, as one increases, the other decreases. This inversely proportional relationship is defined by the speed of light equation:
  • \( c = \lambda u \)
  • Where \( c \) is constant, any increase in wavelength \( \lambda \) must result in a decrease in frequency \( u \), and vice versa.
  • This relationship impacts how we perceive light, as different frequencies correspond to different colors in the visible spectrum.
For example, in our exercise, a wavelength of 600 nm corresponds to a frequency of \( 5 \times 10^{14} \ Hz \). Comprehending this relationship helps when adjusting variables like wavelength or frequency during experiments and when designing optical equipment like lenses and lasers.

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Most popular questions from this chapter

Exposing Photographic Film. The light-sensitive compound on most photographic films is silver bromide, AgBr. A film is "exposed" when the light energy absorbed dissociates this molecule into its atoms. (The actual process is more complex, but the quantitative result does not differ greatly.) The energy of dissociation of AgBr is \(1.00 \times 10^{5} \mathrm{J} / \mathrm{mol} .\) For a photon that is just able to dissociate a molecule of silver bromide, find (a) the photon energy in electron volts; (b) the wavelength of the photon; (c) the frequency of the photon. (d) What is the energy in electron volts of a photon having a frequency of 100 MHz? (e) Light from a firelly can expose photographic film, but the radiation from an FM station broadcasting 50.000 \(\mathrm{W}\) at 100 \(\mathrm{MHz}\) cannot. Explain why this is so.

The photoelectric work function of potassium is 2.3 \(\mathrm{eV}\) If light having a wavelength of 250 nm falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy in electron volts of the most energetic electrons ejected; (c) the speed of these electrons.

When ultraviolet light with a wavelength of 254 nm falls on a clean copper surface, the stopping potential necessary to stop emission of photoelectrons is 0.181 \(\mathrm{V}\) (a) What is the photoelectric threshold wavelength for this copper surface? (b) What is the work function for this surface, and how does your calculated value compare with that given in Table 38.1\(?\)

An x-ray photon is scattered from a free electron (mass \(m )\) at rest. The wavelength of the scattered photon is \(\lambda^{\prime},\) and the final speed of the struck electron is \(v\) . (a) What was the initial wave-length \(\lambda\) of the photon? Express your answer in terms of \(\lambda^{\prime}, v\) and \(m .\) (Hint. Use the relativistic expression for the electron kinetic energy.) (b) Through what angle \(\phi\) is the photon scattered? Express your answer in terms of \(\lambda, \lambda^{\prime},\) and \(m\) . (c) Evaluate your results in parts (a) and (b) for a wavelength of \(5.10 \times 10^{-3}\) nm for the scattered photon and a final electron speed of \(1.80 \times 10^{8} \mathrm{m} / \mathrm{s}\) . Give \(\phi\) in degrees.

A beam of x rays with wavelength 0.0500 nm is Comptonscattered by the electrons in a sample. At what angle from the incident beam should you look to find x rays with a wavelength of (a) \(0.0542 \mathrm{nm} ;\) (b) 0.0521 \(\mathrm{nm}\) ; ( ) 0.0500 \(\mathrm{nm}\) ?

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