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Chapter 38: Problem 24

(a) If the average frequency emitted by a \(120 \mathrm{~W}\) light bulb is \(5.00 \times 10^{14} \mathrm{~Hz}\) and \(10.0 \%\) of the input power is emitted as visible light, approximately how many visible-light photons are emitted per second? (b) At what distance would this correspond to \(1.00 \times 10^{11}\) visible-light photons per \(\mathrm{cm}^{2}\) per second if the light is emitted uniformly in all directions?

Short Answer

Expert verified
The light bulb emits approximately \(3.62 \times 10^{20}\) visible-light photons per second. The distance at which this corresponds to \(1.00 \times 10^{11}\) visible-light photons per \(\mathrm{cm}^{2}\) per second if the light is emitted uniformly in all directions is approximately \(r \approx 3.42 \mathrm{~m}\).

Step by step solution

01

Calculate the energy of one photon

First, calculate the energy of a single photon with the frequency given using the formula \(E = hf\), where \(E\) is the energy, \(h = 6.63 \times 10^{-34} \mathrm{~Js}\) is Planck鈥檚 constant, and \(f = 5.00 \times 10^{14} \mathrm{~Hz}\) is the frequency. This yields \(E = 6.63 \times 10^{-34} \mathrm{~Js} \times 5.00 \times 10^{14} \mathrm{~Hz} = 3.315 \times 10^{-19} \mathrm{~J}\).
02

Calculate the number of visible-light photons emitted per second

Next, calculate the number of photons per second. We know that \(10.0\%\) of the input power (\(120 \mathrm{~W}\)) is emitted as visible light, so the power in visible light is \(0.10 \times 120 \mathrm{~W} = 12 \mathrm{~W} = 12 \mathrm{~J/s}\). By dividing this power by the energy per photon, we find the number of photons per second, \(n = \frac{12 \mathrm{~J/s}}{3.315 \times 10^{-19} \mathrm{~J/photon}} \approx 3.62 \times 10^{20} \mathrm{~photons/s}\).
03

Calculate the distance at which a certain photon density occurs

Finally, solve for the distance from the light bulb at which there are \(1.00 \times 10^{11} \mathrm{~photons/cm}^{2}\mathrm{~s}\). This requires using the inverse square law normalizing for sphere area, where the intensity \(I = \frac{P}{4 \pi r^{2}}\). Solving for \(r\) gives \(r = \sqrt{\frac{P}{4 \pi I}}\). Substituting the given power for visible light (P) and the desired photon density (I) yields \(r = \sqrt{\frac{12 \mathrm{~W}}{4 \pi \times 1.00 \times 10^{11} \mathrm{~photons/cm}^{2}\mathrm{~s} \times 3.315 \times 10^{-19} \mathrm{~J/photon}}}\). After converting \( \mathrm{W}\) to \( \mathrm{J/s}\) and \( \mathrm{cm}^{2}\) to \( \mathrm{m}^{2}\) (remember there are \( \mathrm{10,000 cm}^{2}\) in a \( \mathrm{m}^{2}\)), gives us approximately \(r \approx 3.42 \mathrm{~m}\).

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Key Concepts

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

Planck's Constant
Planck's constant is a fundamental value in quantum mechanics, representing the quantization of energy. It is denoted as 'h' and has a value of approximately 6.63 x 10^-34 Joule seconds (Js). Planck's constant is a crucial factor in the calculation of energy of photons emitted from a light source.

In quantum physics, the energy (E) of a photon is calculated using the formula:
\( E = hf \),
where 'f' is the frequency of the photon. For visible light photons, which are the focus of many calculations, the frequency lies within the range of visible light spectrum. Planck's constant links the energy of a photon with its frequency, allowing us to understand the quantized nature of light.
Inverse Square Law
The inverse square law is a principle used to describe the phenomenon where a specified physical quantity or intensity is inversely proportional to the square of the distance from the source of that physical quantity. In terms of visible light photon emission, as you move further away from the light source, the intensity of light decreases with the square of the distance.

Mathematically, it is expressed as:
\( I = \frac{P}{4 \pi r^2} \),
where 'I' represents the intensity, 'P' is the power of the emitted light, and 'r' is the distance from the source. It's essential when calculating the density of photons at a given distance from a light source, helping us understand how the brightness we perceive diminishes as we get further away.
Visible Light Photon Emission

Quantifying Light's Output

When dealing with visible light photon emission from sources like a light bulb, it's beneficial to quantify how many photons are emitted per second. This emission is often quantified in terms of luminous efficacy, which reflects the light output per watt of electricity consumed.

Applying Planck's constant and the given frequency, one can determine the energy of a single photon and then calculate the number of photons emitted based on the power designated for visible light emission. For instance, if a 120 W light bulb emits 10% of its power as visible light, we can calculate the energy per photon and subsequently determine the number of photons produced per second by the bulb.
Quantum Physics

The Quantum World Explained

Quantum physics is the field of study that deals with the behavior of atoms and particles at the quantum level. As opposed to classical physics, quantum physics recognizes that energy is quantized, and can exist only in discrete amounts, which is where Planck's constant becomes significant. This discrete nature of energy is essential when talking about light, as light can be thought of as a stream of particles called photons, each carrying a quantized amount of energy determined by Planck鈥檚 constant.

Quantum physics rules govern the calculations for the emission of photons from light sources. It helps explain why light can act both as a particle (photon) and a wave (electromagnetic wave), a duality that significantly impacts the way we calculate light emissions and their interactions with the environment.

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Most popular questions from this chapter

X rays are produced in a tube operating at \(24.0 \mathrm{kV}\). After emerging from the tube, x rays with the minimum wavelength produced strike a target and undergo Compton scattering through an angle of \(45.0^{\circ} .\) (a) What is the original x-ray wavelength? (b) What is the wavelength of the scattered x rays? (c) What is the energy of the scattered x rays (in electron volts)?

A photon with wavelength \(\lambda\) is incident on a stationary particle with mass \(M,\) as shown in Fig. \(\mathbf{P 3 8 . 3 9 .}\) The photon is annihilated while an electron-positron pair is produced. The target particle moves off in the original direction of the photon with speed \(V_{M}\). The electron travels with speed \(v\) at angle \(\phi\) with respect to that direction. Owing to momentum conservation, the positron has the same speed as the electron. (a) Using the notation \(\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2}\) and \(\gamma_{M}=\left(1-V_{M}^{2} / c^{2}\right)^{-1 / 2},\) write a relativistic expression for energy conservation in this process. Use the symbol \(m\) for the electron mass. (b) Write an analogous expression for momentum conservation. (c) Eliminate the ratio \(h / \lambda\) between your energy and momentum equations to derive a relationship between \(v / c, V_{M} / c,\) and \(\phi\) in terms of the ratio \(m / M\), keeping \(\gamma\) and \(\gamma_{M}\) as a useful shorthand notation. (d) Consider the case where \(V_{M} \ll c\), and use a binomial expansion to derive an expression for \(\gamma_{M}\) to the first order in \(V_{M}\). Use that result to rewrite your previous result as an expression for \(V_{M}\) in terms of \(c, v, m / M,\) and \(\phi .\) (e) Are there choices of \(v\) and \(\phi\) for which \(V_{M}=0 ?\) (f) Suppose the target particle is a proton. If the electron and positron remain stationary, so that \(v=0\), then with what speed does the proton move, in \(\mathrm{km} / \mathrm{s} ?\) (g) If the electron and positron each have total energy \(5.00 \mathrm{MeV}\) and move with \(\phi=60^{\circ}\), then what is the speed of the proton? (Hint: First solve for \(\gamma\) and \(v .)\) (h) What is the energy of the incident photon in this case?

A beam of photons with just barely enough individual photon energy to create electron-positron pairs undergoes Compton scattering from free electrons before any pair production occurs. (a) Can the Compton-scattered photons create electron-positron pairs? (b) Calculate the momentum magnitude of the photons found at a \(20.0^{\circ}\) angle from the direction of the original beam.

Nuclear fusion reactions at the center of the sun produce gamma-ray photons with energies of about \(1 \mathrm{MeV}\left(10^{6} \mathrm{eV}\right) .\) By contrast, what we see emanating from the sun's surface are visiblelight photons with wavelengths of about \(500 \mathrm{nm}\). A simple model that explains this difference in wavelength is that a photon undergoes Compton scattering many times-in fact, about \(10^{26}\) times, as suggested by models of the solar interior - as it travels from the center of the sun to its surface. (a) Estimate the increase in wavelength of a photon in an average Compton-scattering event. (b) Find the angle in degrees through which the photon is scattered in the scattering event described in part (a). (Hint: A useful approximation is \(\cos \phi \approx 1-\phi^{2} / 2,\) which is valid for \(\phi \ll 1 .\) Note that \(\phi\) is in radians in this expression.) (c) It is estimated that a photon takes about \(10^{6}\) years to travel from the core to the surface of the sun. Find the average distance that light can travel within the interior of the sun without being scattered. (This distance is roughly equivalent to how far you could see if you were inside the sun and could survive the extreme temperatures there. As your answer shows, the interior of the sun is very opaque.)

An ultrashort pulse has a duration of \(9.00 \mathrm{fs}\) and produces light at a wavelength of \(556 \mathrm{nm}\). What are the momentum and momentum uncertainty of a single photon in the pulse?
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