If we had two monochromatic (single-wavelength) light sources-a green one at \(\lambda=0.5 \mu \mathrm{m}\) and a near-infrared one at \(\lambda=1.0 \mu \mathrm{m}-\) each emitting photons at an energy rate of \(1 \mathrm{~W}, 103\) how does the number of photons emerging per second from each source compare? Is it the same number for each because both are \(1 \mathrm{~W}\) sources, or is it a different number-and by what factor, if so?

Short Answer

Expert verified
Question: Compare the number of photons emerging per second from two light sources having different wavelengths (green light at 0.5 micrometers and near-infrared light at 1.0 micrometers) if they emit photons at the same energy rate (1 W). Answer: The ratio of the number of photons emitted per second for green light (Ng) to near-infrared light (Nir) is given by: $$ \dfrac{N_g}{N_{ir}} = \dfrac{E_{ir}}{E_g} $$ After calculating the energy per photon for both wavelengths using Planck's formula and substituting these values, we can determine the ratio and conclude whether the number of photons emerging per second from each source is the same or different. If the calculated ratio is equal to 1, both sources emit the same number of photons per second. Otherwise, the ratio will represent the factor by which their emitted photon numbers per second differ.

Step by step solution

01

Find energy per photon for both wavelengths

We need to calculate the energy of a single photon for both green and near-infrared light. We will use the Planck's formula to do this which is given by: $$ E = \dfrac{hc}{\lambda} $$ where \(E\) is the energy of a single photon, \(h\) is the Planck's constant \((6.626 \times 10^{-34} \, Js)\), \(c\) is the speed of light \((3 \times 10^8 \, m/s)\), and \(\lambda\) is the wavelength. For green light, $$ E_g = \dfrac{(6.626 \times 10^{-34} \, Js)(3 \times 10^8\, m/s)}{0.5 \times 10^{-6}\, m} $$ For near-infrared light, $$ E_{ir} = \dfrac{(6.626 \times 10^{-34} \, Js)(3 \times 10^8\, m/s)}{1.0 \times 10^{-6}\, m} $$
02

Calculate the number of photons emitted per second

As both light sources have the same energy rate (1 W), we can find the number of photons emitted per second for each source by dividing the total power output (P) for each light source by the energy per photon. For green light, $$ N_g = \dfrac{P}{E_g} $$ For near-infrared light, $$ N_{ir} = \dfrac{P}{E_{ir}} $$
03

Compare the number of photons emitted from both sources

To compare the number of photons emitted per second for both sources, find the ratio of the two numbers: $$ \dfrac{N_g}{N_{ir}} = \dfrac{\dfrac{P}{E_g}}{\dfrac{P}{E_{ir}}} $$ As power is the same for both sources, the ratio simplifies to: $$ \dfrac{N_g}{N_{ir}} = \dfrac{E_{ir}}{E_g} $$
04

Conclusion

After calculating the energy per photon for both wavelengths and comparing the number of photons emitted per second, we can conclude whether the number of photons emerging per second from each source is the same or different. If the calculated ratio is equal to 1, it means that both sources emit the same number of photons per second. Otherwise, the ratio will represent the factor by which their emitted photon numbers per second differ.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Understanding Planck's Constant
Planck's constant (\textbf{h}) is a fundamental quantity in quantum mechanics and plays a vital role in understanding the energy of photons. This constant, approximately equal to \(6.626 \times 10^{-34} \, Js\), appears in many equations that bridge the gap between classical and quantum physics.

Discovered by Max Planck, this constant is pivotal in the formula \(E = \frac{hc}{\lambda}\), where \(E\) represents the energy of a photon, \(c\) is the speed of light, and \(\lambda\) is the wavelength of the light. Since \(h\) and \(c\) are constants, the energy of a photon is inversely proportional to its wavelength. This means that light with a shorter wavelength (like green light in the example) has higher energy photons compared to light with a longer wavelength (like near-infrared light).

Understanding Planck's constant is essential in many areas, including photon energy calculation, which we'll explore in the context of monochromatic light sources.
Monochromatic Light Sources
Monochromatic light sources emit light of a single wavelength, such as a laser or a specific LED. These sources are used in a variety of applications such as spectroscopy, communications, and medical procedures. The simplicity of their light spectrum makes it easy to study the properties of light and their interactions with matter.

In our exercise, green light and near-infrared light are used as examples of monochromatic light. While both are expressed in terms of a specific wavelength, they differ significantly in terms of photon energy because their wavelengths are different: \(0.5 \mu m\) for green light and \(1.0 \mu m\) for near-infrared. As a result, when analyzing the number of photons emitted per second from a 1 W source, the different energy levels of these photons play a crucial role.
Photon Energy Calculation
To calculate the energy of individual photons, we use the aforementioned Planck's equation \(E = \frac{hc}{\lambda}\). Here's how it's applied to our exercise. For a monochromatic source of green light with a wavelength of \(0.5 \mu m\), we can compute its photon energy and likewise for the near-infrared light source with a wavelength of \(1.0 \mu m\). Upon calculation, we find that the photon energy of green light is higher due to its shorter wavelength.

After obtaining the energy of a photon for each wavelength, we can calculate the number of photons emitted per second. Since the power output is 1 Watt for both, which is equivalent to 1 Joule per second, we divide this power by the energy of a single photon to determine the number of photons emitted. The key insight here is that the lower the energy per photon, the more photons need to be emitted to achieve the same power output.

In conclusion, the green light source emits fewer photons per second than the near-infrared source based on the calculated ratio. This hands-on approach to photon energy calculation not only reinforces the relevance of Planck's constant but also provides a clear example of how energy and wavelength interact to define the behavior of monochromatic light sources.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Solar photovoltaics are practical for individual homes, but solar thermal is only to be found in large utility-scale installations. What is the practical reason why we should not expect solar thermal installations on peoples' rooftops for electricity generation?

According to the table in Problem 26, San Diego can expect an annual average solar yield of \(5.7 \mathrm{kWh} / \mathrm{m}^{2} /\) day when the panel is tilted to the site latitude and facing south. \(^{116}\) If a household seeks to produce a modest \(8 \mathrm{kWh}\) per day using \(16 \%\) efficient panels, how large will the array need to be? Express as an area in square meters, and in side length for a square of the same area.

If a \(2.5\) electron-volt photon liberates an electron from silicon with a \(1.1 \mathrm{eV}\) band gap, how much kinetic energy does the emerging electron have? Express in both eV and Joules, and then determine the velocity of the electron if the electron mass is \(9 \times 10^{-31} \mathrm{~kg}\).

Which photons are most responsible for heating up a silicon photovoltaic panel in full sun: blue photons or infrared photons (beyond \(1.1 \mu \mathrm{m}\) )?

One way to look at solar payback time time is to note that an installed system will cost something like \(\$ 3,000\) for each \(\mathrm{kW}_{\mathrm{p}}\) (peak capacity), and that you'll produce \(x \mathrm{kWh}\) from that \(1 \mathrm{~kW}_{\mathrm{p}}\) array if your region gets \(x\) hours of full-sun- equivalent on average. Since each kWh of electricity costs something like \(\$ 0.15\), it becomes straightforward to compute the value per day as \(\$ 0.15 x\), and determine how long to match the \$3k investment. The result is independent of the actual array size, depending only on the cost per \(W_{p}\), the solar yield at your location, and the cost of electricity. What would the payback time be, in years, if the cost is \(\$ 3 / W_{p}, 17\) the yield is 6 hours per day of full-sun-equivalent, and electricity in your region costs \(\$ 0.15 / \mathrm{kWh}\) ?

See all solutions

Recommended explanations on Environmental Science Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free