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It is possible to convert radiant energy into electrical energy using photovoltaic cells. Assuming equal efficiency of conversion, would infrared or ultraviolet radiation yield more electrical energy on a per-photon basis?

Short Answer

Expert verified
Ultraviolet radiation yields more electrical energy on a per-photon basis as compared to infrared radiation, as the energy of a photon is inversely proportional to its wavelength and ultraviolet radiation has shorter wavelengths than infrared radiation.

Step by step solution

01

Identify the formulas to calculate energy per photon

We will use the formula for the energy of a photon: Energy of photon (E) = Planck's constant (h) × frequency (ν) = \(h \times \frac{c}{λ}\) where - h = Planck's constant, approximately \(6.626 \times 10^{-34} Js\) - c = speed of light, approximately \(3.00 \times 10^8 m/s\) - λ (lambda) is the wavelength of the radiation. We know that the wavelength of infrared radiation is longer than that of ultraviolet radiation.
02

Compare the wavelengths

Based on the electromagnetic spectrum: - Infrared radiation has a wavelength range of \(700 nm\) to \(1 mm\) (\(10^{-6} m\) to \(10^{-3} m\)) - Ultraviolet radiation has a wavelength range of \(10 nm\) to \(400 nm\) (\(10^{-8} m\) to \(4 \times 10^{-7} m\)) Since the wavelengths of infrared radiation are generally longer than those of ultraviolet radiation, we can assume that λ (infrared) > λ (ultraviolet).
03

Calculate and compare the energy of a photon for infrared and ultraviolet radiation

Using the formula E = \(h \times \frac{c}{λ}\), we can see that the energy of a photon is inversely proportional to its wavelength: - E (infrared) = \(h \times \frac{c}{λ_{infrared}}\) - E (ultraviolet) = \(h \times \frac{c}{λ_{ultraviolet}}\) Since λ (infrared) > λ (ultraviolet), we find that: E (infrared) < E (ultraviolet)
04

Conclusion

Ultraviolet radiation yields more electrical energy on a per-photon basis as compared to infrared radiation.

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Key Concepts

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

Photovoltaic Cells
Photovoltaic cells are amazing devices! They transform light energy directly into electricity. They're key components in solar panels. How do they work?
First, they use a photoelectric effect. This is when light falls onto a material, usually silicon, which releases electrons. These free electrons move and create an electric current. That electric current is what power your gadgets and homes.
Photovoltaic cells have several benefits:
  • They generate clean energy with no pollution.
  • They're renewable, meaning they won't run out.
  • Once installed, they have low maintenance costs.
However, their efficiency depends on the type of light and conditions like the angle of sunlight and temperature. Understanding how different kinds of light affect them is crucial.
Electromagnetic Spectrum
The electromagnetic spectrum is a way to categorize all the different types of light. It includes everything from radio waves to gamma rays. Light can be a wave with various wavelengths. The spectrum arranges them based on these wavelengths.
Here are some common parts of the electromagnetic spectrum:
  • Radio Waves: Used in broadcasting and communications.
  • Microwaves: Great for cooking and satellite transmissions.
  • Infrared: Felt as heat, and used in remote controls.
  • Visible Light: The light we can actually see!
  • Ultraviolet: A bit more energetic, responsible for sunburns.
  • X-rays and Gamma Rays: Used in medical imaging.
All these waves travel at the speed of light but differ in frequency and wavelength. Understanding the spectrum helps us see how different wavelengths carry different energies.
Ultraviolet Radiation
Ultraviolet (UV) radiation is high-energy. It sits between visible light and X-rays on the electromagnetic spectrum. UV rays are invisible to our eyes but have significant effects.
There are a few important things about UV radiation:
  • It has shorter wavelengths (10-400 nm) compared to visible light.
  • More energy means it can cause chemical changes. This is why it tans and can damage skin.
  • It's vital in helping our bodies produce Vitamin D, while too much can harm cells.
For photovoltaic cells, UV radiation has higher energy per photon compared to other types like infrared. This means more potential electricity produced from UV rays. Protecting against excessive UV exposure is important for health.
Infrared Radiation
Infrared radiation is right next to visible light on the electromagnetic spectrum, with longer wavelengths (700 nm to 1 mm) than visible light.
This type of radiation is fascinating because:
  • We feel it as heat; that's how infrared heaters warm spaces.
  • It's used in night vision equipment, allowing us to see in the dark.
  • Infrared photography captures details not seen by the naked eye.
Infrared photons have less energy than ultraviolet ones due to their longer wavelengths. Thus, while they still contribute to electricity production in photovoltaic cells, the output is lower per photon compared to UV light. Still, they have many valuable applications, especially in comfort and technology.

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Most popular questions from this chapter

Write the condensed electron configurations for the following atoms and indicate how many unpaired electrons each has: (a) \(\mathrm{Mg},(\mathbf{b}) \mathrm{Ge},(\mathbf{c}) \mathrm{Br},(\mathbf{d}) \mathrm{V},(\mathbf{e}) \mathrm{Y},(\mathbf{f}) \mathrm{Lu}\)

In the experiment shown schematically below, a beam of neutral atoms is passed through a magnetic field. Atoms that have unpaired electrons are deflected in different directions in the magnetic field depending on the value of the electron spin quantum number. In the experiment illustrated, we envision that a beam of hydrogen atoms splits into two beams. (a) What is the significance of the observation that the single beam splits into two beams? (b) What do you think would happen if the strength of the magnet were increased? (c) What do you think would happen if the beam of hydrogen atoms were replaced with a beam of helium atoms? Why? (d) The relevant experiment was first performed by Otto Stern and Walter Gerlach in \(1921 .\) They used a beam of Ag atoms in the experiment. By considering the electron configuration of a silver atom, explain why the single beam splits into two beams.

List the following types of electromagnetic radiation in order of increasing wavelength: (a) the gamma rays produced by a radioactive nuclide used in medical imaging; (b) radiation from an FM radio station at 93.1 \(\mathrm{MHz}\) on the dial; (c) a radio signal from an AM radio station at 680 \(\mathrm{kHz}\) on the dial; ( d ) the yellow light from sodium vapor streetlights; (e) the red light of a light-emitting diode, such as in a calculator display.

Consider a transition in which the electron of a hydrogen atom is excited from \(n=1\) to \(n=\infty\) . (a) What is the end result of this transition? (b) What is the wavelength of light that must be absorbed to accomplish this process? (c) What will occur if light with a shorter wavelength that in part (b) is used to excite the hydrogen atom? (d) How are the results of parts (b) and (c) related to the plot shown in Exercise 6.88\(?\)

A hydrogen atom orbital has \(n=5\) and \(m_{l}=-2 .\) (a) What are the possible values of \(l\) for this orbital? (b) What are the possible values of \(m_{s}\) for the orbital?

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