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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 vield more electrical energy on a per-photon basis?

Short Answer

Expert verified
On a per-photon basis, ultraviolet radiation would yield more electrical energy than infrared radiation when converted using photovoltaic cells, assuming equal efficiency of conversion. This is because ultraviolet radiation has a higher frequency and therefore higher photon energy compared to infrared radiation as per the formula \(E=hf\).

Step by step solution

01

Understanding the energy of photons

The energy of a photon can be calculated using the formula: \( E = hf \) where \(E\) is the energy of the photon, \(h\) is the Planck's constant (6.63 × 10^-34 Js), and \(f\) is the frequency of the radiation.
02

Comparing the frequency of infrared and ultraviolet radiation

Infrared radiation has a lower frequency and longer wavelength than ultraviolet radiation. This can be confirmed by comparing their positions in the electromagnetic spectrum. In the spectrum, the order is as follows: Radio waves -> Microwaves -> Infrared -> Visible light -> Ultraviolet -> X-rays -> Gamma rays As we move from left to right in the spectrum, the frequency increases and wavelength decreases.
03

Calculating energy of infrared and ultraviolet photons

Due to the lower frequency of infrared radiation compared to ultraviolet radiation, a single photon of infrared radiation will have lower energy than a single photon of ultraviolet radiation. Using the formula mentioned in Step 1, we can conclude that: \(E_{UV} = hf_{UV}\) \(E_{IR} = hf_{IR}\) Since, \(f_{UV} > f_{IR}\), it follows that \(E_{UV} > E_{IR}\).
04

Conclusion

On a per-photon basis, ultraviolet radiation would yield more electrical energy than infrared radiation when converted using photovoltaic cells, assuming equal efficiency of conversion.

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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 devices that convert light into electrical energy. They play a crucial role in harnessing solar power. Made from semiconductor materials, typically silicon, these cells utilize the energy from photons to generate electricity.
When a photon hits the cell, its energy is transferred to an electron in the semiconductor. This energy causes the electron to break free from its usual position, creating an electron-hole pair. The material's electric field directs these free electrons, creating a flow of electric current when the cell is connected to an external circuit.
Photovoltaic cells operate efficiently across different parts of the electromagnetic spectrum; however, they typically perform better with higher-energy photons like those from ultraviolet light. This is because higher energy photons can stimulate more electrons than lower-energy ones like infrared.
Electromagnetic Spectrum
The electromagnetic spectrum is a range of all types of electromagnetic radiation. These radiations include:
  • Radio waves
  • Microwaves
  • Infrared
  • Visible light
  • Ultraviolet
  • X-rays
  • Gamma rays
Each type of radiation in the spectrum has different wavelengths and frequencies.
As we move from radio waves to gamma rays on the spectrum, the frequency increases, and the wavelength decreases. This means ultraviolet light, which is further towards the gamma-ray side compared to infrared, has a higher frequency and hence higher energy per photon.
Understanding the electromagnetic spectrum helps us compare the energy potentials of different types of radiation when used in technologies like photovoltaic cells.
Ultraviolet Radiation
Ultraviolet (UV) radiation is part of the electromagnetic spectrum with higher frequencies than visible light but lower than X-rays. It is divided into three bands:
  • UVA
  • UVB
  • UVC
These bands can penetrate the earth's atmosphere to varying degrees.
UV radiation has enough energy to cause chemical reactions and physical damage to materials, including biological tissues. Thus, it’s often associated with skin damage but also has beneficial uses, like sterilization.
In the context of photovoltaic cells, UV radiation, with its high-energy photons, can potentially generate more electricity than lower-energy infrared radiation. This makes UV an interesting candidate for maximizing electric output from solar cells.
Infrared Radiation
Infrared (IR) radiation is a part of the electromagnetic spectrum. It lies between the visible light and microwave regions. IR is noteworthy for its ability to transfer heat. It's often experienced as warmth from the sun, fires, or heaters.
Infrared light has lower energy per photon than visible or ultraviolet light, which means it produces less electric current per photon in photovoltaic cells.
However, it is abundant and can be used in various applications such as remote controls, thermal imaging, and night-vision technology. While it doesn't match the energy potential of UV radiation in photovoltaic conversion per photon, its ubiquity still makes it useful in certain applications.
Planck's Constant
Planck's constant (\(h\)) is a fundamental constant in physics, crucial to quantum mechanics. Its value is \(6.63 \times 10^{-34} \text{ Js}\). It relates the energy of a photon to its frequency by the formula:\[E = hf\]
This formula shows how energy increases with frequency: when frequency \(f\) is higher, energy \(E\) is greater, assuming \(h\) is constant.
Planck's constant is pivotal in explaining why ultraviolet radiation has more energy per photon than infrared radiation. This concept helps in the evaluation of different radiations for their potential to generate electrical power through devices like photovoltaic cells.

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Most popular questions from this chapter

For each element, count the number of valence electrons, core electrons, and unpaired electrons in the ground state: (a) carbon, (b) phosphorus, (c) neon.

The discovery of hafnium, element number 72, provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements 58-71) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and \(G\). von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), is reacted with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 \mathrm{Write}\) a balanced chemical equation for the reaction. Starting with a 55.4-g sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

Indicate whether energy is emitted or absorbed when the following electronic transitions occur in hydrogen: (a) from \(n=2\) to \(n=6,(b)\) from an orbit of radius \(4.76 \AA\) to one of radius \(0.529 \AA,(c)\) from the \(n=6\) to the \(n=9\) state.

Identify the group of elements that corresponds to each of the following generalized electron configurations: (a) [noble gas] \(n s^{2} n p^{5}\) (b) [noble gas] \(n s^{2}(n-1) d^{2}\) (c) [noble gas] \(n s^{2}(n-1) d^{10} n p^{1}\) (d) [noble gas] \(n s^{2}(n-2) f^{6}\)

(a) The average distance from the nucleus of a 3 s electron in a chlorine atom is smaller than that for a \(3 p\) electron. In light of this fact, which orbital is higher in energy? (b) Would you expect it to require more or less energy to remove a \(3 s\) electron from the chlorine atom, as compared with a \(2 p\) electron? Explain.

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