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Chapter 6: Problem 18

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
The energy of a single photon is inversely proportional to its wavelength. Since ultraviolet radiation has shorter wavelengths than infrared radiation, ultraviolet photons will have more energy than infrared photons. Therefore, ultraviolet radiation would yield more electrical energy on a per-photon basis, assuming equal efficiency in photovoltaic cell conversion.

Step by step solution

01

Recall the relationship between energy, frequency, and wavelength

The energy of a single photon (E) can be calculated using the formula: \[ E = h\nu \] where \(h\) is Planck's constant (\(6.63 \times 10^{-34} Js\)) and \(\nu\) is the frequency of the photon. We can also relate the energy to the wavelength (\(\lambda\)) using the speed of light (c) relationship: \[\nu = \cfrac{c}{\lambda} \]
02

Substitute the wavelength-energy relationship into the energy-frequency relationship

Now that we know both relationships, we can replace \(\nu\) in our energy equation with the wavelength-energy relationship: \[E = h \cdot \cfrac{c}{\lambda} \]
03

Compare infrared and ultraviolet wavelengths

For infrared radiation, the range of wavelengths is about 700 nm to 1 mm, while ultraviolet radiation has wavelengths ranging from 10 nm to 400 nm. Since the energy of a photon is inversely proportional to its wavelength, a photon with a shorter wavelength will have higher energy than one with a longer wavelength.
04

Determine which radiation has more energy per photon

As ultraviolet radiation has shorter wavelengths than infrared radiation, we can conclude that ultraviolet radiation photons will have more energy than those of infrared radiation. Therefore, ultraviolet radiation would yield more electrical energy on a per-photon basis, assuming equal efficiency in photovoltaic cell conversion.

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Most popular questions from this chapter

(a) What is the frequency of radiation that has a wavelength of \(10 \mu \mathrm{m}\), about the size of a bacterium? (b) What is the wavelength of radiation that has a frequency of \(5.50 \times 10^{14} \mathrm{~s}^{-1} ?\) (c) Would the radiations in part (a) or part (b) be visible to the human eye? (d) What distance does electromagnetic radiation travel in \(50.0 \mu\) s?

Calculate the uncertainty in the position of (a) an electron moving at a speed of \((3.00 \pm 0.01) \times 10^{5} \mathrm{~m} / \mathrm{s},(\mathrm{b})\) a neutron moving at this same speed. (The masses of an electron and a neutron are given in the table of fundamental constants in the inside cover of the text.) (c) What are the implications of these calculations to our model of the atom?

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of \(9.38 \times 10^{6} \mathrm{~m} / \mathrm{s}\). What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

Neutron diffraction is an important technique for determining the structures of molecules. Calculate the velocity of a neutron needed to achieve a wavelength of \(0.955 \AA\). (Refer to the inside cover for the mass of the neutron).

Using only a periodic table as a guide, write the condensed electron configurations for the following atoms: (a) Se, (b) \(\mathrm{Rh}\), (c) \(\mathrm{Si}\), (d) \(\mathrm{Hg}\), (e) Hf.
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