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Chapter 6: Problem 22

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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Most popular questions from this chapter

(a) A green laser pointer emits light with a wavelength of 532 nm. What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of 532 -nm photons. What is the energy gap between the ground state and excited state in the laser material?

(a) For an He^ + ion, do the 2 s and 2\(p\) orbitals have the same energy? If not, which orbital has a lower energy? (b) If we add one electron to form the He atom, would your answer to part (a) change?

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\(?\)

Use the de Broglie relationship to determine the wavelengths of the following objects: (a) an 85-kg person skiing at \(50 \mathrm{km} / \mathrm{hr},\) (b) a 10.0 -g bullet fired at \(250 \mathrm{m} / \mathrm{s},\) (c) a lithium atom moving at \(2.5 \times 10^{5} \mathrm{m} / \mathrm{s},(\mathbf{d})\) an ozone \(\left(\mathrm{O}_{3}\right)\) molecule in the upper atmosphere moving at 550 \(\mathrm{m} / \mathrm{s}\) .

The series of emission lines of the hydrogen atom for which \(n_{1}=3\) is called the Paschen series. (a) Determine the region of the electromagnetic spectrum in which the lines of the Paschen series are observed. (b) Calculate the wavelengths of the first three lines in the Paschen series - those for which \(n_{1}=4,5,\) and \(6 .\)
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