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6.1 Electromagnetic energy  (Page 8/27)

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The figure includes three diagrams of waves approaching a flat, horizontal surface that is labeled, “Metal,” from an angle around 45 degrees above and to the left relative to the surface. At the top of the diagram at the center is the label, “E equals h nu.” At the left, a sinusoidal wave reaches the surface and stops. The portion of the diagram near the flat metal surface is labeled, “No electrons ejected,” and the wave is labeled, “700 n m.” To the right, a second similar, more compressed wave, which is labeled, “550 n m,” reaches the flat surface. This time, an arrow extends up and to the right at an angle of approximately 45 degrees. A tiny yellow circle with a negative sign in it is at the center of the arrow shaft. Above this arrow is the equation, “v subscript max equals 2.96 times 10 superscript 5 m per s.” To the far right, a third similar, even more compressed wave, which is labeled “400 n m” reaches the flat surface. This time, an arrow extends up and to the right at an angle of approximately 45 degrees. A tiny yellow circle with a negative sign in it is at the center of the arrow shaft. Above this arrow is the equation “v subscript max equals 6.22 times 10 superscript 5 m per s.”
Photons with low frequencies do not have enough energy to cause electrons to be ejected via the photoelectric effect. For any frequency of light above the threshold frequency, the kinetic energy of ejected electron will be proportional to the energy of the incoming photon.

Calculating the energy of radiation

When we see light from a neon sign, we are observing radiation from excited neon atoms. If this radiation has a wavelength of 640 nm, what is the energy of the photon being emitted?

Solution

We use the part of Planck's equation that includes the wavelength, λ , and convert units of nanometers to meters so that the units of λ and c are the same.

E = h c λ
E = ( 6.626 × 10 −34 J s ) ( 2.998 × 10 8 m s −1 ) ( 640 nm ) ( 1 m 10 9 nm ) E = 3.10 × 10 −19 J

Check your learning

The microwaves in an oven are of a specific frequency that will heat the water molecules contained in food. (This is why most plastics and glass do not become hot in a microwave oven-they do not contain water molecules.) This frequency is about 3 × 10 9 Hz. What is the energy of one photon in these microwaves?

Answer:

2 × 10 −24 J

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Use this simulation program to experiment with the photoelectric effect to see how intensity, frequency, type of metal, and other factors influence the ejected photons.

Photoelectric effect

Identify which of the following statements are false and, where necessary, change the italicized word or phrase to make them true, consistent with Einstein's explanation of the photoelectric effect.

(a) Increasing the brightness of incoming light increases the kinetic energy of the ejected electrons.

(b) Increasing the wavelength of incoming light increases the kinetic energy of the ejected electrons.

(c) Increasing the brightness of incoming light increases the number of ejected electrons.

(d) Increasing the frequency of incoming light can increase the number of ejected electrons.

Solution

(a) False. Increasing the brightness of incoming light has no effect on the kinetic energy of the ejected electrons. Only energy, not the number or amplitude, of the photons influences the kinetic energy of the electrons.

(b) False. Increasing the frequency of incoming light increases the kinetic energy of the ejected electrons. Frequency is proportional to energy and inversely proportional to wavelength. Frequencies above the threshold value transfer the excess energy into the kinetic energy of the electrons.

(c) True. Because the number of collisions with photons increases with brighter light, the number of ejected electrons increases.

(d) True with regard to the threshold energy binding the electrons to the metal. Below this threshold, electrons are not emitted and above it they are. Once over the threshold value, further increasing the frequency does not increase the number of ejected electrons

Check your learning

Calculate the threshold energy in kJ/mol of electrons in aluminum, given that the lowest frequency photon for which the photoelectric effect is observed is 9.87 × 10 14 Hz.

Answer:

3.94 × 10 5 kJ/mol

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Line spectra

Another paradox within the classical electromagnetic theory that scientists in the late nineteenth century struggled with concerned the light emitted from atoms and molecules. When solids, liquids, or condensed gases are heated sufficiently, they radiate some of the excess energy as light. Photons produced in this manner have a range of energies, and thereby produce a continuous spectrum in which an unbroken series of wavelengths is present. Most of the light generated from stars (including our sun) is produced in this fashion. You can see all the visible wavelengths of light present in sunlight by using a prism to separate them. As can be seen in [link] , sunlight also contains UV light (shorter wavelengths) and IR light (longer wavelengths) that can be detected using instruments but that are invisible to the human eye. Incandescent (glowing) solids such as tungsten filaments in incandescent lights also give off light that contains all wavelengths of visible light. These continuous spectra can often be approximated by blackbody radiation curves at some appropriate temperature, such as those shown in [link] .

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Source:  OpenStax, Chemistry. OpenStax CNX. May 20, 2015 Download for free at http://legacy.cnx.org/content/col11760/1.9
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