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Photoelectric Effect and Blackbody Radiation

The Photoelectric Effect

It had been observed since the late 1800s that electrons could be ejected from the clean surface of a metal if irradiated with light. This phenomenon is called the photoelectric effect (Figure 1). Scientists observed that electrons could only be ejected when the frequency of the incident light was greater than some threshold frequency, and that the kinetic energy of the ejected electrons increased with increasing frequency of the incident light.  The electrons in the metal had a certain amount of binding energy keeping them bound. Thus, the incident light needed to have more energy than the binding energy in order to free the electrons.

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.”
Figure 1. 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.

According to classical wave theory, a wave’s energy depends on its intensity (which depends on its amplitude), not its frequency. Therefore, if the photoelectric effect were to be explained in terms of classical wave theory, changing the intensity of the incident light should have determined whether electrons could be ejected, not the frequency. Therefore, a pure classical wave theory explanation of the photoelectric effect was not possible. Scientists did note, however, that the intensity of the light did affect something—the number of electrons ejected in a given time period. In 1905, Albert Einstein was able to resolve the paradox.

Einstein argued that the light striking the metal surface should not be viewed as a wave (as in the view of classical wave theory), but instead as a stream of particles (later called photons) whose energy depends on their frequency according to the equation, E =  hν (where h is a fundamental physical constant known as Planck’s constant, 6.626 × 10-34 J·s). Another common form of this equation is in terms of wavelength using c = νλ, E = \dfrac{hc}{\lambda} (the E in these equations is the energy for a single photon). Einstein explained that electrons were ejected when hit by photons having sufficient energy (a frequency greater than the threshold). The greater the frequency of the incident photon (and thus the greater the energy of the photon), the greater the kinetic energy imparted to the escaping electrons. Einstein argued that, if we think of light consisting of photons, then the intensity of light depends on the number of photons striking the surface times the energy per photon, as opposed to the amplitude of the incoming wave. This explains why the number of ejected electrons increased with increasing intensity of incident light, since the greater the number of incoming photons, the greater the likelihood that they would collide with some of the electrons.

Einstein postulated that each metal has a particular electrostatic attraction for its electrons that must be overcome before an electron can be emitted from its surface. This value is called the binding energy (EBE). If photons of light with energy less than EBE strike a metal surface, no single photon has enough energy to eject an electron, so no electrons are emitted regardless of the intensity of the light. This is an important observation since it showed that the energy of the photons was quantized, or had discrete amounts of energy, and could not be added together to eject a single electron. One photon could eject, at most, one electron. If a photon with energy greater than EBE strikes the metal, then part of its energy is used to overcome the forces that hold the electron to the metal surface, and the excess energy appears as the kinetic energy (EKE) of the ejected electron:

Ephoton = EBE + EKE

Recall from physics that EKE = ½mv2, where m in this case is the mass of the electron (9.11 × 10-31 kg)  and v is the velocity of the electron (in m/s).

Because the photoelectric effect could not be described using wavelike properties, it suggested a particle-like nature of light where the energy of the electromagnetic radiation depends on the frequency (or wavelength) of light, not the intensity of the light. This was the first time that scientists had encountered something that exhibited both wave-like and particle properties. Sometimes it was better to treat electromagnetic radiation as a wave and sometimes as a stream of photons, two seemingly contradictory pictures of reality which independently could not fully explain the phenomena of light, but together they could. It took many years for the scientific community to accept the wave/particle nature of light, and even today the full meaning of wave/particle duality is still debated.

Example 1: 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.

  1. Increasing the brightness (intensity) of incoming light at a fixed frequency increases the kinetic energy of the ejected electrons.
  2. Increasing the wavelength of incoming light increases the kinetic energy of the ejected electrons.
  3. Increasing the brightness (intensity) of incoming light increases the number of ejected electrons.
  4. Increasing the frequency of incoming light can increase the number of ejected electrons.

Solution

  1. False. Increasing the brightness of incoming light at a fixed frequency has no effect on the kinetic energy of the ejected electrons. Only the frequency of light influences the kinetic energy of the electrons.
  2. 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.
  3. True. Because the number of collisions with photons increases with brighter light, the number of ejected electrons increases.
  4. 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. If you carefully explore the PHET demo, you will see that for very high energy photons there are exceptions to this rule.  We will not consider these exceptions in this class.
Example 2: Kinetic Energy of an Ejected Electron

Considering the photoelectric effect, if 331 kJ/mol is the electron binding energy for solid potassium, then:

  1. What minimum amount of energy does a photon need in order to cause a single electron to be emitted?
  2. If solid potassium is irradiated with 300 nm light, what is the maximum kinetic energy of the emitted electrons?

Solution 

(a) \left(\dfrac{331\;\text{kJ}}{\text{mol}\ e^{-}}\right) \left(\dfrac{1000\;\text{J}}{\text{kJ}}\right) \left(\dfrac{1\;\text{mol}\ e^{-}}{6.022 \times 10^{23}\;e^{-}}\right) = 5.50 × 10-19 J/e–

(b) We found the binding energy in (a) as 5.50 × 10-19 J/e–. We will use the 300 nm wavelength to determine the energy of the incoming light. The difference between the energy of the photon and the energy needed to remove the electron is the maximum kinetic energy of the electron.

Ephoton = EBinding Energy + EKE of ejected electrons

\dfrac{hc}{\lambda} = 5.50 × 10-19 J/e– + EKE of ejected electrons

\dfrac{(6.626 \times 10^{-34}\;\text{J}{\cdot}\text{s})(2.998 \times 10^8 \frac{\text{m}}{\text{s}})}{300 \times 10^{-9}\;\text{m}} = 5.50 × 10-19 J/e– + EKE of ejected electrons

EKE of ejected electron = 1.12 × 10-19 J/e–

Check Your Learning

Calculate the threshold energy (which is the same as the binding energy) in kJ/mol of electrons in aluminum, given that the lowest frequency photon for which the photoelectric effect is observed is 9.87 × 1014 Hz.

Answer

394 kJ/mol

Blackbody Radiation and the Ultraviolet Catastrophe

You have just seen that electromagnetic radiation demonstrates properties of particles called photons. The concept of photons was described by Einstein for the purposes of the photoelectric effect, however, several years earlier (1900) Max Planck, in an effort to describe blackbody radiation, developed mathematical expressions that, in hindsight, suggested the particle properties of light to Einstein.  His discovery resolved the apparent contradiction of blackbody radiation, the first phenomenon which caused scientists to recognize the fundamental limitations of classical physics.

A blackbody is a convenient, ideal emitter that approximates the behavior of many materials when heated. A blackbody is “ideal” in the same sense that an ideal gas is a convenient, simple representation of real gases that works well, provided that the pressure is not too high nor the temperature too low. A good approximation of a blackbody that can be used to observe blackbody radiation is a metal oven that can be heated to very high temperatures. The oven has a small hole allowing for the light being emitted within the oven to be observed with a spectrometer, an instrument that can measure the wavelengths and intensities of light. Figure 2 shows the resulting curves for some representative temperatures. Each distribution depends only on a single parameter: the temperature. The wavelength at which the maximum intensity is observed in the blackbody curves, λmax, shift to shorter wavelengths as the temperature increases, reflecting the observation that metals being heated to high temperatures begin to glow a darker red that becomes brighter as the temperature increases, eventually becoming white hot at very high temperatures as the intensities of all of the visible wavelengths become appreciable.

A graph is shown with a horizontal axis labeled, “Wavelength lambda (micrometers)” and a vertical axis labeled, “Intensity I (a r b. units).” The horizontal axis begins at 0 and extends to 3.0 with markings provided every 0.1 micrometer. Similarly, the vertical axis begins at 0 and extends to 10 with markings every 1 unit. Two vertical dashed lines are drawn. The first appears at about 0.39 micrometers and the second at about 0.72 micrometers. To the left of the first of these lines, the label, “Ultraviolet,” appears at the top of the graph. Between these lines, the label, “Visible,” appears at the top of the graph. To the right of the second of these lines, the label, “Infrared,” appears at the top of the graph. To the far right of the graph in open space a purple dot is placed which is labeled, “lambda maximum.” A “Temperature” label is located in a central region of the graph. A blue curve begins on the horizontal axis at about 0.05 micrometers. This curve increases steeply to a maximum value between the two vertical line segments of approximately 9.5 at about 0.55 micrometers. This curve decreases rapidly at first, then tapers off to reach a value of about 1.5 at the far right end of the graph. This blue curve is labeled 6000 K beneath the “Temperature” label. Curves are similarly drawn in green for 5000 K, orange for 4000 K, and red for 3000 K. As the temperature decreases, the height of the peak is lower and shifted right on the graph. The maximum value for the green curve is around 4.5 at 7.2 micrometers. This curve tapers at the right end of the graph to a value around 0.6. The maximum for the orange curve is around 2 at about 0.9 micrometers. This curve tapers at the right end of the graph to a value around 0.2. The maximum for the red curve is around 0.7 at about 1.2 micrometers. This curve tapers at the right end of the graph to a value around 0.1. The entire region under the blue curve that is between the two dashed lines, indicating the visible region, is shaded with vertical bands of color. The colors extending left to right across this region are violet, indigo, blue, green, yellow, orange, and red. A purple dot is placed at the peak of each of the four colored curves. These peaks are connected with a dashed curve.
Figure 2. Blackbody spectral distribution curves are shown for some representative temperatures.

Physicists derived mathematical expressions for the blackbody curves using well-accepted concepts from the theories of classical mechanics and classical electromagnetism. The theoretical expressions as functions of temperature fit the observed experimental blackbody curves well at longer wavelengths, but showed significant discrepancies at shorter wavelengths. Not only did the theoretical curves not show a peak, they absurdly showed the intensity becoming infinitely large as the wavelength became smaller and approached the UV region. This became known as the “ultraviolet catastrophe” because no one could find any problems with the theoretical treatment that could lead to such unrealistic short-wavelength behavior.

Finally, Max Planck derived a theoretical expression for blackbody radiation that fit the experimental observations exactly (within experimental error). Planck developed his theoretical treatment by extending the earlier work that had been based on the premise that the atoms composing the oven vibrated at increasing frequencies (or decreasing wavelengths) as the temperature increased, with these vibrations being the source of the emitted electromagnetic radiation. But where the earlier treatments had allowed the vibrating atoms to have any energy values obtained from a continuous set of energies (perfectly reasonable, according to classical physics), Planck found that by restricting the vibrational energies to discrete values for each frequency, he could derive an expression for blackbody radiation that correctly had the intensity dropping rapidly for the short wavelengths in the UV region: E = hν. We used this equation above to calculate the energy of a photon, but it originated with Planck in his description of blackbody radiation. The quantity h is a constant now known as Planck’s constant, in his honor. Although Planck was pleased he had resolved the blackbody radiation paradox, he was disturbed that to do so, he needed to assume the vibrating atoms required quantized energies, which he was unable to explain. The value of Planck’s constant is very small, 6.626 × 10−34 Joule seconds (J·s), which helps explain why energy quantization had not been observed previously in macroscopic phenomena.

Example 3: 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 = \dfrac{hc}{\lambda} = \dfrac{(6.626 \times 10^{-34}\;\text{J}{\cdot}\text{s})(2.998 \times 10^{8}\;\text{ms}^{-1})}{(640\;\text{nm})\left(\frac{1\;\text{m}}{10^{9}\;\text{nm}}\right)} = 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 × 109 Hz. What is the energy of one photon in these microwaves?

Answer

2 × 10−24 J

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