What is the wavelength, in nm, of a photon with energy (a) 0.30 eV, (b) 3.0 eV, and (c) 30 eV? For each, is this wavelength visible, ultraviolet, or infrared light?
Ch 38: Quantization
Knight Calc5th EditionPhysics for Scientists and EngineersISBN: 9780137344796Not the one you use?Change textbook
Chapter 38, Problem 4
Photoelectrons are observed when a metal is illuminated by light with a wavelength less than 388 nm. What is the metal's work function?
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Identify the key concept: The photoelectric effect relates the energy of incident photons to the work function of the metal. The work function (Φ) is the minimum energy required to eject an electron from the metal surface.
Use the equation for the energy of a photon: \( E = \frac{hc}{\lambda} \), where \( h \) is Planck's constant (\( 6.626 \times 10^{-34} \, \text{J·s} \)), \( c \) is the speed of light (\( 3.00 \times 10^8 \; \text{m/s} \)), and \( \lambda \) is the wavelength of the light in meters.
Convert the given wavelength from nanometers to meters: \( \lambda = 388 \, \text{nm} = 388 \times 10^{-9} \, \text{m} \).
Substitute the values of \( h \), \( c \), and \( \lambda \) into the photon energy equation to calculate the energy of the photon: \( E = \frac{(6.626 \times 10^{-34})(3.00 \times 10^8)}{388 \times 10^{-9}} \). This energy corresponds to the threshold energy for ejecting electrons, which is equal to the work function \( \Phi \).
Conclude that the work function \( \Phi \) of the metal is equal to the calculated photon energy, as the wavelength given represents the threshold wavelength for photoemission.

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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Photoelectric Effect
The photoelectric effect is the phenomenon where electrons are emitted from a material, typically a metal, when it absorbs light of sufficient energy. This effect demonstrates the particle nature of light, where photons collide with electrons, providing them enough energy to escape the material's surface. The threshold wavelength, below which photoemission occurs, is crucial for determining the work function of the metal.
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Work Function
The work function is the minimum energy required to remove an electron from the surface of a metal. It is a characteristic property of each metal and is typically expressed in electron volts (eV). The work function can be calculated using the equation: φ = hc/λ, where φ is the work function, h is Planck's constant, c is the speed of light, and λ is the wavelength of the incident light.
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Photon Energy
Photon energy is the energy carried by a single photon, which can be calculated using the formula E = hc/λ. In this equation, E represents the energy, h is Planck's constant, c is the speed of light, and λ is the wavelength of the light. For the photoelectric effect to occur, the energy of the incoming photons must be greater than or equal to the work function of the metal.
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Related Practice
Textbook Question
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Textbook Question
A photoelectric-effect experiment finds a stopping potential of 1.56 V when light of 200 nm is used to illuminate the cathode. From what metal is the cathode made?
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Electrons in a photoelectric-effect experiment emerge from an aluminum surface with a maximum kinetic energy of 1.30 eV. What is the wavelength of the light?
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A 100 W incandescent lightbulb emits about 5 W of visible light. (The other 95 W are emitted as infrared radiation or lost as heat to the surroundings.) The average wavelength of the visible light is about 600 nm, so make the simplifying assumption that all the light has this wavelength. How many visible-light photons does the bulb emit per second?
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