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Ch 38: Photons: Light Waves Behaving as Particles
Young & Freedman Calc - University Physics 14th Edition
Young & Freedman Calc14th EditionUniversity PhysicsISBN: 9780321973610Not the one you use?Change textbook
All textbooksYoung & Freedman Calc 14th EditionCh 38: Photons: Light Waves Behaving as ParticlesProblem 6
Chapter 38, Problem 6

The photoelectric threshold wavelength of a tungsten surface is 272272 nm. Calculate the maximum kinetic energy of the electrons ejected from this tungsten surface by ultraviolet radiation of frequency 1.45×10151.45\(\times\)10^{15} Hz. Express the answer in electron volts.

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Step 1: Understand the photoelectric effect equation, which is given by \( K_{\text{max}} = h \nu - \phi \), where \( K_{\text{max}} \) is the maximum kinetic energy of the ejected electrons, \( h \) is Planck's constant, \( \nu \) is the frequency of the incident radiation, and \( \phi \) is the work function of the material.
Step 2: Calculate the work function \( \phi \) using the threshold wavelength \( \lambda_{\text{threshold}} \). The relationship is \( \phi = \frac{hc}{\lambda_{\text{threshold}}} \), where \( c \) is the speed of light. Substitute \( h = 6.626 \times 10^{-34} \, \text{J·s} \), \( c = 3.00 \times 10^8 \, \text{m/s} \), and \( \lambda_{\text{threshold}} = 272 \, \text{nm} \) (convert to meters: \( 272 \, \text{nm} = 272 \times 10^{-9} \, \text{m} \)).
Step 3: Calculate the energy of the incident ultraviolet radiation using \( E = h \nu \), where \( \nu = 1.45 \times 10^{15} \, \text{Hz} \). Substitute \( h = 6.626 \times 10^{-34} \, \text{J·s} \) and \( \nu \) into the equation.
Step 4: Use the photoelectric effect equation \( K_{\text{max}} = h \nu - \phi \) to find the maximum kinetic energy of the ejected electrons. Subtract the work function \( \phi \) (calculated in Step 2) from the energy of the incident radiation \( h \nu \) (calculated in Step 3).
Step 5: Convert the maximum kinetic energy \( K_{\text{max}} \) from joules to electron volts (eV). Use the conversion factor \( 1 \, \text{eV} = 1.602 \times 10^{-19} \, \text{J} \). Divide \( K_{\text{max}} \) in joules by \( 1.602 \times 10^{-19} \) to express the result in electron volts.

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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 when it absorbs light or electromagnetic radiation. This effect demonstrates the particle nature of light, where photons transfer energy to electrons. The energy of the incoming photons must exceed a certain threshold to liberate electrons, which is determined by the material's work function.
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Threshold Wavelength

The threshold wavelength is the maximum wavelength of light that can cause the photoelectric effect in a given material. It is inversely related to the energy of the photons; shorter wavelengths correspond to higher energy. For tungsten, a threshold wavelength of 272 nm indicates that photons with wavelengths longer than this will not have enough energy to eject electrons.
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Kinetic Energy of Ejected Electrons

The kinetic energy of ejected electrons in the photoelectric effect can be calculated using the equation KE = E_photon - φ, where E_photon is the energy of the incoming photon and φ is the work function of the material. The energy of the photon can be determined from its frequency using E_photon = h * f, where h is Planck's constant. The result is typically expressed in electron volts (eV), a common unit for measuring energy at the atomic scale.
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Related Practice
Textbook Question

The photoelectric work function of potassium is 2.32.3 eV. If light that has a wavelength of 190190 nm falls on potassium, find the stopping potential in volts.

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Textbook Question

What would the minimum work function for a metal have to be for visible light (380380–750750 nm) to eject photoelectrons?

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Textbook Question

A laser used to weld detached retinas emits light with a wavelength of 652652 nm in pulses that are 20.020.0 ms in duration. The average power during each pulse is 0.6000.600 W. How much energy is in each pulse in joules? In electron volts?

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Textbook Question

The photoelectric work function of potassium is 2.32.3 eV. If light that has a wavelength of 190190 nm falls on potassium, find the kinetic energy, in electron volts, of the most energetic electrons ejected.

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Textbook Question

A photon has momentum of magnitude 8.24×10−288.24\(\times\)10^{-28} kg-m/s. What is the wavelength of this photon? In what region of the electromagnetic spectrum does it lie?

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Textbook Question

A photon has momentum of magnitude 8.24×10−288.24\(\times\)10^{-28} kg-m/s. What is the energy of this photon? Give your answer in joules and in electron volts.

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