Skip to ContentGo to accessibility pageKeyboard shortcuts menu
OpenStax Logo
Physics

Critical Thinking Items

PhysicsCritical Thinking Items
Search for key terms or text.

Critical Thinking Items

21.1 Planck and Quantum Nature of Light

10 .
How is the frequency of a blackbody’s emission influenced by temperature?
  1. The frequency is independent of temperature.
  2. The frequency scales linearly with temperature.
  3. The frequency scales with T2.
  4. The frequency scales with T4.
11.

Why does the intensity shown in the blackbody radiation graph decrease after its peak frequency is achieved?

A graph of blackbody radiation that shows curves of the intensity of E M radiation versus wavelength in nanometers.
  1. Because after reaching the peak frequency, the photons created at a particular frequency are too many for energy intensity to continue to decrease.
  2. Because after reaching the peak frequency, the photons created at a particular frequency are too few for energy intensity to continue to decrease.
  3. Because after reaching the peak frequency, the photons created at a particular frequency are too many for energy intensity to continue to increase.
  4. Because after reaching the peak frequency, the photons created at a particular frequency are too few for energy intensity to continue to increase.
12 .
Photography development laboratories generally use red lighting to allow workers to see without damaging the developing photographs. Why might this be the case?
  1. Red light carries less damaging energy than other wavelengths of visible light.
  2. Red light carries more energy than other wavelengths of visible light and can pass through without causing damage.
  3. The greater frequency of red light compared with other visible light causes fewer vibrations in the developing photos.
  4. The shorter wavelength of red light compared with other visible light causes fewer vibrations in the developing photos
13.

Why is it assumed that a perfect absorber of light (like a blackbody) must also be a perfect emitter of light?

  1. To achieve electrostatic equilibrium with its surroundings
  2. To achieve thermal equilibrium with its surroundings
  3. To achieve mechanical equilibrium with its surroundings
  4. To achieve chemical equilibrium with its surroundings

21.2 Einstein and the Photoelectric Effect

14.

Light is projected onto a semi-conductive surface. If the intensity is held constant but the frequency of light is increased, what will happen?

  1. As frequency is increased, electrons will stop being ejected from the surface.
  2. As frequency is increased, electrons will begin to be ejected from the surface.
  3. As frequency is increased, it will have no effect on the electrons being ejected as the intensity is the same.
  4. As frequency is increased, the rate at which the electrons are being ejected will increase.
15.

Why is it important to consider what material to use when designing a light meter? Consider the worked example from Section 21-2 for assistance.

  1. A light meter should contain material that responds only to high frequency light.
  2. A light meter should contain material that responds to low frequency light.
  3. A light meter should contain material that has high binding energy.
  4. A light meter should contain a material that does not show any photoelectric effect.
16.

Why does overexposure to UV light often result in sunburn when overexposure to visible light does not? This is why you can get burnt even on a cloudy day.

  1. UV light carries less energy than visible light and can penetrate our body.
  2. UV light carries more energy than visible light, so it cannot break bonds at the cellular level.
  3. UV light carries more energy than visible light and can break bonds at the cellular level.
  4. UV light carries less energy than visible light and cannot penetrate the human body.
17.

If you pick up and shake a piece of metal that has electrons in it free to move as a current, no electrons fall out. Yet if you heat the metal, electrons can be boiled off. Explain both of these facts as they relate to the amount and distribution of energy involved with shaking the object as compared with heating it.

  1. Thermal energy is added to the metal at a much higher rate than energy added due to shaking.
  2. Thermal energy is added to the metal at a much lower rate than energy added due to shaking.
  3. If the thermal energy added is below the binding energy of the electrons, they may be boiled off.
  4. If the mechanical energy added is below the binding energy of the electrons, they may be boiled off.

21.3 The Dual Nature of Light

18 .
In many macroscopic collisions, a significant thermal energy term is required to conserve energy. Why is no such term required in the collision depicted in the figure?

Collision of an electron with a photon of energy E equal to h f is shown. The electron is represented as a spherical ball and the photon as an ellipse enclosing a wave. After collision the energy of the photon becomes E prime equal to h f prime and the final energy of an electron K E sub e is equal to E minus E prime. The direction of electron and photon before and after collision is represented by arrows.

  1. The figure depicts Rayleigh scattering, in which all terms are thermal already.
  2. The figure depicts Compton scattering, which is on the molecular level where all thermal energy exists as kinetic energy.
  3. The figure depicts a perfectly elastic collision, so no thermal term is required.
  4. The figure depicts Raman scattering, so all thermal energy is stored as vibrations in the particles.
19.

In what region of the electromagnetic spectrum will photons be most effective in accelerating a solar sail?

  1. ultraviolet rays
  2. infrared rays
  3. X-rays
  4. gamma rays
20.

True or false—Electron microscopes can resolve images that are smaller than the images resolved by light microscopes.

  1. false
  2. true
21 .
How would observations of Compton scattering change if ultraviolet light were used in place of X-rays?
  1. Ultraviolet light carries less energy than X-rays. As a result, Compton scattering would be easier to detect.
  2. Ultraviolet light carries less energy than X-rays. As a result, Compton scattering would be more difficult to detect.
  3. Ultraviolet light carries more energy than X-rays. As a result, Compton scattering would be easier to detect.
  4. Ultraviolet light has higher energy than X-rays. As a result, Compton scattering would be more difficult to detect.
PreviousNext
Order a print copy
Citation/Attribution

This book may not be used in the training of large language models or otherwise be ingested into large language models or generative AI offerings without OpenStax's permission.

Want to cite, share, or modify this book? This book uses the Creative Commons Attribution License and you must attribute Texas Education Agency (TEA). The original material is available at: https://www.texasgateway.org/book/tea-physics . Changes were made to the original material, including updates to art, structure, and other content updates.

Attribution information
  • If you are redistributing all or part of this book in a print format, then you must include on every physical page the following attribution:
    Access for free at https://openstax.org/books/physics/pages/1-introduction
  • If you are redistributing all or part of this book in a digital format, then you must include on every digital page view the following attribution:
    Access for free at https://openstax.org/books/physics/pages/1-introduction
Citation information

© Feb 19, 2025 Texas Education Agency (TEA). The OpenStax name, OpenStax logo, OpenStax book covers, OpenStax CNX name, and OpenStax CNX logo are not subject to the Creative Commons license and may not be reproduced without the prior and express written consent of Rice University.