Principles of Physics: A Calculus-Based Text
5th Edition
ISBN: 9781133104261
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 28, Problem 1CQ
(a)
To determine
Identify the two major flaws of Rayleigh–Jeans law for blackbody
(b)
To determine
Explain how Planck’s law deals with the flaws that are in Rayleigh–Jeans law.
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A blackbody is an object with a radiation spectrum that is dependent solely on its tempera-
ture. A blackbody spectrum (or spectral radiancy curve) is described by the Planck Radiation
Law.
(a)
i. Sketch the spectral radiancy curves for blackbodies with temperatures of T = 4000 K
and T = 6000 K, respectively. Describe the main differences between the two
curves in terms of the appropriate physical laws defined as a function of tempera-
ture.
ii. What is the wavelength at peak intensity for each blackbody? State the part of
the electromagnetic spectrum to which each wavelength belongs.
(b) Use the Planck Radiation Law to determine the power radiated per unit area between
the wavelengths A 500 nanometres and λ = 503 nanometres for the T 6000 K
blackbody. What fraction of the blackbody's radiancy lies in this wavelength range?
=
The photoelectric equation for the kinetic energy of a photoelectron is, following Einstein, E <
hf – W, whereh is Planck's constant, f is the frequency of the light, and W is the work-function.
Sodium has W = 3.2×10-19 J. When sodium is illuminated by monochromatic light of a particular
frequency, electrons are emitted with speeds up to 8 x 105 ms-1.
a) Calculate the wavelength of the light.
b) Calculate the stopping potential.
Planck hypothesized that the blackbody radiation has discrete energy. Calculate the energy of a photon in Joule and electron volts. The frequency of that photon is 50 MHz.
a) 2.608 x 10-7J and 3.315 x 10-26 eV
b) 3.315 x 10-26 J and 2.608 x 10-7 eV
c) 2.608 x 10-26J and 3.315 x 10-7eV
d) 3.315 x 10-7J and 2.608 x 10-26 eV
Chapter 28 Solutions
Principles of Physics: A Calculus-Based Text
Ch. 28.1 - Prob. 28.1QQCh. 28.2 - Prob. 28.2QQCh. 28.2 - Prob. 28.3QQCh. 28.2 - Prob. 28.4QQCh. 28.5 - Prob. 28.5QQCh. 28.5 - Prob. 28.6QQCh. 28.6 - Prob. 28.7QQCh. 28.10 - Prob. 28.8QQCh. 28.10 - Prob. 28.9QQCh. 28.13 - Prob. 28.10QQ
Ch. 28 - Prob. 1OQCh. 28 - Prob. 2OQCh. 28 - Prob. 3OQCh. 28 - Prob. 4OQCh. 28 - Prob. 5OQCh. 28 - Prob. 6OQCh. 28 - Prob. 7OQCh. 28 - Prob. 8OQCh. 28 - Prob. 9OQCh. 28 - Prob. 10OQCh. 28 - Prob. 11OQCh. 28 - Prob. 12OQCh. 28 - Prob. 13OQCh. 28 - Prob. 14OQCh. 28 - Prob. 15OQCh. 28 - Prob. 16OQCh. 28 - Prob. 17OQCh. 28 - Prob. 18OQCh. 28 - Prob. 1CQCh. 28 - Prob. 2CQCh. 28 - Prob. 3CQCh. 28 - Prob. 4CQCh. 28 - Prob. 5CQCh. 28 - Prob. 6CQCh. 28 - Prob. 7CQCh. 28 - Prob. 8CQCh. 28 - Prob. 9CQCh. 28 - Prob. 10CQCh. 28 - Prob. 11CQCh. 28 - Prob. 12CQCh. 28 - Prob. 13CQCh. 28 - Prob. 14CQCh. 28 - Prob. 15CQCh. 28 - Prob. 16CQCh. 28 - Prob. 17CQCh. 28 - Prob. 18CQCh. 28 - Prob. 19CQCh. 28 - Prob. 20CQCh. 28 - Prob. 1PCh. 28 - Prob. 2PCh. 28 - Prob. 3PCh. 28 - Prob. 4PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 13PCh. 28 - Prob. 14PCh. 28 - Prob. 15PCh. 28 - Prob. 16PCh. 28 - Prob. 17PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Prob. 20PCh. 28 - Prob. 21PCh. 28 - Prob. 22PCh. 28 - Prob. 23PCh. 28 - Prob. 24PCh. 28 - Prob. 25PCh. 28 - Prob. 26PCh. 28 - Prob. 27PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 31PCh. 28 - Prob. 32PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - Prob. 37PCh. 28 - Prob. 38PCh. 28 - Prob. 39PCh. 28 - Prob. 40PCh. 28 - Prob. 41PCh. 28 - Prob. 42PCh. 28 - Prob. 43PCh. 28 - Prob. 44PCh. 28 - Prob. 45PCh. 28 - Prob. 46PCh. 28 - Prob. 47PCh. 28 - Prob. 48PCh. 28 - Prob. 49PCh. 28 - Prob. 50PCh. 28 - Prob. 51PCh. 28 - Prob. 52PCh. 28 - Prob. 53PCh. 28 - Prob. 54PCh. 28 - Prob. 55PCh. 28 - Prob. 56PCh. 28 - Prob. 57PCh. 28 - Prob. 58PCh. 28 - Prob. 59PCh. 28 - Prob. 60PCh. 28 - Prob. 61PCh. 28 - Prob. 62PCh. 28 - Prob. 63PCh. 28 - Prob. 64PCh. 28 - Prob. 65PCh. 28 - Prob. 66PCh. 28 - Prob. 67PCh. 28 - Prob. 68PCh. 28 - Prob. 69PCh. 28 - Prob. 70PCh. 28 - Prob. 71PCh. 28 - Prob. 72PCh. 28 - Prob. 73PCh. 28 - Prob. 74P
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- In an experiment on the photoelectric effect, a metal is illuminated by visible light of different wavelengths. A photoelectron has a maximum kinetic energy of 0.9 eV when red light of wavelength 640 nm is used. With blue light of wavelength 420 nm, the maximum kinetic energy of the photoelectron is 1.9 eV. Use this information to calculate an experimental value for the Planck constant h. [arrow_forwardConsider a black body of surface area 22.0 cm² and temperature 5700 K. (a) How much power does it radiate? 131675.5 W (b) At what wavelength does it radiate most intensely? 508.421 nm (c) Find the spectral power per wavelength at this wavelength. Remember that the Planck intensity is "intensity per unit wavelength", with units of W/m³, and "power per unit wavelength" is equal to that intensity times the surface area, with units of W/m 131.5775 Your response differs significantly from the correct answer. Rework your solution from the beginning and check each step carefully. W/marrow_forwardWhen you talk about FM radio stations and talk about the frequency it emits at, you are talking about the frequency of the electromagnetic wave that leaves the station. Traditionally radio stations are measured in MHz, where 1MHz = 106 Hz. If a FM radio station is at 93.8 MHz, what is the energy in Joules of the photon associated with this radio wave? Planck's Constant is 6.63 x 10-34 J*sarrow_forward
- For a black body, the temperature and the wavelength of the emission maximum, Amax, are related by Wein's Law, expressed as: T/°C λmax/nm Values of Amax from a small pinhole in an electrically heated container were determined at a series of temperatures. The results are given below. Deduce the value of Planck's constant. 1000 2181 c = 3.00 x 108 m/s 1500 1600 λmaxT = 2000 1240 k= 1.38 x 10-34 J-S hc 4.965k 2500 1035 3000 878 3500 763arrow_forwardA) Calculate the de Broglie wavelength of a neutron (mn = 1.67493×10-27 kg) moving at one six hundredth of the speed of light (c/600). Enter at least 4 significant figures. (I got the answer 949.4 pm but it is wrong, please help) B) Calculate the velocity of an electron (me = 9.10939×10-31 kg) having a de Broglie wavelength of 230.1 pm.arrow_forwardIn a particular photoelectric experiment, a stopping potential of 2.1 V is measured when ultraviolet light with a wavelength of 290 nm is incident on a metal. Given the light of speed c = 3.0 x 108 m/s and Planck constant h = 6.625 x 10-34 J. s or 4.14 × 10-15 eV.s %3D (a) Describe and illustrate the photoelectric experiment and explain why it cannot be explained by classical physics. (b) Using the same setup and metal, determine the stopping potential if blue light with a wavelength of 440 nm is used, instead of the ultraviolet light. (c) Using the same setup and metal, describe what happened if a red light with a wavelength of 620 nm is used, instead of the ultraviolet light.arrow_forward
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