Fundamentals of Physics, Binder Ready Version
10th Edition
ISBN: 9781118230640
Author: Halliday, David; Resnick, Robert; Walker, Jearl
Publisher: WILEY
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Chapter 38, Problem 81P
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Electron’s de Broglie wavelength in nanometers is
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In a photoelectric experiment it is found that a stopping potential of 1.00 V is needed to stop all the electrons when incident light of wavelength 225 nm is used and 1.5 V is needed for light of
wavelength 207 nm.
From these data determine Planck's constant. (Enter your answer, in eV s, to at least four significant figures.)
4.2367e-15 X ev s
From these data determine the work function (in eV) of the metal.
4.6
X ev
PART A: A metal surface is illuminated with photons with a frequency f=1.5×10^15 Hz. The stopping potential for electrons photoemitted from the surface is 3.6 V. What is the work function of the metal?
Answer= 2.6 eV
PART B: A certain metal has a work function ϕ. What is the maximum photon wavelength that will produce photoemission? Express your answer in terms of ϕ,Planck's constant h, and the speed of light c.
*Please answer Part B*
Consider an electron with kinetic energy E = 10 eV. Calculate the de Broglie wavelength of the electron. Express your answer in Angstroms. 1 Angstrom = 1 x 10-10 m.
Chapter 38 Solutions
Fundamentals of Physics, Binder Ready Version
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- A sodium atom in one of the states labeled “Lowest excited levels” in Fig. remains in that state, on average, for 1.6 * 10-8 s before it makes a transition to the ground state, emitting a photon with wavelength 589.0 nm and energy 2.105 eV. What is the uncertainty in energy of that excited state? What is the wavelength spread of the corresponding spectral line?arrow_forwardDe Broglie postulated that the relationship λ =h/p is valid for relativistic particles. What is the de Broglie wavelength for a (relativistic) electron having a kinetic energy of 3.00 MeV?arrow_forwardThrough what potential difference ΔVΔV must electrons be accelerated (from rest) so that they will have the same wavelength as an x-ray of wavelength 0.130 nmnm? Use 6.626×10−34 J⋅sJ⋅s for Planck's constant, 9.109×10−31 kgkg for the mass of an electron, and 1.602×10−19 CC for the charge on an electron. Express your answer using three significant figures. =89.0 V Through what potential difference ΔVΔV must electrons be accelerated so they will have the same energy as the x-ray in Part A? Use 6.626×10−34 J⋅sJ⋅s for Planck's constant, 3.00×108 m/sm/s for the speed of light in a vacuum, and 1.602×10−19 CC for the charge on an electron. Express your answer using three significant figures. Second question is what I need help on! Thanks!arrow_forward
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