Fundamentals of Physics Extended
10th Edition
ISBN: 9781118230725
Author: David Halliday, Robert Resnick, Jearl Walker
Publisher: Wiley, John & Sons, Incorporated
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Chapter 38, Problem 81P
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Electron’s de Broglie wavelength in nanometers is
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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.
Answer= λ =hc/ϕ
PART C: Electrons emitted from a metal surface with a work function ϕ = 2.8 eV have a corresponding stopping potential of V0 = 3.6 V. If a metal with a work functionϕnew = 2.2 eV is illuminated by the same wavelength of light, what will be the new stopping potential? Express your answer with the appropriate units.
*Please answer Part C*
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
Find the de Broglie wavelength À for an electron moving at a speed of 1.00 × 106 m/s. (Note that this speed is low enough that
the classical momentum formula p = mv is still valid.) Recall that the mass of an electron is me = 9.11 × 10-³1 kg, and
Planck's constant is h = 6.626 × 10-34 J.s.
Chapter 38 Solutions
Fundamentals of Physics Extended
Ch. 38 - Prob. 1QCh. 38 - Prob. 2QCh. 38 - Prob. 3QCh. 38 - Prob. 4QCh. 38 - Prob. 5QCh. 38 - Prob. 6QCh. 38 - Prob. 7QCh. 38 - Prob. 8QCh. 38 - Prob. 9QCh. 38 - Prob. 10Q
Ch. 38 - Prob. 11QCh. 38 - Prob. 12QCh. 38 - Prob. 13QCh. 38 - Prob. 14QCh. 38 - Prob. 15QCh. 38 - Prob. 16QCh. 38 - Prob. 1PCh. 38 - Prob. 2PCh. 38 - Prob. 3PCh. 38 - Prob. 4PCh. 38 - Prob. 5PCh. 38 - Prob. 6PCh. 38 - Prob. 7PCh. 38 - Prob. 8PCh. 38 - Prob. 9PCh. 38 - Prob. 10PCh. 38 - Prob. 11PCh. 38 - Prob. 12PCh. 38 - Prob. 13PCh. 38 - Prob. 14PCh. 38 - Prob. 15PCh. 38 - Prob. 16PCh. 38 - Prob. 17PCh. 38 - Prob. 18PCh. 38 - Prob. 19PCh. 38 - Prob. 20PCh. 38 - Prob. 21PCh. 38 - Prob. 22PCh. 38 - Prob. 23PCh. 38 - Prob. 24PCh. 38 - Prob. 25PCh. 38 - Prob. 26PCh. 38 - Prob. 27PCh. 38 - Prob. 28PCh. 38 - Prob. 29PCh. 38 - Prob. 30PCh. 38 - Prob. 31PCh. 38 - Prob. 32PCh. 38 - Prob. 33PCh. 38 - Prob. 34PCh. 38 - Prob. 35PCh. 38 - Prob. 36PCh. 38 - Prob. 37PCh. 38 - Prob. 38PCh. 38 - Prob. 39PCh. 38 - Prob. 40PCh. 38 - Prob. 41PCh. 38 - Prob. 42PCh. 38 - Prob. 43PCh. 38 - Prob. 44PCh. 38 - Prob. 45PCh. 38 - Prob. 46PCh. 38 - Prob. 47PCh. 38 - Prob. 48PCh. 38 - Prob. 49PCh. 38 - Prob. 50PCh. 38 - Prob. 51PCh. 38 - Prob. 52PCh. 38 - Prob. 53PCh. 38 - Prob. 54PCh. 38 - Prob. 55PCh. 38 - Prob. 56PCh. 38 - Prob. 57PCh. 38 - Prob. 58PCh. 38 - Prob. 59PCh. 38 - Prob. 60PCh. 38 - Prob. 61PCh. 38 - Prob. 62PCh. 38 - Prob. 63PCh. 38 - Prob. 64PCh. 38 - Prob. 65PCh. 38 - Prob. 66PCh. 38 - Prob. 67PCh. 38 - Prob. 68PCh. 38 - Prob. 69PCh. 38 - Prob. 70PCh. 38 - Prob. 71PCh. 38 - Prob. 72PCh. 38 - Prob. 73PCh. 38 - Prob. 74PCh. 38 - Prob. 75PCh. 38 - Prob. 76PCh. 38 - Prob. 77PCh. 38 - Prob. 78PCh. 38 - Prob. 79PCh. 38 - Prob. 80PCh. 38 - Prob. 81PCh. 38 - Prob. 82PCh. 38 - Prob. 83PCh. 38 - Prob. 84PCh. 38 - Prob. 85PCh. 38 - Prob. 86PCh. 38 - Prob. 87PCh. 38 - Prob. 88PCh. 38 - Prob. 89PCh. 38 - Prob. 90P
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- An electron is accelerated through a potential difference of 26 000 V. What is the de Broglie wavelength of the electron (in m)?arrow_forwardThe diameter of an atomic nucleus is about 10 fm (1 fm = 10-15 m). What is the kinetic energy, in MeV, of a proton with a de Broglie wavelength of 10 fm?arrow_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_forward
- 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 264 nm is used and 2.3 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.) eV s From these data determine the work function (in eV) of the metal. eVarrow_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
- What is the de Broglie wavelength for an electron with speed (a) v = 0.469c and (b) v = 0.958c? (Hint: Use the correct relativistic expression for linear momentum if necessary.)arrow_forwardUsing the Broglie"s relationship, the velocity of an electron with a (mass e = 9.11×10-31 kg) having a wavelength of 700 nm is?arrow_forwardWhat is the de Broglie wavelength of a particle with a momentum of 1.11 x 10-10 kg·m/s? (h = 6.626 x 10^-34 J·s)arrow_forward
- The Kinetic Energy (K.E.), of a beam of electrons, accelerated through a potential V, equals the energy of a photon of wavelength 5460 nm. Find the de Broglie wavelength associated with this beam of electrons.arrow_forwardLight of wavelength 350 nm falls on a potassium surface, and the photoelectrons have amaximum kinetic energy of 1.3 eV.What is the work function of potassium?The speed of light is 3 × 108 m/s and Planck’sconstant is 6.63 × 10−34 J · s.Answer in units of eV. What is the threshold frequency for potassium?Answer in units of Hz.arrow_forward1.2. When a photoelectric cell with a copper cathode is illuminated by a Hg arc lamp, with wavelength 245 nm, the reverse potential across the cell (i.e. the voltage that just stops the fastest emitted electrons) is measured to be 0.181 V. Calculate the work function and threshold frequency of copper.arrow_forward
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