FUNDAMENTALS OF PHYSICS EXTEND 11E
11th Edition
ISBN: 9781119813293
Author: Halliday
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 EXTEND 11E
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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- 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*arrow_forwardFind 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.arrow_forwardAn electron is accelerated through a potential difference of 26 000 V. What is the de Broglie wavelength of the electron (in m)?arrow_forward
- The 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_forwardIn 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_forward
- 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
- Using 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_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_forward
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