University Physics with Modern Physics (14th Edition)
14th Edition
ISBN: 9780321973610
Author: Hugh D. Young, Roger A. Freedman
Publisher: PEARSON
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Chapter 38, Problem 38.9DQ
To determine
The reason for the photocurrent to have the same value for large positive values of
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The photoelectric equation for the kinetic energy of a photoelectron is, following Einstein, E ≤ hf − W, where h is Planck’s constant, f is the frequency of the light and W is the work-function.When cesium is illuminated by monochromatic ultraviolet light of wavelength 365 nm, electronsare emitted. The stopping potential is found to be 1.3 V.
a) Calculate the work function W of cesium.b) Calculate the maximum speed of the electrons.
UV radiation having a wavelength of 120 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum velocity of the ejected photoelectrons? No need to use relativistic formulas in this case, so you can just use the standard formula KE =12??2=12mv2.
A metal surface is illuminated with light of different wavelengths and the corresponding stopping
potentials of the photoelectrons are shown in the Table below.
A (A) 5581 61566731
V (V)1,05 0,86 0,67
Using the Table, determine the photoelectric threshold wavelength (in Å).
Chapter 38 Solutions
University Physics with Modern Physics (14th Edition)
Ch. 38.1 - Silicon films become better electrical conductors...Ch. 38.2 - Prob. 38.2TYUCh. 38.3 - Prob. 38.3TYUCh. 38.4 - Prob. 38.4TYUCh. 38 - Prob. 38.1DQCh. 38 - Prob. 38.2DQCh. 38 - Prob. 38.3DQCh. 38 - Prob. 38.4DQCh. 38 - Prob. 38.5DQCh. 38 - Prob. 38.6DQ
Ch. 38 - Prob. 38.7DQCh. 38 - Prob. 38.8DQCh. 38 - Prob. 38.9DQCh. 38 - Prob. 38.10DQCh. 38 - Prob. 38.11DQCh. 38 - Prob. 38.12DQCh. 38 - Prob. 38.13DQCh. 38 - Prob. 38.14DQCh. 38 - Prob. 38.15DQCh. 38 - Prob. 38.16DQCh. 38 - Prob. 38.17DQCh. 38 - Prob. 38.1ECh. 38 - Prob. 38.2ECh. 38 - Prob. 38.3ECh. 38 - Prob. 38.4ECh. 38 - Prob. 38.5ECh. 38 - Prob. 38.6ECh. 38 - Prob. 38.7ECh. 38 - Prob. 38.8ECh. 38 - Prob. 38.9ECh. 38 - Prob. 38.10ECh. 38 - Prob. 38.11ECh. 38 - Prob. 38.12ECh. 38 - Prob. 38.13ECh. 38 - Prob. 38.14ECh. 38 - Prob. 38.15ECh. 38 - Prob. 38.16ECh. 38 - Prob. 38.17ECh. 38 - Prob. 38.18ECh. 38 - Prob. 38.19ECh. 38 - Prob. 38.20ECh. 38 - Prob. 38.21ECh. 38 - An electron and a positron are moving toward each...Ch. 38 - Prob. 38.23ECh. 38 - Prob. 38.24ECh. 38 - Prob. 38.25ECh. 38 - Prob. 38.26PCh. 38 - Prob. 38.27PCh. 38 - Prob. 38.28PCh. 38 - Prob. 38.29PCh. 38 - Prob. 38.30PCh. 38 - Prob. 38.31PCh. 38 - Prob. 38.32PCh. 38 - Prob. 38.33PCh. 38 - Prob. 38.34PCh. 38 - Prob. 38.35PCh. 38 - Prob. 38.36PCh. 38 - Prob. 38.37PCh. 38 - Prob. 38.38PCh. 38 - Prob. 38.39PCh. 38 - Prob. 38.40CPCh. 38 - Prob. 38.41PPCh. 38 - Prob. 38.42PPCh. 38 - Prob. 38.43PPCh. 38 - Prob. 38.44PPCh. 38 - Prob. 38.45PP
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- The work function for potassium is 2.26 eV. What is the cutoff frequency when this metal is used as photoelectrode? What is the stopping potential when for the emitted electrons when this photo electrode is exposed to radiation of frequency 1200 THz?arrow_forwardIn the interpretation of the photoelectric effect, how is it known that an electron does not absorb more than one photon?arrow_forwardIf the work function of a metal is 3.2 eV, what is the maximum wavelength that a photon can have to eject a photoelectron from this metal surface?arrow_forward
- In a photoelectric experiments, a graph of the light frequency f is plotted against the maximum kinetic energy Kmax of the photoelectron as shown in Figure below fx10“Hz 4.83 K (eV) max Based on the graph, for the light of frequency 7.14 x1014 Hz, calculate a. the threshold wavelength, b. the maximum speed of the photoelectron. (Given c =3.00x10® m s1, h =6.63 x 1034ª J s, me=9.11 x 1031 kg, mp=1.67 x 1027 kg and e=1.60x1019 C)arrow_forwardDuring an experiment 8.7×10¹4 Hz light shines on a metal plate and the Photoelectric Effect occurs. It is determined that the stopping potential is 2.3 V. Determine the momentum (in N's) a photon has at the given frequency & determine the cutoff frequency (fo). p= fo=arrow_forwardUV radiation having a wavelength of 84 nm falls on gold metal, to which electrons are bound by 4.82 eV. What is the maximum velocity of the ejected photoelectrons? No need to use relativistic formulas in this case, so you can just use the standard formula KE =12mv2. The correct answer is 1.87E6 m/s how do I get that?arrow_forward
- Q. 12: Photoelectric emission takes place (a). when incident wavelength is greater than threshold wavelength when incident wavelength is less than threshold wavelength (b) (c) (d) when incident frequency is greater than threshold frequency at any frequencyarrow_forwardThe graph below shows the maximum kinetic energy of emitted photoelectrons as a function of the energy of the photons that are incident on a particular surface. From this graph, determine the following: a) the work function (in eV) 2) the threshold frequency (in Hz) Thank you for the help!arrow_forwardExample: Light of frequency of 1.00 x 10¹5 Hz illuminates a sodium surface. The ejected photoelectrons are found to have a maximum kinetic energy of 1.78 eV. Find the threshold frequency for this metal. Given: Solution: KEmax = (1.78 eV)(1.60 x 10-19 J/eV) KEmax hf - hft ; ft= = = 2.85 x 10-1⁹ J hf-KEmax h (6.63 x 10-34 Js) (1.00 x 10¹5 Hz)-(2.85 x 10-¹⁹ J) 6.63 x 10-34 Js f= 1.00 x 1015 Hz f₁= 5.70 X 1014 Hz Activity: Solve the following problems. Use the rubrics as your guide in presenting your solution (refer to attachment A.1 at the next page). 1. In the photoelectric effect, it is found that incident photons with energy 5.00 eV will produce electrons with a maximum kinetic energy 3.00 eV. What is the threshold frequency of this material?arrow_forward
- The energy conservation principle that applies to the photoelectric experiment is Ephoton = (KE)electron + W, where W is the “work function” for the metal. (The work function is the minimum energy required to eject an electron from the metal surface.) The work function for calcium metal is 4.60 10−19 J. If calcium is irradiated with 400-nm photons, what is the de Broglie wavelength of the resulting photoelectron beam?arrow_forwardA 90nm radiation falls on a photoelectric surface, and the work function of photoelectrode is 4.3 ev. The Plank's constant is 6.625 x 1034 J.s and mass of electron is 9.11 x 1031 kg. Determine ) The maximum kinetic energy of the electrons in Joules: () The velocity of the photoelectrons in m/s:| (C) The stopping potential in Volts. when the same surface is illuminated with light of wavelength 205 nmarrow_forwardRadiation with a frequency of 7.52 × 1014 Hz illuminates a photoelectric surface in a photoelectric cell. If the work function of this surface is 2.20 eV, what stopping voltage would be required to reduce the current through this cell to zero?arrow_forward
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