PHYSICS FOR SCIEN & ENGNR W/MOD MAST
4th Edition
ISBN: 9780134112039
Author: GIANCOLI
Publisher: PEARSON
expand_more
expand_more
format_list_bulleted
Concept explainers
Question
Chapter 37, Problem 19P
To determine
The work function of the metal.
Expert Solution & Answer
Want to see the full answer?
Check out a sample textbook solutionStudents have asked these similar questions
(i) How does one explain the emission of electrons from a photosensitive surface with the help of Einstein’s photoelectric equation?
(ii) The work function of the following metals is given : Na = 2.75 eV, K = 2.3 eV, Mo = 4.17 eV and Ni 5.15 eV. Which of these metals will not cause photoelectric emission for radiation of wavelength 3300 A from a laser source placed 1 m away from these metals? What happens if the laser source is brought nearer and placed 50 cm away?
(4) (i) Light shining on a metal surface produces photoelectrons with a maximum kinetic energy of 2.0 eV. The light intensity is then doubled. Now what is the maximum kinetic energy of the photoelectrons, in eV?
(ii) The detector in an ordinary digital camera is made of silicon. This detector works by the photoelectric effect. The longest wavelength of light that an ordinary digital camera can detect has a wavelength of 1 micron (where 1 micron = 10^-6 m). What is the work function of silicon, in eV?
(iii) Infrared cameras don't use detectors made of silicon. For an infrared camera to detect infrared radiation with a wavelength of 22 microns, its detector must be made of a dierent material. What is the maximum possible work function of this material, in eV?
(c) Light of wavelength 550 nm falls on the surface of metal of work function 3.5 eV.
(i)
Calculate the maximum kinetic energy of the electrons emitted
Calculate the cut-off frequency.
(ii)
Chapter 37 Solutions
PHYSICS FOR SCIEN & ENGNR W/MOD MAST
Ch. 37.2 - Prob. 1AECh. 37.2 - Prob. 1BECh. 37.4 - Prob. 1CECh. 37.7 - Prob. 1DECh. 37.7 - Prob. 1EECh. 37.11 - Prob. 1FECh. 37 - Prob. 1QCh. 37 - Prob. 2QCh. 37 - Prob. 3QCh. 37 - Prob. 4Q
Ch. 37 - Prob. 5QCh. 37 - Prob. 6QCh. 37 - Prob. 7QCh. 37 - Prob. 8QCh. 37 - Prob. 9QCh. 37 - Prob. 10QCh. 37 - Prob. 11QCh. 37 - Prob. 12QCh. 37 - Prob. 13QCh. 37 - Prob. 14QCh. 37 - Prob. 15QCh. 37 - Prob. 16QCh. 37 - Prob. 17QCh. 37 - Prob. 18QCh. 37 - Prob. 19QCh. 37 - Prob. 20QCh. 37 - Prob. 21QCh. 37 - Prob. 22QCh. 37 - Prob. 23QCh. 37 - Prob. 24QCh. 37 - Prob. 25QCh. 37 - Prob. 26QCh. 37 - Prob. 27QCh. 37 - Prob. 28QCh. 37 - Prob. 1PCh. 37 - Prob. 2PCh. 37 - Prob. 3PCh. 37 - Prob. 4PCh. 37 - Prob. 5PCh. 37 - Prob. 6PCh. 37 - Prob. 7PCh. 37 - Prob. 8PCh. 37 - Prob. 9PCh. 37 - Prob. 10PCh. 37 - Prob. 11PCh. 37 - Prob. 12PCh. 37 - Prob. 13PCh. 37 - Prob. 14PCh. 37 - Prob. 15PCh. 37 - Prob. 16PCh. 37 - Prob. 17PCh. 37 - Prob. 18PCh. 37 - Prob. 19PCh. 37 - Prob. 20PCh. 37 - Prob. 21PCh. 37 - Prob. 22PCh. 37 - Prob. 23PCh. 37 - Prob. 24PCh. 37 - Prob. 25PCh. 37 - Prob. 26PCh. 37 - Prob. 27PCh. 37 - Prob. 28PCh. 37 - Prob. 29PCh. 37 - Prob. 30PCh. 37 - Prob. 31PCh. 37 - Prob. 32PCh. 37 - Prob. 33PCh. 37 - Prob. 34PCh. 37 - Prob. 35PCh. 37 - Prob. 36PCh. 37 - Prob. 37PCh. 37 - Prob. 38PCh. 37 - Prob. 39PCh. 37 - Prob. 40PCh. 37 - Prob. 41PCh. 37 - Prob. 42PCh. 37 - Prob. 43PCh. 37 - Prob. 44PCh. 37 - Prob. 45PCh. 37 - Prob. 46PCh. 37 - Prob. 47PCh. 37 - Prob. 48PCh. 37 - Prob. 49PCh. 37 - Prob. 50PCh. 37 - Prob. 51PCh. 37 - Prob. 52PCh. 37 - Prob. 53PCh. 37 - Prob. 54PCh. 37 - Prob. 55PCh. 37 - Prob. 56PCh. 37 - Prob. 57PCh. 37 - Prob. 58PCh. 37 - Prob. 59PCh. 37 - Prob. 60PCh. 37 - Prob. 61PCh. 37 - Prob. 62PCh. 37 - Prob. 63PCh. 37 - Prob. 64PCh. 37 - Prob. 65PCh. 37 - Prob. 66PCh. 37 - Prob. 67PCh. 37 - Prob. 68PCh. 37 - Prob. 69PCh. 37 - Prob. 70PCh. 37 - Prob. 71PCh. 37 - Prob. 72GPCh. 37 - Prob. 73GPCh. 37 - Prob. 74GPCh. 37 - Prob. 75GPCh. 37 - Prob. 76GPCh. 37 - Prob. 77GPCh. 37 - Prob. 78GPCh. 37 - Prob. 79GPCh. 37 - Prob. 80GPCh. 37 - Prob. 81GPCh. 37 - Prob. 82GPCh. 37 - Prob. 83GPCh. 37 - Prob. 84GPCh. 37 - Prob. 85GPCh. 37 - Prob. 86GPCh. 37 - Prob. 87GPCh. 37 - Prob. 88GPCh. 37 - Prob. 89GPCh. 37 - Prob. 90GPCh. 37 - Prob. 91GPCh. 37 - Prob. 92GPCh. 37 - Prob. 93GPCh. 37 - Show that the wavelength of a particle of mass m...Ch. 37 - Prob. 95GPCh. 37 - Prob. 96GPCh. 37 - Prob. 97GPCh. 37 - Prob. 98GPCh. 37 - Prob. 99GPCh. 37 - Prob. 100GP
Knowledge Booster
Learn more about
Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, physics and related others by exploring similar questions and additional content below.Similar questions
- (b) (i) Calculate the de Broglie wavelength of an electron having a mass of 9.11 x 1031 kg and a charge of 1.602 x 10-19 J with a Kinetic energy of 135 eV. The value of the Planck's constant is equal to 6.63 * 10-34 Js. (ii) Assume that an electron is moving along the x-axis with a speed of 3.66 x 106 m/s and with a precision of 0.50%. Calculate the minimum uncertainty (as allowed by the uncertainty principle in quantum theory) with which the position of the electron along the X-axis simultaneously can be measured with the speed?arrow_forward7) White light has wavelengths that range from 380 nm to 750 nm. This light strikes a metal that has a work function of 2.28 eV. (a) What is the maximum kinetic energy (in joules) of the electrons that are emitted from the metal? (b) For what wavelengths will no electrons be emitted? range ofarrow_forward(b) Calculate the de Broglie wavelength of an electron having a mass of 9.11 x 10-31 kg and a charge of 1.602 x 10-19 J with a Kinetic energy of 110 eV. The value of the Planck’s constant is equal to 6.63 * 10-34 Js.arrow_forward
- (c) The energy of an ultraviolet light is 3.28 eV. (i) What is its wavelength? (Given: h=6.63✕10-34 Js ; e=1.602✕10-19 C). (ii) Based on the de Broglie's hypothesis, determine the velocity of the electron. (Given: h=6.63✕10-34 Js ; me=9.11✕10-31 kg).arrow_forward(b) Evaluate the ratio of the de Broglie wavelength of electron to that of proton when (m₂=9.11 × 10-3¹ kg, mp=1.67 × 10-27 kg) (i) both have the same kinetic energy. (ii) The electron kinetic energy is 1000 eV, and the proton kineticarrow_forward(i) Monochromatic light of frequency 6.0 × 1014 Hz is produced by a laser. The power emitted is 2.0 × 10-3 W. Estimate the number of photons emitted per second on an average by the source. (ii) Draw a plot showing the variation of photoelectric current versus the intensity of incident radiation on a given photosensitive surface.arrow_forward
- (i) Define the term ‘threshold frequency’ as used in photoelectric effect. (ii) Plot a graph showing the variation of photoelectric current as a function of anode potential for two light beams having the same frequency but different intensities I1 and I2 (I1 > I2 ).arrow_forward(a) If a photon and an electron each have the same energy of 20.0 eV, find the wavelength of each. (b) If a photon and an electron each have the same wavelength of 250 nm, find the energy of each. (c) You want to study an organic molecule that is about 250 nm long using either a photon or an electron microscope. Approximately what wavelength should you use and which probe, the electron or the photon, is likely to damage the molecule the least?arrow_forward3-47. Data for stopping potential versus wavelength for the photoelectric effect using sodium are λ nm V₁ V 200 300 400 500 600 2.06 1.05 0.41 0.03 4.20arrow_forward
- (5) The total power output from a star is 4.5 x 1026 W. Assuming that all the emitted radiation has a wavelength λ = 550 nm, calculate the number of photons that are emitted per second.arrow_forwardIn the photoelectric effect, electrons are ejected from a metal surface when light strikes it. A certain minimum energy, Emin, is required to eject an electron. Any energy absorbed beyond that minimum gives kinetic energy to the ejected electron. It is found that when light at a wavelength of 540 nm falls on a cesium surface, an electron is ejected with a kinetic energy of 2.60 x 10-20 J. When the wavelength is 400 nm, the kinetic energy is 1.54 x 10-19 J. Calculate Emin for cesium in joules Calculate the longest wavelength, in nanometers, that will eject electrons from cesium.arrow_forward(b) Electromagnetic radiations having 400 nm wavelength falls on the surface of potassium, it resulted into the electrons emission with a K.E. of 1.79*105 Jmol'. Calculate the minimum energy required to remove an electron removal from potassium. Also determine the maximum wavelength needed for the emission of a photoelectron.arrow_forward
arrow_back_ios
SEE MORE QUESTIONS
arrow_forward_ios
Recommended textbooks for you
- Modern PhysicsPhysicsISBN:9781111794378Author:Raymond A. Serway, Clement J. Moses, Curt A. MoyerPublisher:Cengage Learning
Modern Physics
Physics
ISBN:9781111794378
Author:Raymond A. Serway, Clement J. Moses, Curt A. Moyer
Publisher:Cengage Learning