COLLEGE PHYSICS:STRAT.APP.TECH.UPD.-PKG
3rd Edition
ISBN: 9780134568430
Author: Knight
Publisher: PEARSON
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Chapter 28, Problem 5CQ
a.
To determine
To explain: The reason, why the curve become horizontal for
b.
To determine
To explain: The reason why the current does not immediately drop to zero for
c.
To determine
To explain: The reason why current is zero for
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Figure Q28.5 shows the typical photoelectric behavior of a metal as the anode-cathode potential difference ΔV is varied.
a. Why do the curves become horizontal for ΔV ? 1V? Shouldn’t the current increase as the potential difference increases? Explain.
b. Why doesn’t the current immediately drop to zero for ΔV < 0 V? Shouldn’t ΔV < 0 V prevent the electrons from reaching the anode? Explain.
c. The current is zero for ΔV < -2.0 V . Where do the electrons go? Are no electrons emitted if ΔV < -2.0 V? Or if they are, why is there no current? Explain.
Two light sources are used in a photoelectric experiment to determine the work function for a particular metal surface. When green light from a mercury lamp (? = 546.1 nm) is used, a stopping potential of 0.838 V reduces the photocurrent to zero.
(a) Based on this measurement, what is the work function for this metal?answer in eV(b) What stopping potential would be observed when using light from a red lamp (? = 693.0 nm)?answer in V
The work function of lithium is W0 = 2.93 eV.
(a) What is the maximum wavelength of incident light for which photoelectrons will be released from lithium? m(b) What is the minimum frequency of incident light, called the cutoff frequency, for which photoelectrons will be released from lithium? Hz(c) What is the maximum kinetic energy of photoelectrons emitted by lithium if 8.30-eV photons strike its surface? eV
Chapter 28 Solutions
COLLEGE PHYSICS:STRAT.APP.TECH.UPD.-PKG
Ch. 28 - Prob. 1CQCh. 28 - Prob. 2CQCh. 28 - Prob. 3CQCh. 28 - Prob. 4CQCh. 28 - Prob. 5CQCh. 28 - Prob. 6CQCh. 28 - Prob. 7CQCh. 28 - Prob. 8CQCh. 28 - Prob. 9CQCh. 28 - Prob. 10CQ
Ch. 28 - Prob. 11CQCh. 28 - Prob. 12CQCh. 28 - Prob. 13CQCh. 28 - Prob. 14CQCh. 28 - Prob. 15CQCh. 28 - Prob. 16CQCh. 28 - Prob. 17CQCh. 28 - Prob. 18CQCh. 28 - Prob. 19CQCh. 28 - Prob. 20CQCh. 28 - Prob. 21CQCh. 28 - Prob. 22CQCh. 28 - Prob. 23CQCh. 28 - Prob. 24CQCh. 28 - Prob. 25CQCh. 28 - Prob. 26MCQCh. 28 - Prob. 27MCQCh. 28 - Prob. 28MCQCh. 28 - Prob. 29MCQCh. 28 - Prob. 30MCQCh. 28 - Prob. 31MCQCh. 28 - Prob. 32MCQCh. 28 - Prob. 33MCQCh. 28 - Prob. 34MCQCh. 28 - Prob. 35MCQCh. 28 - Prob. 1PCh. 28 - Prob. 2PCh. 28 - Prob. 3PCh. 28 - Prob. 4PCh. 28 - Prob. 5PCh. 28 - Prob. 6PCh. 28 - Prob. 7PCh. 28 - Prob. 8PCh. 28 - Prob. 9PCh. 28 - Prob. 10PCh. 28 - Prob. 11PCh. 28 - Prob. 12PCh. 28 - Prob. 13PCh. 28 - Prob. 14PCh. 28 - Prob. 15PCh. 28 - Prob. 16PCh. 28 - Prob. 17PCh. 28 - Prob. 18PCh. 28 - Prob. 19PCh. 28 - Prob. 20PCh. 28 - Prob. 21PCh. 28 - Prob. 22PCh. 28 - Prob. 23PCh. 28 - Prob. 24PCh. 28 - Prob. 25PCh. 28 - Prob. 26PCh. 28 - Prob. 27PCh. 28 - Prob. 28PCh. 28 - Prob. 29PCh. 28 - Prob. 30PCh. 28 - Prob. 31PCh. 28 - Prob. 32PCh. 28 - Prob. 33PCh. 28 - Prob. 34PCh. 28 - Prob. 35PCh. 28 - Prob. 36PCh. 28 - Prob. 37PCh. 28 - Prob. 38PCh. 28 - Prob. 39PCh. 28 - Prob. 40PCh. 28 - Prob. 41PCh. 28 - Prob. 42PCh. 28 - Prob. 43PCh. 28 - Prob. 44PCh. 28 - Prob. 45PCh. 28 - Prob. 46PCh. 28 - Prob. 47PCh. 28 - Prob. 48PCh. 28 - Prob. 49PCh. 28 - Prob. 50GPCh. 28 - Prob. 51GPCh. 28 - Prob. 52GPCh. 28 - Prob. 53GPCh. 28 - Prob. 54GPCh. 28 - Prob. 55GPCh. 28 - Prob. 56GPCh. 28 - Prob. 57GPCh. 28 - Prob. 58GPCh. 28 - Prob. 59GPCh. 28 - Prob. 60GPCh. 28 - Prob. 61GPCh. 28 - Prob. 62GPCh. 28 - Prob. 63GPCh. 28 - Prob. 64GPCh. 28 - Prob. 65GPCh. 28 - Prob. 66GPCh. 28 - Prob. 67GPCh. 28 - Prob. 68GPCh. 28 - Prob. 69GPCh. 28 - Prob. 70GPCh. 28 - Prob. 71GPCh. 28 - Prob. 72GPCh. 28 - Prob. 73GPCh. 28 - Prob. 74GPCh. 28 - Prob. 75GPCh. 28 - Prob. 76GPCh. 28 - Prob. 77GPCh. 28 - Prob. 78GPCh. 28 - Prob. 79GPCh. 28 - Prob. 80GPCh. 28 - Prob. 81GPCh. 28 - Prob. 82GPCh. 28 - Prob. 83GPCh. 28 - Prob. 84GPCh. 28 - Prob. 85MSPPCh. 28 - Prob. 86MSPPCh. 28 - Prob. 87MSPPCh. 28 - Prob. 88MSPP
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- A 400-nm laser beam is projected onto a calcium electrode. The power of the laser beam is 2.00 mW and the work function of calcium is 2.31 eV. (a) How many photoelectrons per second are ejected? (b) What net power is carried away by photoelectrons?arrow_forwardThe 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_forwardA 600-nm light falls on a photoelectric surface and electrons with the maximum kinetic energy of 0.17 eV are emitted. Determine (a) the work function and (b) the cutoff frequency of the surface. (c) What is the stopping potential when the surface is illuminated with light of wavelength 400 nm?arrow_forward
- A laser with a power output of 2.00 mW at a 400-nm wavelength is used to project a beam of light onto a calcium photoelectrode. (a) How many photoelectrons leave the calcium surface per second? (b) What power is carried away by ejected photoelectrons, given that the work function of calcium is 2.31 eV? (c) Calculate the photocurrent. (d) If the photoelectrode suddenly becomes electrically insulated and the setup of two electrodes in the circuit suddenly starts to act like a 2.00-pF capacitor, how long will current flow before the capacitor voltage stops it?arrow_forwardA physicist is watching a 15-kg orangutan at a zoo swing lazily in a tire at the end of a rope. He (the physicist) notices that each oscillation takes 3.00 s and hypothesizes that the energy is quantized. (a) What is the difference in energy in joules between allowed oscillator states? (b) What is the value of n for a state where the energy is 5.00 J? (c) Can the quantization be observed?arrow_forwardThe work function for indium is 4.12 eV. (a) Convert the value of the work function from electron volts to joules. J(b) Find the cutoff frequency for indium. Hz(c) What maximum wavelength of light incident on indium releases photoelectrons from the indium's surface? nm(d) If light of energy 8.64 eV is incident on indium, what is the maximum kinetic energy of the ejected photoelectrons? Give the answer in electron volts. eV(e) For photons of energy 8.64 eV, what stopping potential would be required to arrest the current of photoelectrons? Varrow_forward
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