In a photocell, ultraviolet (UV) light provides enough energy to some electrons in barium metal to eject them from the surface at high speed. To measure the maximum energy of the electrons, another plate above the barium surface is kept at a negative enough potential that the emitted electrons are slowed down and stopped, and return to the barium surface. See Fig. 17-46. If the plate voltage is –3.02 V (compared to the barium) when the fastest electrons are stopped, what was the speed of these electrons when they were emitted? Plate UV light V=-3.02 V V=0 FIGURE 17–46 Barium Problem 80.

In a photocell, ultraviolet (UV) light provides enough energy to some electrons in barium metal to eject them from the surface at high speed. To measure the maximum energy of the electrons, another plate above the barium surface is kept at a negative enough potential that the emitted electrons are slowed down and stopped, and return to the barium surface. See Fig. 17-46. If the plate voltage is –3.02 V (compared to the barium) when the fastest electrons are stopped, what was the speed of these electrons when they were emitted? Plate UV light V=-3.02 V V=0 FIGURE 17–46 Barium Problem 80.

College Physics

11th Edition

ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

Chapter1: Units, Trigonometry. And Vectors

Section: Chapter Questions

Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...

See similar textbooks

Similar questions

b): A parallel-plate air-filled capacitor having area 40 cm² and plate spacing 1.0 mm is charged to a potential difference of 600 V. Find (a) the capacitance, (b) the magnitude of the charge on each plate, (c) the stored energy, (d) the electric field between the plates, and (e) the energy density between the plates. c): A capacitor is charged until its stored energy is 4.00 J. A second capacitor is the
(22%) Problem 6: An air-filled capacitor has capacitance Co. When it is connected to a battery with EMF Vo it has charge Qo and stored energy Uo. Once the capacitor has charged, then the switch is opened, and finally a slab of material with dielectric constant K is inserted into the gap. -Vo + Vo d DV 20% Part (a) Which of the following physical quantities does not change when the dielectric material is inserted? Grade = 100% Correct Answver Student Final Submission Feedback Not available until end date The charge stored on the capacitor plates. Correct! Grade Summary Deduction for Final Submission 0% Deductions for Incorrect Submissions, Hints and Feedback [?] 0% Student Grade = 100 - 0 - 0 = 100% Submission History All Date times are displaved in Eastern Standard Time.Red submission date times indicate late work. Date Time Answer Hints Feedback 1 Feb 27, 2021 7:44 PM The charge stored on the capacitor plates. E * 20% Part (b) Enter an expression for the potential difference between…
Please help with with problem 19
Compute the work done on V=1 cm3 of acetone when the electric field intensity is increased from E1-0 to E₂=104 V/m at the temperature T=20°C. The relative dielectric constant of the acetone is &=21.4 (the dielectric constant for the open space is &o=8.854∙10-¹² C²s²/(kg m³).
Find the total energy stored (in units of m) in the group of capacitors shown below given that the potential difference Nab Is equal to 19.9 V7 50 uF 10 ul HH 04:07 AR
If neededBoltzmann Constant=1.38×10-23 J/K = 8.62×10-5eV/KUnit charge= 1.602 ×10-5 CPlanck’s constant= 6.626×10−34 J.sec=4.136×10−15 eV.sec
A point particle with charge q = -200 nC and mass m = 3.0x10-6 g is released from rest at the midpoint of a parallel-plate capacitor. The capacitor has plate separation d = 3.0 mm. The negative electrode is at x = 0 and the positive electrode is at x = d. The voltage across the capcitor has been measured to be AVC = 2.5 V. In this problem, ignore the weight of the particle. Give all numerical answers to two significant figures. a) What direction does the particle move? b) What is the electric field strength inside the capacitor, in V/m?
Two parallel-plate capacitors, 5.2 uF each, are connected in series to a 11 V battery. One of the capacitors is then squeezed so that its plate separation is halved. Because of the squeezing, (a) how much additional charge is transferred to the capacitors by the battery and (b) what is the increase in the sum of the charges: the charge on the positive plate of one capacitor plus the charge on the positive plate of the other capacitor? Give your answers in Coulombs.
The parallel plates in a capacitor, with a plate area of 6.70 cm2 and an air-filled separation of 2.60 mm, are charged by a 7.70 V battery. They are then disconnected from the battery and pulled apart (without discharge) to a separation of 8.50 mm. Neglecting fringing, find (d) the work required to separate the plates.
The thickness of the cell membrane is 5 nm, the potential inside the cell is - 65 mV, the potential outside the cell is +10 mV. The electric field strength in the membrane is (kV/m) (give the modulus of magnitude)
The two plates of a parallel -plate capacitor are separated by a distance 8.4mm (assume that the lower plate is at the origin). Lower plate is maintained at potential 7.3V and the upper plate at 2.0V. Assuming negligible fringing effect at the edges, determine the potential 6.72mm. O 0.190 O -0.505 O 0.695 O -0.631 O-6.625
Rank the following five capacitors from greatest to smallest capacitance, noting any cases of equality. (a) a 20-uF capacitor with a 4-V potential difference between its plates (b) a 30-uF capacitor with charges of magnitude 90 µC on each plate (c) a capacitor with charges of magnitude 80 µC on its plates, differing by 2 V in potential, (d) a 10-pF capacitor storing energy 125 µ) (e) a capacitor storing energy 250 µJ with a 10-V potential difference O 1. b> a = c>d = e O 2. c> b >a >d>e O 3. e >a > b>d>c O 4. e>d> a >b>c