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Photoelectrons from a material with a binding energy of 2.71 eV are ejected by 420-nm photons. Once ejected, how long does it take these electrons to travel 2.50 cm to a detection device?

$8.44 \times 10^{-8} \mathrm{s}$

Physics 103

Chapter 29

Introduction to Quantum Physics

Quantum Physics

Cornell University

University of Winnipeg

McMaster University

Lectures

02:51

Quantum mechanics (QM) is …

10:58

In physics, the photoelect…

01:24

Photoelectrons are ejected…

What is the maximum veloci…

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Calculate the binding ener…

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A laser with a power outpu…

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How fast must an electron …

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What is the binding energy…

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An atom (not a hydrogen at…

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A hydrogen atom, initially…

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(a) Calculate the number o…

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Electrons in a photoelectr…

07:30

Integrated ConceptsA l…

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Suppose that light of inte…

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Two different monochromati…

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How many photons per secon…

So we're given a binding energy of 2.71 electron volts and we have these electrons that are being ejected by 420 nano meter photons, which could be written as 420 attempts, 10 to the negative nine meters. And we want to know how long it'll take to travel 2.5 centimeters. I'm gonna write that as my delta X 2.5 centimeters is 0.0 to 5 meters, so are cool here is to solve for kinetic energy because if we know our kinetic energy that we can use that to solve four my speed. And if I know how fast we're moving, then I know how long it'll take to travel that 0.0 to 5 meters. So our kinetic energy can be found by replacing my frequency with my speed of light divided by my wavelength. So my speed of what here is going to be three times 10 to the eighth meters per second. I'm also going to be using planks constant in election volts. So which is 4.14 times 10 to the negative 15 electron volts seconds. Soap lugging all that in I have 4.14 times 10 to the negative 15. I have my frequency as my speed of light divided by my wavelength minus my work function, which is 2.71 again. This is my binding energy and I get a kinetic energy of 0.24 electron volts. But if we want to sell for velocity, we're gonna want to convert that to Jules. So one electron volts is 1.6 times 10 to the negative 19. Jules, this gives us about 3.95 times 10 to the negative. 20 Jules again are kinetic energy is 1/2 M v squared. So solving for velocity, I can multiply by two and divide by the mass of an electron in this case, because that's what being ejected and square root to find. My velocity. So this is two times 3.95 trying to tend to the negative 20 divided by the mass of an electron which is 9.11 times 10th in the negative 31 gives me a velocity of about 2.94 times 10 to the fifth meters per second. This movie incredibly fast. So you could block out some space here if we know how fast it's moving. Then my this distance traveled, divided by my speed will give me my time. So can. That was 0.0 to 5. Divided by 2.94 gives me a time of approximately 8.4 times 10 to the negative eight seconds.

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