Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Chapter 28, Problem 32P
To determine

The cut off wavelength, λ0=1240nmV

Expert Solution & Answer
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Answer to Problem 32P

It is shown that the cut off wavelength, λ0=1240nmV

Explanation of Solution

Given info: Cut off wavelength, λ0=1240nmV

Formula used:

The energy of photon with shortest wavelength must equivalent the max kinetic energy of an electron:

  hf0=hcλ0=eV or λ0=hceV

Where,

  h is Planck’s constant, 6.63×1034Js

  c is speed of light, 3.00×108m/s

  e is charge on electron, 1eV=1.6×1019J/eV

Calculation:

Substitute the known and given values in expression λ0=hceV

  λ0=(6.63×1034Js)(3.00×108m/s)(109nm/m)(1.60×1019J/eV)(1e)V

  λ0=(1.24×103Vnm)V

  λ0=1240nmV

Conclusion:

The given condition is shown.

Chapter 28 Solutions

Physics: Principles with Applications

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