Physics: Principles with Applications
Physics: Principles with Applications
6th Edition
ISBN: 9780130606204
Author: Douglas C. Giancoli
Publisher: Prentice Hall
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Question
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Chapter 28, Problem 40P
To determine

The wavelength of He-Ne laser.

Expert Solution & Answer
Check Mark

Answer to Problem 40P

  633nm

Explanation of Solution

Given info:

The laser of He-Ne .

Formula used:

  λ=(1.24×103eVnm)E

Where, λ is a wavelength and E is energy in eV.

Calculation:

Transition from the E3' state to the E2' state releases photons with energy 1.96eV .

The wavelength is calculated by

  λ=(1.24×103eVnm)ΔE=(1.24×103eVnm)(1.96eV)=633nm

Conclusion:

The required value is 633nm .

Chapter 28 Solutions

Physics: Principles with Applications

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