EBK PHYSICS FOR SCIENTISTS AND ENGINEER
EBK PHYSICS FOR SCIENTISTS AND ENGINEER
6th Edition
ISBN: 9781319321710
Author: Mosca
Publisher: VST
Question
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Chapter 37, Problem 38P
To determine

The wavelength of the photon.

Expert Solution & Answer
Check Mark

Answer to Problem 38P

The wavelength for H2 , HD and D2 are 4.43μm , 3.88μm and 3.18μm respectively.

Explanation of Solution

Given:

The effective force constant is 580N/m .

Formula used:

The expression for wavelength is given by,

  λ=hcE0

The expression for energy is given by,

  E0=h2πKμ

Calculation:

The energy for H2 is calculated as,

  E0,H2=h2πKμ=6.634× 10 34J/s2π 580N/m 3.3× 10 27 =(( 4.41× 10 20 J)( 1eV 1.6× 10 19 J ))=0.28eV

The wavelength of H2 is calculated as,

  λH2=hcE0=1.24× 10 6eVm0.28eV=(( 4.43× 10 6 m)( 10 6 μm 1m ))=4.43μm

The energy for HD is calculated as,

  E0,HD=h2πKμ=6.634× 10 34J/s2π 580N/m 4( 3.3× 10 27 ) 3 =(( 5.09× 10 20 J)( 1eV 1.6× 10 19 J ))=0.32eV

The wavelength of HD is calculated as,

  λHD=hcE0=1.24× 10 6eVm0.32eV=(( 3.88× 10 6 m)( 10 6 μm 1m ))=3.88μm

The energy for D2 is calculated as,

  E0,D2=h2πKμ=6.634× 10 34J/s2π 580N/m 2( 3.3× 10 27 )=(( 6.24× 10 20 J)( 1eV 1.6× 10 19 J ))=0.39eV

The wavelength of D2 is calculated as,

  λD2=hcE0=1.24× 10 6eVm0.39eV=(( 3.88× 10 6 m)( 10 6 μm 1m ))=3.18μm

Conclusion:

Therefore, the wavelength for H2 , HD and D2 are 4.43μm , 3.88μm and 3.18μm respectively.

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