The wavelength of the photon.

Question

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Answer 1

Question

Chapter 37, Problem 38P

To determine

The wavelength of the photon.

Expert Solution & Answer

Answer to Problem 38P

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

Explanation of Solution

Given:

The effective force constant is 580 N/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 −34 J/s2π 580 N/m 3.3× 10 −27 =(( 4.41× 10 −20 J)( 1 eV 1.6× 10 −19 J ))=0.28 eV

The wavelength of H2 is calculated as,

λH2=hcE0=1.24× 10 −6 eV⋅m0.28 eV=(( 4.43× 10 −6 m)( 10 6 μm 1 m ))=4.43 μm

The energy for HD is calculated as,

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

The wavelength of HD is calculated as,

λHD=hcE0=1.24× 10 −6 eV⋅m0.32 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.88 μm

The energy for D2 is calculated as,

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

The wavelength of D2 is calculated as,

λD2=hcE0=1.24× 10 −6 eV⋅m0.39 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.18 μm

Conclusion:

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

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Answer 2

Question

Chapter 37, Problem 38P

To determine

The wavelength of the photon.

Expert Solution & Answer

Answer to Problem 38P

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

Explanation of Solution

Given:

The effective force constant is 580 N/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 −34 J/s2π 580 N/m 3.3× 10 −27 =(( 4.41× 10 −20 J)( 1 eV 1.6× 10 −19 J ))=0.28 eV

The wavelength of H2 is calculated as,

λH2=hcE0=1.24× 10 −6 eV⋅m0.28 eV=(( 4.43× 10 −6 m)( 10 6 μm 1 m ))=4.43 μm

The energy for HD is calculated as,

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

The wavelength of HD is calculated as,

λHD=hcE0=1.24× 10 −6 eV⋅m0.32 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.88 μm

The energy for D2 is calculated as,

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

The wavelength of D2 is calculated as,

λD2=hcE0=1.24× 10 −6 eV⋅m0.39 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.18 μm

Conclusion:

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

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Answer 3

Question

Chapter 37, Problem 38P

To determine

The wavelength of the photon.

Expert Solution & Answer

Answer to Problem 38P

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

Explanation of Solution

Given:

The effective force constant is 580 N/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 −34 J/s2π 580 N/m 3.3× 10 −27 =(( 4.41× 10 −20 J)( 1 eV 1.6× 10 −19 J ))=0.28 eV

The wavelength of H2 is calculated as,

λH2=hcE0=1.24× 10 −6 eV⋅m0.28 eV=(( 4.43× 10 −6 m)( 10 6 μm 1 m ))=4.43 μm

The energy for HD is calculated as,

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

The wavelength of HD is calculated as,

λHD=hcE0=1.24× 10 −6 eV⋅m0.32 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.88 μm

The energy for D2 is calculated as,

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

The wavelength of D2 is calculated as,

λD2=hcE0=1.24× 10 −6 eV⋅m0.39 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.18 μm

Conclusion:

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

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Answer 4

Question

Chapter 37, Problem 38P

To determine

The wavelength of the photon.

Expert Solution & Answer

Answer to Problem 38P

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

Explanation of Solution

Given:

The effective force constant is 580 N/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 −34 J/s2π 580 N/m 3.3× 10 −27 =(( 4.41× 10 −20 J)( 1 eV 1.6× 10 −19 J ))=0.28 eV

The wavelength of H2 is calculated as,

λH2=hcE0=1.24× 10 −6 eV⋅m0.28 eV=(( 4.43× 10 −6 m)( 10 6 μm 1 m ))=4.43 μm

The energy for HD is calculated as,

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

The wavelength of HD is calculated as,

λHD=hcE0=1.24× 10 −6 eV⋅m0.32 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.88 μm

The energy for D2 is calculated as,

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

The wavelength of D2 is calculated as,

λD2=hcE0=1.24× 10 −6 eV⋅m0.39 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.18 μm

Conclusion:

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

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Answer 5

Chapter 37, Problem 38P

To determine

The wavelength of the photon.

Expert Solution & Answer

Answer to Problem 38P

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

Explanation of Solution

Given:

The effective force constant is 580 N/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 −34 J/s2π 580 N/m 3.3× 10 −27 =(( 4.41× 10 −20 J)( 1 eV 1.6× 10 −19 J ))=0.28 eV

The wavelength of H2 is calculated as,

λH2=hcE0=1.24× 10 −6 eV⋅m0.28 eV=(( 4.43× 10 −6 m)( 10 6 μm 1 m ))=4.43 μm

The energy for HD is calculated as,

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

The wavelength of HD is calculated as,

λHD=hcE0=1.24× 10 −6 eV⋅m0.32 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.88 μm

The energy for D2 is calculated as,

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

The wavelength of D2 is calculated as,

λD2=hcE0=1.24× 10 −6 eV⋅m0.39 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.18 μm

Conclusion:

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

Answer 6

Chapter 37, Problem 38P

To determine

The wavelength of the photon.

Expert Solution & Answer

Answer to Problem 38P

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

Explanation of Solution

Given:

The effective force constant is 580 N/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 −34 J/s2π 580 N/m 3.3× 10 −27 =(( 4.41× 10 −20 J)( 1 eV 1.6× 10 −19 J ))=0.28 eV

The wavelength of H2 is calculated as,

λH2=hcE0=1.24× 10 −6 eV⋅m0.28 eV=(( 4.43× 10 −6 m)( 10 6 μm 1 m ))=4.43 μm

The energy for HD is calculated as,

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

The wavelength of HD is calculated as,

λHD=hcE0=1.24× 10 −6 eV⋅m0.32 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.88 μm

The energy for D2 is calculated as,

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

The wavelength of D2 is calculated as,

λD2=hcE0=1.24× 10 −6 eV⋅m0.39 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.18 μm

Conclusion:

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

Answer 7

Chapter 37, Problem 38P

To determine

The wavelength of the photon.

Answer 8

Expert Solution & Answer

Answer to Problem 38P

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

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given:

The effective force constant is 580 N/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 −34 J/s2π 580 N/m 3.3× 10 −27 =(( 4.41× 10 −20 J)( 1 eV 1.6× 10 −19 J ))=0.28 eV

The wavelength of H2 is calculated as,

λH2=hcE0=1.24× 10 −6 eV⋅m0.28 eV=(( 4.43× 10 −6 m)( 10 6 μm 1 m ))=4.43 μm

The energy for HD is calculated as,

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

The wavelength of HD is calculated as,

λHD=hcE0=1.24× 10 −6 eV⋅m0.32 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.88 μm

The energy for D2 is calculated as,

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

The wavelength of D2 is calculated as,

λD2=hcE0=1.24× 10 −6 eV⋅m0.39 eV=(( 3.88× 10 −6 m)( 10 6 μm 1 m ))=3.18 μm

Conclusion:

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

Answer 12

Ch. 37 - Prob. 1P Ch. 37 - Prob. 2P Ch. 37 - Prob. 3P Ch. 37 - Prob. 4P Ch. 37 - Prob. 5P Ch. 37 - Prob. 6P Ch. 37 - Prob. 7P Ch. 37 - Prob. 8P Ch. 37 - Prob. 9P Ch. 37 - Prob. 10P

The wavelength of the photon.

The wavelength of the photon.

Answer to Problem 38P

Explanation of Solution

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Chapter 37 Solutions