The wavelength of the photon.

Question

Want to see more full solutions like this?

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Students have asked these similar questions

In a block of Cobalt/iron alloy, it is suspected that the Cobalt (Z 27) is very poorly mixed with the iron (Z 26). Given that the ionization energy of hydrogen is 13.6 eV predict the energies of the K absorption edges of the constituents of the alloy. = =

Kl is a salt with lattice constant of a = 0.716 nm. Suppose one measures the powder diffraction from a cubic crystal of Kl using Cu Ka X-radiation having wavelength of 2 = 0.154 nm. The positions of the first nine reflections to occur at 20 are as the following. Assume n=1 intensity 20 a) Fill Table#1 and do peak assignment. i.e find (hkl) 20 0 d, = na/2 sin0 h² + k² + 1² = a²ldu (hkl) 21.8° 25.200 36.000 42.50 44.50 51.750 56.8° 58.450 64.65° b) From the allowed (hkl) planes that you have obtained, identify the expected crystal structure?

a) Starting from a general fornula of desity of States. g(k) dk= Vkdk 272 Show that g(w) dw= 3x Vw°dw for Phonons under the assumption thait the phonoh dre aicoustic, meaning U = does not depend 3. on W or k. 3N b) In a solid of N atoms, there are vibrational modes(phonons) with the highat frequercy Wp.

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

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

Want to see more full solutions like this?

Chapter 37 Solutions