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

2. 1. Tube Rating Charts Name: Directions: For the given information state if the technique is safe or unsafe and why. 60 Hertz Stator Operation Effective Focal Spot Size- 0.6 mm Peak Kilovolts MA 2 150 140 130 120 110 100 90 80 70 2501 60 50 40 30 .01 .02 .04.06 .1 .2 .4.6 1 8 10 Maximum Exposure Time In Seconds Is an exposure of 80 kVp, 0.1 second and 200 mA within the limits of the single phase, 0.6 mm focal spot tube rating chart above? Is an exposure of 100 kVp, 0.9 second and 150 mA within the limits of the single phase, 0.6 mm focal spot tube rating chart above?

Q: You have a CO2 laser resonator (λ = 10.6 μm). It has two curved mirrors with R₁=10m, R2= 8m, and mirror separation /= 5m. Find: R2-10 m tl Z-O 12 R1-8 m 1. Confocal parameter. b= 21w2/2 =√1 (R1-1)(R2-1)(R1+R2-21)/R1+R2-21) 2. Beam waist at t₁ & t2- 3. Waist radius (wo). 4. 5. The radius of the laser beam outside the resonator and about 0.5m from R₂- Divergence angle. 6. Radius of curvature for phase front on the mirrors R₁ & R2-

No chatgpt pls will upvote

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