The energy of the waves from highest to lowest.

Question

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Answer 1

Question

Chapter 6, Problem 38Q

To determine

The energy of the waves from highest to lowest.

Expert Solution & Answer

Answer to Problem 38Q

The energy of the waves will be arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

Explanation of Solution

Given data:

Wave I: λ = 3m.

Wave II: λ = 2m.

Wave III: λ = 6m.

Wave IV: λ = 6m.

Wave V: λ = 3m.

Wave VI: λ = 2m.

Formula used:

The energy of the waves are calculated by the following formula:

E=hνc=λν.

Calculation:

Having the known values, the energy of the wave is calculated as:

E=hνc=λν

Where, E is the energy of the wave.

h is the Planck’s constant; h=6.63×10−34Js

c is the velocity of light; c=3×108ms−1

Wave I:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1Wave II:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1Wave III:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1Wave IV:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1

Wave V:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1

Wave VI:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1.

Conclusion:

Thus, the energy of the waves are arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

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Answer 2

Question

Chapter 6, Problem 38Q

To determine

The energy of the waves from highest to lowest.

Expert Solution & Answer

Answer to Problem 38Q

The energy of the waves will be arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

Explanation of Solution

Given data:

Wave I: λ = 3m.

Wave II: λ = 2m.

Wave III: λ = 6m.

Wave IV: λ = 6m.

Wave V: λ = 3m.

Wave VI: λ = 2m.

Formula used:

The energy of the waves are calculated by the following formula:

E=hνc=λν.

Calculation:

Having the known values, the energy of the wave is calculated as:

E=hνc=λν

Where, E is the energy of the wave.

h is the Planck’s constant; h=6.63×10−34Js

c is the velocity of light; c=3×108ms−1

Wave I:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1Wave II:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1Wave III:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1Wave IV:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1

Wave V:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1

Wave VI:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1.

Conclusion:

Thus, the energy of the waves are arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

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Answer 3

Question

Chapter 6, Problem 38Q

To determine

The energy of the waves from highest to lowest.

Expert Solution & Answer

Answer to Problem 38Q

The energy of the waves will be arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

Explanation of Solution

Given data:

Wave I: λ = 3m.

Wave II: λ = 2m.

Wave III: λ = 6m.

Wave IV: λ = 6m.

Wave V: λ = 3m.

Wave VI: λ = 2m.

Formula used:

The energy of the waves are calculated by the following formula:

E=hνc=λν.

Calculation:

Having the known values, the energy of the wave is calculated as:

E=hνc=λν

Where, E is the energy of the wave.

h is the Planck’s constant; h=6.63×10−34Js

c is the velocity of light; c=3×108ms−1

Wave I:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1Wave II:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1Wave III:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1Wave IV:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1

Wave V:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1

Wave VI:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1.

Conclusion:

Thus, the energy of the waves are arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

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Answer 4

Question

Chapter 6, Problem 38Q

To determine

The energy of the waves from highest to lowest.

Expert Solution & Answer

Answer to Problem 38Q

The energy of the waves will be arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

Explanation of Solution

Given data:

Wave I: λ = 3m.

Wave II: λ = 2m.

Wave III: λ = 6m.

Wave IV: λ = 6m.

Wave V: λ = 3m.

Wave VI: λ = 2m.

Formula used:

The energy of the waves are calculated by the following formula:

E=hνc=λν.

Calculation:

Having the known values, the energy of the wave is calculated as:

E=hνc=λν

Where, E is the energy of the wave.

h is the Planck’s constant; h=6.63×10−34Js

c is the velocity of light; c=3×108ms−1

Wave I:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1Wave II:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1Wave III:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1Wave IV:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1

Wave V:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1

Wave VI:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1.

Conclusion:

Thus, the energy of the waves are arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

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Answer 5

Chapter 6, Problem 38Q

To determine

The energy of the waves from highest to lowest.

Expert Solution & Answer

Answer to Problem 38Q

The energy of the waves will be arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

Explanation of Solution

Given data:

Wave I: λ = 3m.

Wave II: λ = 2m.

Wave III: λ = 6m.

Wave IV: λ = 6m.

Wave V: λ = 3m.

Wave VI: λ = 2m.

Formula used:

The energy of the waves are calculated by the following formula:

E=hνc=λν.

Calculation:

Having the known values, the energy of the wave is calculated as:

E=hνc=λν

Where, E is the energy of the wave.

h is the Planck’s constant; h=6.63×10−34Js

c is the velocity of light; c=3×108ms−1

Wave I:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1Wave II:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1Wave III:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1Wave IV:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1

Wave V:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1

Wave VI:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1.

Conclusion:

Thus, the energy of the waves are arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

Answer 6

Chapter 6, Problem 38Q

To determine

The energy of the waves from highest to lowest.

Expert Solution & Answer

Answer to Problem 38Q

The energy of the waves will be arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

Explanation of Solution

Given data:

Wave I: λ = 3m.

Wave II: λ = 2m.

Wave III: λ = 6m.

Wave IV: λ = 6m.

Wave V: λ = 3m.

Wave VI: λ = 2m.

Formula used:

The energy of the waves are calculated by the following formula:

E=hνc=λν.

Calculation:

Having the known values, the energy of the wave is calculated as:

E=hνc=λν

Where, E is the energy of the wave.

h is the Planck’s constant; h=6.63×10−34Js

c is the velocity of light; c=3×108ms−1

Wave I:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1Wave II:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1Wave III:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1Wave IV:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1

Wave V:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1

Wave VI:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1.

Conclusion:

Thus, the energy of the waves are arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

Answer 7

Chapter 6, Problem 38Q

To determine

The energy of the waves from highest to lowest.

Answer 8

Expert Solution & Answer

Answer to Problem 38Q

The energy of the waves will be arranged as:

10×10−26>6.63×10−26>3.3×10−26II=VI>I=V>III=IV.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given data:

Wave I: λ = 3m.

Wave II: λ = 2m.

Wave III: λ = 6m.

Wave IV: λ = 6m.

Wave V: λ = 3m.

Wave VI: λ = 2m.

Formula used:

The energy of the waves are calculated by the following formula:

E=hνc=λν.

Calculation:

Having the known values, the energy of the wave is calculated as:

E=hνc=λν

Where, E is the energy of the wave.

h is the Planck’s constant; h=6.63×10−34Js

c is the velocity of light; c=3×108ms−1

Wave I:λ=3mν=cλ=3× 1083=108s−1E=hνE=(6.63×10−34)( 108)=6.63×10−26Js−1Wave II:λ=2mν=cλ=3× 1082=15×107s−1E=hνE=(6.63×10−34)(15× 107)=10×10−26Js−1Wave III:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1Wave IV:λ=6mν=cλ=3× 1086=0.5×108s−1E=hνE=(6.63×10−34)(0.5× 108)=3.3×10−26Js−1