6.14 For photon with the following energies, calculate the wavelength and identify the region of the spectrum they are from. (a) 3.5 × 10 − 20 J, (b) 8.7 × 10 − 26 J, (c) 7.1 × 10 − 17 J, (d) 5.5 × 10 − 27 J

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Answer 1

Textbook Question

Chapter 6, Problem 6.17PAE

6.14 For photon with the following energies, calculate the wavelength and identify the region of the spectrum they are from. (a) 3.5 × 10 − 20 J, (b) 8.7 × 10 − 26 J, (c) 7.1 × 10 − 17 J, (d) 5.5 × 10 − 27 J

(a)

Expert Solution

Interpretation Introduction

To determine:

Find out the wavelength and identify the region of the spectrum of electromagnetic wave, whose energy of one photon is 3.5×10−20J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=3.5×10−20J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)3.5× 10 −20J=5.686×10−6m=5.7μm

From the electromagnetic spectrum this wave length lies in the IR region.

(b)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies whose one photon energy is 8.7×10−26J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=8.7×10−26J

Putting these values in above equation you get.

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)8.7× 10 −26J=2.287=2.3m

From the electromagnetic spectrum this wave length lies in the region of radio wave.

(c)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies, whose one photon energy is 7.1×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=7.1×10−17J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)7.1× 10 −17J=2.803×10−9m=2.803nm

From the spectrum this wave length lies in the region y-ray. So it belongs to y-rays.

(d)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies. Whose one photon energy is 5.5×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=5.5×10−27J

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)5.5× 10 −27J=36.18m

This wave length lies in the region of radio wave. So it lies in the radio wave region.

Conclusion

Wave length is inversely proportional to energy of a photon.

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Answer 2

Textbook Question

Chapter 6, Problem 6.17PAE

6.14 For photon with the following energies, calculate the wavelength and identify the region of the spectrum they are from. (a) 3.5 × 10 − 20 J, (b) 8.7 × 10 − 26 J, (c) 7.1 × 10 − 17 J, (d) 5.5 × 10 − 27 J

(a)

Expert Solution

Interpretation Introduction

To determine:

Find out the wavelength and identify the region of the spectrum of electromagnetic wave, whose energy of one photon is 3.5×10−20J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=3.5×10−20J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)3.5× 10 −20J=5.686×10−6m=5.7μm

From the electromagnetic spectrum this wave length lies in the IR region.

(b)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies whose one photon energy is 8.7×10−26J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=8.7×10−26J

Putting these values in above equation you get.

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)8.7× 10 −26J=2.287=2.3m

From the electromagnetic spectrum this wave length lies in the region of radio wave.

(c)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies, whose one photon energy is 7.1×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=7.1×10−17J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)7.1× 10 −17J=2.803×10−9m=2.803nm

From the spectrum this wave length lies in the region y-ray. So it belongs to y-rays.

(d)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies. Whose one photon energy is 5.5×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=5.5×10−27J

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)5.5× 10 −27J=36.18m

This wave length lies in the region of radio wave. So it lies in the radio wave region.

Conclusion

Wave length is inversely proportional to energy of a photon.

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Answer 3

Textbook Question

Chapter 6, Problem 6.17PAE

6.14 For photon with the following energies, calculate the wavelength and identify the region of the spectrum they are from. (a) 3.5 × 10 − 20 J, (b) 8.7 × 10 − 26 J, (c) 7.1 × 10 − 17 J, (d) 5.5 × 10 − 27 J

(a)

Expert Solution

Interpretation Introduction

To determine:

Find out the wavelength and identify the region of the spectrum of electromagnetic wave, whose energy of one photon is 3.5×10−20J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=3.5×10−20J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)3.5× 10 −20J=5.686×10−6m=5.7μm

From the electromagnetic spectrum this wave length lies in the IR region.

(b)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies whose one photon energy is 8.7×10−26J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=8.7×10−26J

Putting these values in above equation you get.

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)8.7× 10 −26J=2.287=2.3m

From the electromagnetic spectrum this wave length lies in the region of radio wave.

(c)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies, whose one photon energy is 7.1×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=7.1×10−17J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)7.1× 10 −17J=2.803×10−9m=2.803nm

From the spectrum this wave length lies in the region y-ray. So it belongs to y-rays.

(d)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies. Whose one photon energy is 5.5×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=5.5×10−27J

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)5.5× 10 −27J=36.18m

This wave length lies in the region of radio wave. So it lies in the radio wave region.

Conclusion

Wave length is inversely proportional to energy of a photon.

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Answer 4

Textbook Question

Chapter 6, Problem 6.17PAE

6.14 For photon with the following energies, calculate the wavelength and identify the region of the spectrum they are from. (a) 3.5 × 10 − 20 J, (b) 8.7 × 10 − 26 J, (c) 7.1 × 10 − 17 J, (d) 5.5 × 10 − 27 J

(a)

Expert Solution

Interpretation Introduction

To determine:

Find out the wavelength and identify the region of the spectrum of electromagnetic wave, whose energy of one photon is 3.5×10−20J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=3.5×10−20J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)3.5× 10 −20J=5.686×10−6m=5.7μm

From the electromagnetic spectrum this wave length lies in the IR region.

(b)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies whose one photon energy is 8.7×10−26J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=8.7×10−26J

Putting these values in above equation you get.

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)8.7× 10 −26J=2.287=2.3m

From the electromagnetic spectrum this wave length lies in the region of radio wave.

(c)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies, whose one photon energy is 7.1×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=7.1×10−17J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)7.1× 10 −17J=2.803×10−9m=2.803nm

From the spectrum this wave length lies in the region y-ray. So it belongs to y-rays.

(d)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies. Whose one photon energy is 5.5×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=5.5×10−27J

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)5.5× 10 −27J=36.18m

This wave length lies in the region of radio wave. So it lies in the radio wave region.

Conclusion

Wave length is inversely proportional to energy of a photon.

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Answer 5

Textbook Question

Answer 6

Textbook Question

Answer 7

(a)

Expert Solution

Interpretation Introduction

To determine:

Find out the wavelength and identify the region of the spectrum of electromagnetic wave, whose energy of one photon is 3.5×10−20J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=3.5×10−20J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)3.5× 10 −20J=5.686×10−6m=5.7μm

From the electromagnetic spectrum this wave length lies in the IR region.

(b)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies whose one photon energy is 8.7×10−26J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=8.7×10−26J

Putting these values in above equation you get.

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)8.7× 10 −26J=2.287=2.3m

From the electromagnetic spectrum this wave length lies in the region of radio wave.

(c)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies, whose one photon energy is 7.1×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=7.1×10−17J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)7.1× 10 −17J=2.803×10−9m=2.803nm

From the spectrum this wave length lies in the region y-ray. So it belongs to y-rays.

(d)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies. Whose one photon energy is 5.5×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=5.5×10−27J

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)5.5× 10 −27J=36.18m

This wave length lies in the region of radio wave. So it lies in the radio wave region.

Conclusion

Wave length is inversely proportional to energy of a photon.

Answer 8

(a)

Expert Solution

Interpretation Introduction

To determine:

Find out the wavelength and identify the region of the spectrum of electromagnetic wave, whose energy of one photon is 3.5×10−20J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=3.5×10−20J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)3.5× 10 −20J=5.686×10−6m=5.7μm

From the electromagnetic spectrum this wave length lies in the IR region.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Interpretation Introduction

To determine:

Find out the wavelength and identify the region of the spectrum of electromagnetic wave, whose energy of one photon is 3.5×10−20J.

Answer 13

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=3.5×10−20J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)3.5× 10 −20J=5.686×10−6m=5.7μm

From the electromagnetic spectrum this wave length lies in the IR region.

Answer 14

(b)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies whose one photon energy is 8.7×10−26J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=8.7×10−26J

Putting these values in above equation you get.

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)8.7× 10 −26J=2.287=2.3m

From the electromagnetic spectrum this wave length lies in the region of radio wave.

Answer 15

(b)

Expert Solution

Answer 16

(b)

Expert Solution

Answer 17

Expert Solution

Answer 18

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies whose one photon energy is 8.7×10−26J.

Answer 19

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=8.7×10−26J

Putting these values in above equation you get.

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)8.7× 10 −26J=2.287=2.3m

From the electromagnetic spectrum this wave length lies in the region of radio wave.

Answer 20

(c)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies, whose one photon energy is 7.1×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=7.1×10−17J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)7.1× 10 −17J=2.803×10−9m=2.803nm

From the spectrum this wave length lies in the region y-ray. So it belongs to y-rays.

Answer 21

(c)

Expert Solution

Answer 22

(c)

Expert Solution

Answer 23

Expert Solution

Answer 24

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies, whose one photon energy is 7.1×10−17J.

Answer 25

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=7.1×10−17J

Putting these values in above equation you get

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)7.1× 10 −17J=2.803×10−9m=2.803nm

From the spectrum this wave length lies in the region y-ray. So it belongs to y-rays.

Answer 26

(d)

Expert Solution

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies. Whose one photon energy is 5.5×10−17J.

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=5.5×10−27J

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)5.5× 10 −27J=36.18m

This wave length lies in the region of radio wave. So it lies in the radio wave region.

Conclusion

Wave length is inversely proportional to energy of a photon.

Answer 27

(d)

Expert Solution

Answer 28

(d)

Expert Solution

Answer 29

Expert Solution

Answer 30

Interpretation Introduction

To determine:

Find out the wave length and identify the region in spectrum lies. Whose one photon energy is 5.5×10−17J.

Answer 31

Explanation of Solution

λ=( hCE)h=6.634×10−34JsC=3×108m/sE=5.5×10−27J

λ=( 6.634× 10 −34 Js)×( 3× 10 8 m/s)5.5× 10 −27J=36.18m

This wave length lies in the region of radio wave. So it lies in the radio wave region.

Answer 32

Conclusion

Wave length is inversely proportional to energy of a photon.

Answer 33

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6.14 For photon with the following energies, calculate the wavelength and identify the region of the spectrum they are from. (a) 3.5 × 10 − 20 J, (b) 8.7 × 10 − 26 J, (c) 7.1 × 10 − 17 J, (d) 5.5 × 10 − 27 J

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