College Physics Volume 2
College Physics Volume 2
2nd Edition
ISBN: 9781319115111
Author: Roger Freedman, Todd Ruskell, Philip R. Kesten, David L. Tauck
Publisher: W. H. Freeman
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Chapter 22, Problem 47QAP
To determine

(a)

The wavelength and frequency of photon that have the energy 2.33×1019J ?

Expert Solution
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Explanation of Solution

Calculation:

Wavelength
  λ=hcEphoton=6.63×1034×3×1082.33×1019=8.57×107m=857nm

Frequency
  f=Ephotonh=2.33×10196.63×1034=3.51×1014Hz

Conclusion:

Wavelength =857nm

Frequency =3.51×1014Hz

To determine

(b)

The wavelength and frequency of photon that have the energy 4.50×1019J ?

Expert Solution
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Explanation of Solution

Calculation:

Wavelength
  λ=hcEphoton=6.63×1034×3×1084.5×1019=4.42×107m=442nm

Frequency
  f=Ephotonh=4.50×10196.63×1034=6.79×1014Hz

Conclusion:

Wavelength =442nm

Frequency =6.79×1014Hz

To determine

(c)

The wavelength and frequency of photon that have the energy 3.20×1019J ?

Expert Solution
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Explanation of Solution

Calculation:

Wavelength
  λ=hcEphoton=6.63×1034×3×1083.20×1019=6.22×107m=622nm

Frequency
  f=Ephotonh=3.20×10196.63×1034=4.83×1014Hz

Conclusion:

Wavelength =622nm

Frequency =4.83×1014Hz

To determine

(d)

The wavelength and frequency of photon that have the energy 8.55×1019J ?

Expert Solution
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Explanation of Solution

Calculation:

Wavelength
  λ=hcEphoton=6.63×1034×3×1088.55×1019=2.33×107m=233nm

Frequency
  f=Ephotonh=8.55×10196.63×1034=1.29×1015Hz

Conclusion:

Wavelength =233nm

Frequency =1.29×1015Hz

To determine

(e)

The wavelength and frequency of photon that have the energy 63.3eV ?

Expert Solution
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Explanation of Solution

Calculation:

Energy in joules,
  Ephoton=63.3×1.60×1019=1.01×1017J

Wavelength
  λ=hcEphoton=6.63×1034×3×1081.01×1017=1.96×108m=19.6nm

Frequency
  f=Ephotonh=1.01×10176.63×1034=1.53×1016Hz

Conclusion:

Wavelength =19.6nm

Frequency =1.53×1016Hz

To determine

(f)

The wavelength and frequency of photon that have the energy 8.77eV ?

Expert Solution
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Explanation of Solution

Calculation:

Energy in joules,
  Ephoton=8.77×1.60×1019=1.40×1018J

Wavelength
  λ=hcEphoton=6.63×1034×3×1081.40×1018=1.42×108m=14.2nm

Frequency
  f=Ephotonh=1.40×10186.63×1034=2.12×1015Hz

Conclusion:

Wavelength =14.2nm

Frequency =2.12×1015Hz

To determine

(g)

The wavelength and frequency of photon that have the energy 1.98eV ?

Expert Solution
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Explanation of Solution

Calculation:

Energy in joules,
  Ephoton=1.98×1.60×1019=3.17×1019J

Wavelength
  λ=hcEphoton=6.63×1034×3×1083.17×1019=6.28×107m=628nm

Frequency
  f=Ephotonh=3.17×10196.63×1034=4.78×1014Hz

Conclusion:

Wavelength =628nm

Frequency =4.78×1014Hz

To determine

(h)

The wavelength and frequency of photon that have the energy 4.55eV ?

Expert Solution
Check Mark

Explanation of Solution

Calculation:

Energy in joules,
  Ephoton=4.55×1.60×1019=7.28×1019J

Wavelength
  λ=hcEphoton=6.63×1034×3×1087.28×1019=2.73×107m=273nm

Frequency
  f=Ephotonh=7.28×10196.63×1034=1.10×1015Hz

Conclusion:

Wavelength =273nm

Frequency =1.10×1015Hz

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