Glencoe Physics: Principles and Problems, Student Edition
Glencoe Physics: Principles and Problems, Student Edition
1st Edition
ISBN: 9780078807213
Author: Paul W. Zitzewitz
Publisher: Glencoe/McGraw-Hill
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Chapter 28, Problem 41A

a.

To determine

To calculate: Number of photons emitted per second.

a.

Expert Solution
Check Mark

Answer to Problem 41A

Number of photons per second that are emitted = 2_ .

Explanation of Solution

Given:

A laser beam’s power equals the photon energy times the number of photons per second that are emitted.

Wavelength of first laser = 840 nm

Wavelength of second laser = 427 nm

Formula used:

Energy change is given by,

  ΔE=hcλ

Where h is Planck’s constant, c is speed of light and λ is wavelength.

Calculation:

Energy change in 840 nm laser is calculated as:

  ΔE=hcλ                   [hc=1240 eVnm]      =1240 eVnm840 nm     =1.48 eV

Energy change in 427 nm laser is calculated as:

  ΔE=hcλ                   [hc=1240 eVnm]      =1240 eVnm427 nm     =2.90 eV

Ratio of energy change in each photon:

  =1.48 eV2.90 eV= 0.5

Number of photons per second = 10.5=2

Conclusion:

Number of photons per second that are emitted = 2_ .

b.

To determine

To calculate: Number of photons emitted per second.

b.

Expert Solution
Check Mark

Answer to Problem 41A

Number of photons per second = 2.1×1016_

Explanation of Solution

Given:

Power of laser beam = 5.0 mW

Wavelength of laser beam = 840 nm

Formula used:

Energy change is given by,

  ΔE=hcλ

Where h is Planck’s constant, c is speed of light and λ is wavelength.

Number of photons per second is given by,

  n=PE

Where P is power and E is energy.

Calculation:

Energy change in 840 nm laser is calculated as:

  ΔE=hcλ                   [hc=1240 eVnm]      =(1240 eVnm)(1.60×1019 J/eV)840 nm      =2.4×1019 J

Number of photons per second is calculate as:

  n=PE    =5.0×103 J/s2.4×1019 J/photon    = 2.1×1016 photon/s

Conclusion:

Number of photons per second = 2.1×1016_ .

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