Physics
Physics
3rd Edition
ISBN: 9780073512150
Author: Alan Giambattista, Betty Richardson, Robert C. Richardson Dr.
Publisher: McGraw-Hill Education
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Chapter 27, Problem 91P
To determine

The number of photons emitted per second.

Expert Solution & Answer
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Answer to Problem 91P

The number of photons emitted per second is 2.2×1014photons/s_.

Explanation of Solution

Write the expression for the number of photons emitted per second by the laser.

    N=EEphoton                                                                                                                (I)

Here, N is the number of photons emitted per second by the laser, E is the total energy of the pulse, Ephoton is the energy of a photon.

Write the expression for the energy of a photon.

    Ephoton=hcλ                                                                                                              (II)

Here, h is the Planck’s constant, c is the speed of the light, λ is the wavelength of the light.

Write the expression for the energy density of the pulse.

    u=ε0Erms2                                                                                                            (III)

Here, u is the energy density of the pulse, ε0 permittivity of free space, Erms is the rms electric field.

Write the expression for the energy density of the pulse in terms of total energy.

    u=EV                                                                                                                  (IV)

Here, E is the total energy, V is the volume of the cylindrical pulse.

Use equation (III) in (IV) and solve for E.

    E=ε0Erms2V                                                                                                             (V)

Write the expression for the volume of the cylindrical pulse.

    V=length×area

Write the expression for the length of the cylinder.

    l=cΔt                                                                                                                   (VI)

Here, l is the length of the cylinder, Δt is the time interval.

Write the expression for the area of the cylinder.

    A=π(d2)2                                                                                                         (VII)

Here, d is the diameter of the cylinder, A is the area of the cylinder.

Substitute equation (VI) and (VII) in the expression for the volume of the cylinder.

    V=cΔt×π(d2)2                                                                                              (VIII)

Substitute equation (V), (II) and equation (VIII) in equation(I),

    N=ε0Erms2Vhcλ=λε0Erms2(14cΔtπd2)hc                                                                                (IX)

Rearrange the above equation to find number of photons per second.

    NΔt=λε0Erms2(πd2)4h                                                                                     (X)

Conclusion:

Substitute 640nm for λ, 8.854×1012C2/(Nm2) for ε0, 120V/m for Erms and 1.5mm for d and 6.626×1034Js for h in equation (X) to find NΔt.

    NΔt=π(640nm)[8.854×1012C2/(Nm2)](120V/m)2(1.5mm)24(6.626×1034Js)=π(640nm×1m109nm)[8.854×1012C2/(Nm2)](120V/m)2(1.5mm×1m1000mm)24(6.626×1034Js)=2.2×1014photons/s

Therefore, the number of photons emitted per second is 2.2×1014photons/s_ .

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Chapter 27 Solutions

Physics

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