Chapter 27, Problem 89P

To determine

The number of photons emitted per second.

Answer to Problem 89P

The number of photons emitted per second is $\underset{\_}{2.2\times {10}^{14}\text{photons}/\text{s}}$.

Explanation of Solution

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

    $N=\frac{E}{E_{\text{photon}}}$                                                                                                                (I)

Here, $N$ is the number of photons emitted per second by the laser, $E$ is the total energy of the pulse, $E_{\text{photon}}$ is the energy of a photon.

Write the expression for the energy of a photon.

    $E_{\text{photon}}=\frac{hc}{\lambda }$                                                                                                              (II)

Here, $h$ is the Planck’s constant, $c$ is the speed of the light, $\lambda$ is the wavelength of the light.

Write the expression for the energy density of the pulse.

    $\langle u\rangle ={\epsilon }_0{E}_{\text{rms}}^2$                                                                                                            (III)

Here, $\langle u\rangle$ is the energy density of the pulse, ${\epsilon }_0$ permittivity of free space, $E_{\text{rms}}$ is the rms electric field.

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

    $\langle u\rangle =\frac{E}{V}$                                                                                                                  (IV)

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

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

    $E={\epsilon }_0{E}_{\text{rms}}^{2}V$                                                                                                             (V)

Write the expression for the volume of the cylindrical pulse.

    $V=\text{length×area}$

Write the expression for the length of the cylinder.

    $l=c\Delta t$                                                                                                                   (VI)

Here, $l$ is the length of the cylinder, $\Delta t$ is the time interval.

Write the expression for the area of the cylinder.

    $A=\pi {\left(\frac{d}{2}\right)}^2$                                                                                                         (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.

    $V=c\Delta t\times \pi {\left(\frac{d}{2}\right)}^2$                                                                                              (VIII)

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

    $\begin{array}{c}N=\frac{{\epsilon }_0{E}_{\text{rms}}^{2}V}{\frac{hc}{\lambda }}\\ =\frac{\lambda {\epsilon }_0{E}_{\text{rms}}^2\left(\frac{1}{4}c\Delta t\pi d^2\right)}{hc}\end{array}$                                                                                (IX)

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

    $\frac{N}{\Delta t}=\frac{\lambda {\epsilon }_0{E}_{\text{rms}}^2\left(\pi d^2\right)}{4h}$                                                                                     (X)

Conclusion:

Substitute $640\text{nm}$ for $\lambda$, $8.854\times {10}^{-12}{\text{C}}^2/\left(\text{N}\cdot {\text{m}}^2\right)$ for ${\epsilon }_0$, $120\text{V}/\text{m}$ for $E_{\text{rms}}$ and $1.5\text{mm}$ for $d$ and $6.626\times {10}^{-34}\text{J}\cdot \text{s}$ for $h$ in equation (X) to find $\frac{N}{\Delta t}$.

    $\begin{array}{c}\frac{N}{\Delta t}=\frac{\pi \left(640\text{nm}\right)\left[8.854\times {10}^{-12}{\text{C}}^2/\left(\text{N}\cdot {\text{m}}^2\right)\right]{\left(120\text{V}/\text{m}\right)}^2{\left(1.5\text{mm}\right)}^2}{4\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}\\ =\frac{\pi \left(640\text{nm}\times \frac{1\text{m}}{{10}^9\text{nm}}\right)\left[8.854\times {10}^{-12}{\text{C}}^2/\left(\text{N}\cdot {\text{m}}^2\right)\right]{\left(120\text{V}/\text{m}\right)}^2{\left(1.5\text{mm}\times \frac{1\text{m}}{1000\text{mm}}\right)}^2}{4\left(6.626\times {10}^{-34}\text{J}\cdot \text{s}\right)}\\ =2.2\times {10}^{14}\text{photons}/\text{s}\end{array}$

Therefore, the number of photons emitted per second is $\underset{\_}{2.2\times {10}^{14}\text{photons}/\text{s}}$ .