Chapter 27, Problem 93P

(a)

To determine

The energy and momentum of one laser photon if the laser wavelength is $670\text{ nm}$ .

Answer to Problem 93P

The energy of the photon is $1.9\text{ eV}$ and its momentum is $9.9\times {10}^{-28}\text{ kg}\cdot \text{m/s}$ .

Explanation of Solution

Write the equation for the energy of the photon.

  $E=\frac{hc}{\lambda }$                                                                                                                     (I)

Here, $E$ is the energy of the photons, $h$ is the Planck’s constant, $c$ is the speed of light in vacuum, and $\lambda$ is the wavelength of the photon.

Write the equation for the energy of the momentum of the photon in terms of energy.

  $E=pc$

Here, $p$ is the momentum of the photon.

Rewrite the above equation for $p$ .

  $p=\frac{E}{c}$                                                                                                                      (II)

Conclusion:

The value of $hc$ is $1240\text{ eV}\cdot \text{nm}$ and the value of $c$ is $3.00\times {10}^8\text{ m/s}$ .

Substitute $1240\text{ eV}\cdot \text{nm}$ for $hc$ and $670\text{ nm}$ for $\lambda$ in equation (I) to find $E$ .

  $\begin{array}{c}E=\frac{1240\text{ eV}\cdot \text{nm}}{670\text{ nm}}\\ =1.85\text{ eV}\\ \approx 1.9\text{ eV}\end{array}$

Substitute $1.85\text{ eV}$ for $E$ and $3.00\times {10}^8\text{ m/s}$ for $c$ in equation (II) to find $p$ .

  $\begin{array}{c}p=\frac{1.85\text{ eV}\frac{1.6\times {10}^{-19}\text{ J}}{1\text{ eV}}}{3.00\times {10}^8\text{ m/s}}\\ =9.9\times {10}^{-28}\text{ kg}\cdot \text{m/s}\end{array}$

Therefore, the energy of the photon is $1.9\text{ eV}$ and its momentum is $9.9\times {10}^{-28}\text{ kg}\cdot \text{m/s}$ .

(b)

To determine

The number of photons per second emitted by the laser.

Answer to Problem 93P

The number of photons per second emitted by the laser is $3\times {10}^{15}\text{ photons/s}$ .

Explanation of Solution

The number of photons per second emitted by the laser is given by the ratio of power of the laser to the energy per photon.

Write the equation for the rate of photon emission.

  $\text{rate}=\frac{P}{E}$                                                                                                                (III)

Conclusion:

Given that the output power of the laser pointer is $1\text{ mW}$ .

Substitute $1\text{ mW}$ for $P$ and $1.85\text{ eV}$ for $E$ in equation (III) to find the photon emission rate.

  $\begin{array}{c}\text{rate}=\frac{\left(1\text{ mW}\frac{1\text{ W}}{{10}^3\text{ mW}}\right)}{\left(1.85\text{ eV}\frac{1.6\times {10}^{-19}\text{ J}}{1\text{ eV}}\right)}\\ =\frac{1\times {10}^{-3}\text{ W}}{2.96\times {10}^{-19}\text{ J}}\\ =3\times {10}^{15}\text{ photons/s}\end{array}$

Therefore, number of photons per second emitted by the laser is $3\times {10}^{15}\text{ photons/s}$ .

(c)

To determine

The average force on the laser due to the momentum carried away by the emitted photons.

Answer to Problem 93P

The average force on the laser due to the momentum carried away by the emitted photons is $3\times {10}^{-12}\text{ N}$ .

Explanation of Solution

The impulse momentum theorem states that the impulse is equal to the change in momentum. The impulse is the product of the average force applied and the time interval during which the force acts.

Write the mathematical expression for the impulse-momentum theorem.

  $\Delta p=F_{\text{av}}\Delta t$

Here, $\Delta p$ is the change in momentum, $F_{\text{av}}$ is the average force applied and $\Delta t$ is the time interval.

Rewrite the above equation for $F_{\text{av}}$ .

  $F_{\text{av}}=\frac{\Delta p}{\Delta t}$                                                                                                               (IV)

Use equation (II) to write the expression for the change in momentum.

  $\Delta p=\frac{\Delta E}{c}$                                                                                                                 (V)

Write the equation for $\Delta t$ .

  $\Delta t=\frac{\Delta E}{P}$                                                                                                                (VI)

Put equations (V) and (VI) in equation (IV).

  $\begin{array}{c}{F}_{\text{av}}=\frac{\Delta E/c}{\Delta E/P}\\ =\frac{P}{c}\end{array}$                                                                                                          (VII)

Conclusion:

Substitute $1\text{ mW}$ for $P$ and $3.00\times {10}^8\text{ m/s}$ for $c$ in equation (VII) to find $F_{\text{av}}$ .

  $\begin{array}{c}{F}_{\text{av}}=\frac{\left(1\text{ mW}\frac{1\text{ W}}{{10}^3\text{ mW}}\right)}{3.00\times {10}^8\text{ m/s}}\\ =\frac{1\times {10}^{-3}\text{ W}}{3.00\times {10}^8\text{ m/s}}\\ =3\times {10}^{-12}\text{ N}\end{array}$

Therefore, the average force on the laser due to the momentum carried away by the emitted photons is $3\times {10}^{-12}\text{ N}$ .