/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A laser produces a 5.00 -mW beam... [FREE SOLUTION] | 91影视

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Chapter 30: Problem 46

A laser produces a 5.00 -mW beam of light, consisting of photons with a wavelength of \(632.8 \mathrm{nm}\). (a) How many photons are emitted by the laser each second? (b) The laser beam strikes a black surface and is absorbed. What is the change in the momentum of each photon that is absorbed? (c) What force does the laser beam exert on the black surface?

Short Answer

Expert verified
(a) \(1.59 \times 10^{16}\) photons/s, (b) \(1.047 \times 10^{-27} \text{ kg} \cdot \text{m/s}\), (c) \(1.67 \times 10^{-11} \text{ N}\).

Step by step solution

01

Calculating Photon Energy

First, we determine the energy of each photon using its wavelength. The energy of a photon is given by the equation \[ E = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \) is Planck鈥檚 constant, \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light, and \( \lambda = 632.8 \text{ nm} = 632.8 \times 10^{-9} \text{ m} \). Substituting the values, we find:\[ E = \frac{(6.626 \times 10^{-34}) \times (3.00 \times 10^8)}{632.8 \times 10^{-9}} \approx 3.14 \times 10^{-19} \text{ J} \].
02

Calculating Number of Photons Per Second

To find the number of photons emitted per second, use the formula \[ n = \frac{P}{E} \]where \( P = 5.00 \times 10^{-3} \text{ W} \) is the power of the laser, and \( E \) is the energy of one photon. Inputting the values we have:\[ n = \frac{5.00 \times 10^{-3}}{3.14 \times 10^{-19}} \approx 1.59 \times 10^{16} \text{ photons/second} \].
03

Calculating Change in Photon Momentum

The change in momentum of a photon when it is absorbed is equal to its initial momentum since it is reduced to zero. The momentum of a photon is given by \[ p = \frac{h}{\lambda} \].Thus, substitute the known values:\[ p = \frac{6.626 \times 10^{-34}}{632.8 \times 10^{-9}} \approx 1.047 \times 10^{-27} \text{ kg} \cdot \text{m/s} \].
04

Calculating Force Exerted on Surface

The force exerted on the surface is the rate of change of momentum of the photons. Thus, the force can be determined by \[ F = \frac{\Delta p \cdot n}{\text{time}} \].Since we consider the emission per second, \( \Delta p = p \) and \( n = 1.59 \times 10^{16} \) from Step 2, substitute the values:\[ F = (1.047 \times 10^{-27}) \cdot (1.59 \times 10^{16}) \approx 1.67 \times 10^{-11} \text{ N} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Photon Energy Calculation
When it comes to understanding photon energy, an important equation to know is that the energy of a photon is calculated using its wavelength. This is given by the formula:
\[ E = \frac{hc}{\lambda} \]
Here, \( h \) is Planck鈥檚 constant \( (6.626 \times 10^{-34} \text{ J} \cdot \text{s}) \), \( c \) is the speed of light \( (3.00 \times 10^8 \text{ m/s}) \), and \( \lambda \) is the wavelength. For example, if a laser emits light at a wavelength of \( 632.8 \text{ nm} \), you can determine the energy of each photon it emits by substituting the known values.
- Plug in \( \lambda = 632.8 \times 10^{-9} \text{ m} \) into the formula.- Calculate the value of \( E \) which results in approximately \( 3.14 \times 10^{-19} \text{ J} \).
This calculation is essential in finding out how much energy each individual photon carries.
Momentum of Photons
Photon momentum might sound kind of strange since photons don't have mass, but they do have momentum, thanks to their wave-like properties. The momentum \( p \) of a photon is given by:
\[ p = \frac{h}{\lambda} \]
The substitution of the values of Planck's constant \( h \) and the wavelength \( \lambda \) helps us calculate the photon's momentum.
- For the wavelength of \( 632.8 \text{ nm} \), convert it to meters (\( 632.8 \times 10^{-9} \text{ m} \)).- Putting this into the equation gives a momentum of approximately \( 1.047 \times 10^{-27} \text{ kg} \cdot \text{m/s} \).
It's key to note that when a photon hits and is absorbed by a surface, it actively transfers its momentum, creating a force.
Force Exerted by Light
Finally, let's tackle how light can exert force. When photons hit a surface and are absorbed, their momentum is transferred, exerting a force.
To find this force, we use the equation:
\[ F = \frac{\Delta p \cdot n}{\text{time}} \]where \( \Delta p \) is the change in momentum and \( n \) is the number of photons.
- Each photon's momentum is \( 1.047 \times 10^{-27} \text{ kg} \cdot \text{m/s} \).- Earlier, we calculated that \( 1.59 \times 10^{16} \) photons are hitting the surface every second.
Multiply these together to find:- The force exerted, which is approximately \( 1.67 \times 10^{-11} \text{ N} \).
Even though the force from a single photon is tiny, the collective impact of many photons can be significant, defying initial expectations since light interacts with objects in surprising ways!

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Most popular questions from this chapter

(a) Which has the greater momentum, a photon of red light or a photon of blue light? Explain. (b) Calculate the momentum of a photon of red light \(\left(f=4.0 \times 10^{14} \mathrm{Hz}\right)\) and a photon of blue light \(\left(f=7.9 \times 10^{14} \mathrm{Hz}\right)\).

Find the kinctic energy of an electron whose de Broglie wavelength is 1.5 A.

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