/*! 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 4 A helium-neon laser emits red li... [FREE SOLUTION] | 91Ó°ÊÓ

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A helium-neon laser emits red light at wavelength \(\lambda=633 \mathrm{nm}\) in a beam of diameter \(3.5 \mathrm{~mm}\) and at an energy-emission rate of \(5.0 \mathrm{~mW}\). A detector in the beam's path totally absorbs the beam. At what rate per unit area does the detector absorb photons?

Short Answer

Expert verified
The detector absorbs photons at a rate of approximately \(1.65 \times 10^{21}\) photons per second per square meter.

Step by step solution

01

Calculate the Frequency of the Laser Light

To find the photon absorption rate, we first need to calculate the frequency of the laser light using its wavelength. The formula to calculate frequency \( f \) from wavelength \( \lambda \) is:\[ f = \frac{c}{\lambda} \]where \( c = 3.00 \times 10^8 \) m/s is the speed of light and \( \lambda = 633 \times 10^{-9} \) m. Plugging in the values, we get:\[ f = \frac{3.00 \times 10^8}{633 \times 10^{-9}} \approx 4.74 \times 10^{14} \text{ Hz} \]
02

Calculate the Energy of a Single Photon

The energy \( E \) of a single photon is given by the equation:\[ E = hf \]where \( h = 6.63 \times 10^{-34} \text{ Jâ‹…s} \) is Planck's constant. Now substituting the values we calculated previously:\[ E = (6.63 \times 10^{-34}) \times (4.74 \times 10^{14}) \approx 3.14 \times 10^{-19} \text{ J} \]
03

Calculate the Number of Photons Absorbed Per Second

The rate of energy emission by the laser is given as \( 5.0 \text{ mW} \) or \( 5.0 \times 10^{-3} \text{ W} \). Since \( 1 \text{ W} = 1 \text{ J/s} \), the number of photons \( n \) emitted per second is given by:\[ n = \frac{\text{Power}}{E} = \frac{5.0 \times 10^{-3}}{3.14 \times 10^{-19}} \approx 1.59 \times 10^{16} \text{ photons/s} \]
04

Calculate the Area of the Beam

The beam has a diameter of \( 3.5 \text{ mm} \) or \( 3.5 \times 10^{-3} \text{ m} \). The area \( A \) of the beam is calculated using the formula for the area of a circle \( A = \pi r^2 \), where \( r \) is the radius:\[ r = \frac{3.5 \times 10^{-3}}{2} = 1.75 \times 10^{-3} \text{ m} \]\[ A = \pi \times (1.75 \times 10^{-3})^2 \approx 9.62 \times 10^{-6} \text{ m}^2 \]
05

Calculate the Rate of Photon Absorption Per Unit Area

Now that we have the number of photons absorbed per second and the area of the beam, the rate per unit area \( R \) is:\[ R = \frac{n}{A} = \frac{1.59 \times 10^{16}}{9.62 \times 10^{-6}} \approx 1.65 \times 10^{21} \text{ photons/(s m}^2\text{)} \]

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Key Concepts

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

Wavelength of Light
Light, being an electromagnetic wave, has a characteristic called wavelength, denoted as \( \lambda \). It represents the distance between two consecutive peaks of the wave—measured in meters (m). In this exercise, we are dealing with red light emitted by a helium-neon laser, which has a wavelength of 633 nanometers (nm). To better understand, note that 1 nanometer equals \( 10^{-9} \) meters, making light wavelength incredibly small and often invisible to the human eye.
When light interacts with materials, its wavelength is a crucial factor that determines how it behaves. Wavelength affects various properties like the color of light in visible spectra. In essence, red light has longer wavelengths than blue light, placing it on the lower energy end of the visible light spectrum.
Understanding wavelength is key because it helps us further discern light's behavior when it travels through different mediums, or when it's interacting with detectors, as seen in the laser beam exercise.
Photon Energy
Photon energy is the amount of energy carried by a single photon, and it's essential for understanding light-matter interactions. This energy can be calculated using the equation \( E = hf \), where \( E \) is the energy of the photon, \( h \) is Planck's constant \((6.63 \times 10^{-34} \text{ J}\cdot\text{s})\), and \( f \) is the frequency of the light.
In our exercise, we calculated the photon energy to be approximately \( 3.14 \times 10^{-19} \text{ J} \) for the red light emitted by the laser. This energy represents the quantum of energy exchanged when a photon is absorbed by a detector. The lower the wavelength, the higher the frequency—and hence, the greater the energy of the photon.
Photon energy is a key concept in understanding not only how light is absorbed and emitted but also other phenomena such as photoelectric effects and quantum mechanics.
The higher the photon energy, the more impactful is the interaction it can have with matter. Knowing this, we can predict and analyze various light behaviors and their influence on various technologies, including but not limited to laser engineering and optical devices.
Frequency Calculation
Frequency, symbolized by \( f \), is the number of wavelengths passing a point per second. It is measured in Hertz (Hz), with one Hz equaling one cycle per second. To calculate this, you can use the formula \( f = \frac{c}{\lambda} \), where \( c \) is the speed of light \((3.00 \times 10^8 \text{ m/s})\).
For light with a wavelength of 633 nm, as used in the laser example, the frequency comes out to approximately \( 4.74 \times 10^{14} \text{ Hz} \). This high frequency is typical for visible light, part of the broader electromagnetic spectrum.
Understanding frequency helps in determining many of light's properties, including photon energy and color. The concept anchors many everyday technologies—from radio transmissions to microwave ovens—since they all operate based on electromagnetic wave frequencies.
  • Higher frequencies mean higher energy photons.
  • Frequency is inversely related to wavelength: as one increases, the other decreases.
Knowing light's frequency aids in envisioning its interaction with materials, ensuring this exercise not only measures but also predicts photon absorption rates.
Beam Area Calculation
To calculate the beam area of the laser light, we must consider the geometry of the beam. Assuming a circular cross-section, the area \( A \) is given by the formula \( A = \pi r^2 \), where \( r \) is the radius of the circle.
For our red laser beam, the diameter is 3.5 mm. To find the radius, simply divide the diameter by two: \( r = \frac{3.5 \text{ mm}}{2} = 1.75 \text{ mm} \) or \( 1.75 \times 10^{-3} \text{ m} \) after converting to meters.
The area, therefore, is calculated as \( A = \pi \times (1.75 \times 10^{-3})^2 \approx 9.62 \times 10^{-6} \text{ m}^2 \). This area is crucial because it determines the concentration of the light beam intersecting with the detector.
A smaller beam concentrates more energy over a smaller area, increasing the potential rate of photon absorption. Calculating beam area allows the practice of understanding how size affects interaction rates, which is important in engineering and scientific contexts involving lasers and other optical systems.

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Most popular questions from this chapter

(a) Let \(n=a+i b\) be a complex number, where \(a\) and \(b\) are real (positive or negative) numbers. Show that the product \(n n^{*}\) is always a positive real number. (b) Let \(m=c+i d\) be another complex number. Show that \(|n m|=|n||m|\)

You will find in Chapter 39 that electrons cannot move in definite orbits within atoms, like the planets in our solar system. To see why, let us try to "observe" such an orbiting electron by using a light microscope to measure the electron's presumed orbital position with a precision of, say, \(10 \mathrm{pm}\) (a typical atom has a radius of about \(100 \mathrm{pm}\) ). The wavelength of the light used in the microscope must then be about \(10 \mathrm{pm}\). (a) What would be the photon energy of this light? (b) How much energy would such a photon impart to an electron in a head-on collision? (c) What do these results tell you about the possibility of "viewing" an atomic electron at two or more points along its presumed orbital path? (Hint: The outer electrons of atoms are bound to the atom by energies of only a few electron-volts.)

Using the classical equations for momentum and kinetic energy, show that an electron's de Broglie wavelength in nanometers can be written as \(\lambda=1.226 / \sqrt{K}\), in which \(K\) is the electron's kinetic energy in electron-volts.

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