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A helium-neon laser emits red light at wavelength \(\lambda=633 \mathrm{~nm}\) in a beam of diameter \(3.0 \mathrm{~mm}\) 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 rate of photon absorption per unit area is approximately \( 2.25 \times 10^{21} \text{ photons/m}^2 \text{ per second} \).

Step by step solution

01

Convert wavelength and energy emission rate to standard units

First, convert the given wavelength from nanometers to meters for easier calculations: \( 633 \text{ nm} = 633 \times 10^{-9} \text{ m} \). Similarly, the energy emission rate in milliwatts should be converted to watts: \( 5.0 \text{ mW} = 5.0 \times 10^{-3} \text{ W} \).
02

Calculate photon energy

The energy of one photon can be calculated using the formula \( E = \frac{hc}{\lambda} \), where \( h = 6.626 \times 10^{-34} \text{ J} \cdot \text{s} \) is Planck's constant, and \( c = 3.0 \times 10^8 \text{ m/s} \) is the speed of light. Substitute \( \lambda = 633 \times 10^{-9} \text{ m} \) to find: \[ E = \frac{(6.626 \times 10^{-34})(3.0 \times 10^{8})}{633 \times 10^{-9}} \approx 3.141 \times 10^{-19} \text{ J} \].
03

Calculate the rate of photons emitted

Using the energy emission rate of the laser, \( 5.0 \times 10^{-3} \text{ W} \), divide by the energy per photon calculated in Step 2:\[ \text{Number of photons per second} = \frac{5.0 \times 10^{-3}}{3.141 \times 10^{-19}} \approx 1.59 \times 10^{16} \text{ photons/s} \].
04

Determine beam area

Calculate the cross-sectional area of the laser beam using its diameter. The formula for the area \( A \) of a circle is \( A = \pi \left( \frac{d}{2} \right)^2 \), where \( d = 3.0 \times 10^{-3} \text{ m} \). So:\[ A = \pi \left( \frac{3.0 \times 10^{-3}}{2} \right)^2 \approx 7.07 \times 10^{-6} \text{ m}^2 \].
05

Calculate absorption rate per unit area

Finally, determine the absorption rate per unit area by dividing the number of photons absorbed per second by the beam area from Step 4:\[ \frac{1.59 \times 10^{16}}{7.07 \times 10^{-6}} \approx 2.25 \times 10^{21} \text{ photons/m}^2 \text{ per second} \].

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Key Concepts

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

Helium-Neon Laser
Helium-Neon lasers, commonly referred to as He-Ne lasers, are popular for their simplicity and the clarity of the laser beam they produce at the wavelength of 633 nanometers. These lasers work by exciting helium and neon gases in a sealed tube.
The emitted red light is coherent and monochromatic, which means it has a single color and wavelength that does not shift over time. This makes it perfect for a variety of applications like holography, barcode scanning, and in scientific research.
A crucial point about the He-Ne laser is the reliability and precision it provides due to low divergence, which means the laser beam spreads out very little as it travels further from its source. This property allows the He-Ne laser to maintain its focus over long distances, making it excellent for precise measurements and experiments.
Photon Energy Calculation
Calculating the energy of a photon is essential when working with laser light. Photon energy is directly related to the wavelength of light. The relationship is given by the formula:
\[ E = \frac{hc}{\lambda} \]
where:
  • \( E \) is the energy of the photon in joules.
  • \( h \) is Planck's constant, approximately \(6.626 \times 10^{-34} \text{ J} \cdot \text{s}\).
  • \( c \) is the speed of light, roughly \( 3.0 \times 10^8 \text{ m/s} \).
  • \( \lambda \) is the wavelength of the light in meters.
By calculating this for a He-Ne laser with a wavelength of 633 nanometers, you determine each photon's energy as approximately \(3.141 \times 10^{-19} \text{ J}\). This energy calculation is fundamental for determining how much power each photon carries through the laser beam.
Beam Cross-Sectional Area
The beam cross-sectional area is an important concept when dealing with lasers because it affects the concentration of light and energy that can be delivered to a certain spot. For a circular laser beam, the area is calculated using the diameter of the beam.
The formula to find the cross-sectional area \( A \) is:
\[ A = \pi \left( \frac{d}{2} \right)^2 \]
where \( d \) is the beam's diameter.
For instance, with a He-Ne laser beam having a diameter of 3.0 millimeters (converted to 0.003 meters), the area is approximately \(7.07 \times 10^{-6} \text{ m}^2\). This knowledge helps in determining how energy from the laser is distributed over a surface, critical for applications requiring precise targeting or energy density assessments.
Photons per Second
The concept of photons per second is important when dealing with the power of a laser beam. It refers to how many photons are emitted by the laser every second, which correlates with the laser's energy output.
This is calculated by dividing the laser's total energy emission rate by the energy per photon. If a He-Ne laser emits \(5.0 \times 10^{-3} \text{ W}\) of power, and each photon's energy is \(3.141 \times 10^{-19} \text{ J}\), the number of photons emitted per second can be calculated as:
\[ \frac{5.0 \times 10^{-3}}{3.141 \times 10^{-19}} \approx 1.59 \times 10^{16} \text{ photons/s}\]
Knowing the number of photons per second helps in evaluating the laser's capability to illuminate or heat a target area, and allows for precise control in applications like cutting, welding, or medical procedures.

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Most popular questions from this chapter

An ultraviolet lamp emits light of wavelength \(400 \mathrm{~nm}\) at the rate of 400 W. An infrared lamp emits light of wavelength \(700 \mathrm{~nm}\), also at the rate of \(400 \mathrm{~W}\). (a) Which lamp emits photons at the greater rate and (b) what is that greater rate?

Gamma rays of photon energy \(0.511 \mathrm{MeV}\) are directed onto an aluminum target and are scattered in various directions by loosely bound electrons there. (a) What is the wavelength of the incident gamma rays? (b) What is the wavelength of gamma rays scattered at \(60.0^{\circ}\) to the incident beam? (c) What is the photon energy of the rays scattered in this direction?

A photon undergoes Compton scattering off a stationary free electron. The photon scatters at \(90.0^{\circ}\) from its initial direction; its initial wavelength is \(4.00 \times 10^{-12} \mathrm{~m}\). What is the electron's kinetic energy?

The beam emerging from a \(1.5 \mathrm{~W}\) argon laser \((\lambda=515 \mathrm{~nm})\) has a diameter \(d\) of \(3.5 \mathrm{~mm}\). The beam is focused by a lens system with an effective focal length \(f_{\mathrm{L}}\) of \(2.5 \mathrm{~mm}\). The focused beam strikes a totally absorbing screen, where it forms a circular diffraction pattern whose central disk has a radius \(R\) given by \(1.22 f_{\mathrm{L}} \lambda / d\). It can be shown that \(84 \%\) of the incident energy ends up within this central disk. At what rate are photons absorbed by the screen in the central disk of the diffraction pattern?

An electron moves through a region of uniform electric potential of \(-200 \mathrm{~V}\) with a (total) energy of \(500 \mathrm{eV}\). What are its (a) kinetic energy (in electron-volts), (b) momentum, (c) speed, (d) de Broglie wavelength, and (e) angular wave number?

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