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How many photons per second are emitted by a 7.50-mW CO\(_2\) laser that has a wavelength of 10.6 \(\mu\)m?

Short Answer

Expert verified
The laser emits approximately \( 3.99 \times 10^{16} \) photons per second.

Step by step solution

01

Identify and Note Down the Given Information

Firstly, we need to note the given values. The power of the laser is \( P = 7.50 \, \text{mW} = 7.50 \times 10^{-3} \, \text{W} \). The wavelength of the laser is \( \lambda = 10.6 \, \mu\text{m} = 10.6 \times 10^{-6} \, \text{m} \).
02

Calculate the Energy of a Single Photon

The energy \( E \) of a single photon can be calculated using the formula: \( E = \frac{hc}{\lambda} \), where \( h = 6.63 \times 10^{-34} \, \text{J s} \) is Planck's constant and \( c = 3.00 \times 10^{8} \, \text{m/s} \) is the speed of light. Substituting the values, we get \( E = \frac{6.63 \times 10^{-34} \times 3.00 \times 10^{8}}{10.6 \times 10^{-6}} \approx 1.88 \times 10^{-20} \, \text{J} \).
03

Calculate the Number of Photons Emitted Per Second

The number of photons emitted per second is calculated by dividing the power of the laser by the energy of a single photon: \( N = \frac{P}{E} = \frac{7.50 \times 10^{-3}}{1.88 \times 10^{-20}} \approx 3.99 \times 10^{16} \).
04

Conclusion

The laser emits approximately \( 3.99 \times 10^{16} \) photons per second.

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Key Concepts

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

Planck's Constant
Planck's Constant is a fundamental constant used in quantum mechanics that relates the energy of a photon to its frequency. Denoted by the symbol \( h \), its value is approximately \( 6.63 \times 10^{-34} \, \text{J s} \). This tiny value highlights the minute energies involved in individual photons compared to everyday objects.
In the formula \( E = \frac{hc}{\lambda} \), Planck's Constant \( h \) is used along with the speed of light \( c \) and the wavelength \( \lambda \) to determine the energy \( E \) of a single photon. The interplay of these constants allows us to translate the seemingly abstract concept of wave properties into tangible energy values. Understanding this relationship is crucial to grasp topics like the photoelectric effect and quantum theory.
Key points about Planck's Constant include:
  • Bridges a gap between the energy and frequency of electromagnetic waves.
  • Essential in explaining how energy is quantized in quantum mechanics.
  • Integral in calculations involving photon emissions in lasers and other devices.
Understanding and correctly applying Planck's Constant can profoundly impact how we compute and conceptualize energy transformations at the microscopic level.
Photon Energy
Photon Energy refers to the energy carried by a single photon, the basic unit of light and other electromagnetic radiation. It is a direct result of the frequency and wavelength associated with that photon.
The relationship between these properties is given by the formula \( E = \frac{hc}{\lambda} \). By substituting the constants for Planck's Constant \( h \) and the speed of light \( c \), as well as the wavelength \( \lambda \), you can calculate the energy \( E \) of a photon in joules. This formula highlights how the energy of a photon is inversely proportional to its wavelength. Therefore, shorter wavelengths imply higher energy photons.
Important aspects of calculating Photon Energy include:
  • The equation's dependence on accurate measurements of the listed constants.
  • How varying wavelengths affect energy levels.
  • Its role in technologies like lasers and solar panels, where precise energy levels are crucial.
Mastering the understanding of photon energy is fundamental in both theoretical physics and applied technologies, making it a key concept for students and scientists alike.
Laser Power
Laser Power is a measure of the total energy output rate of a laser, typically given in watts (W) or milliwatts (mW). It indicates how much energy is delivered over a period, often per second.
In the context of photon emission, laser power directly relates to how many photons are emitted by the laser each second. By knowing the laser power and the energy per photon, one can determine the emission rate using the formula \( N = \frac{P}{E} \), where \( N \) is the number of photons per second, \( P \) is the power of the laser, and \( E \) is the energy of a single photon.
Considerations when dealing with Laser Power:
  • Specified in either watts or a fraction thereof, such as milliwatts, for precision in measurement.
  • Critical in applications like communications, manufacturing, and medical treatments.
  • Influences the capability of a laser to perform diverse functions based on the number of photons it emits.
Understanding laser power helps clarify how laser systems work and optimize their usage in both experimental and practical applications, emphasizing its significance in both scientific inquiry and industry practices.
Wavelength
Wavelength is the distance between successive crests of a wave, especially points in a sound wave or electromagnetic wave. For photons, and particularly in the case of laser emissions, wavelength is inherently tied to both the energy formula and photon emission rates.
The wavelength of a laser, noted usually in micrometers (\( \mu m \)) or meters (\( m \)), provides critical details about the laser's color and the energy each photon carries. A laser's wavelength directly determines the energy per photon through the equation \( E = \frac{hc}{\lambda} \), illustrating the inverse relationship between wavelength and energy.
Central points regarding wavelength include:
  • A shorter wavelength corresponds to a higher energy photon.
  • Essential in determining a laser's efficiency, effectiveness, and specific application scope.
  • Varies across regions like ultraviolet, visible, infrared, impacting its practical applications.
Grasping how wavelength influences photon properties is vital for interpreting the behavior of light in numerous scientific and technological contexts, strengthening one's understanding of electromagnetic theories and practices.

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Most popular questions from this chapter

In the Bohr model of the hydrogen atom, what is the de Broglie wavelength of the electron when it is in (a) the \(n\) = 1 level and (b) the \(n\) = 4 level? In both cases, compare the de Broglie wavelength to the circumference 2\(\pi{r_n}\) of the orbit.

To investigate the structure of extremely small objects, such as viruses, the wavelength of the probing wave should be about one-tenth the size of the object for sharp images. But as the wavelength gets shorter, the energy of a photon of light gets greater and could damage or destroy the object being studied. One alternative is to use electron matter waves instead of light. Viruses vary considerably in size, but 50 nm is not unusual. Suppose you want to study such a virus, using a wave of wavelength 5.00 nm. (a) If you use light of this wavelength, what would be the energy (in eV) of a single photon? (b) If you use an electron of this wavelength, what would be its kinetic energy (in eV)? Is it now clear why matter waves (such as in the electron microscope) are often preferable to electromagnetic waves for studying microscopic objects?

(a) For one-electron ions with nuclear charge Z, what is the speed of the electron in a Bohr-model orbit labeled with \(n\)? Give your answer in terms of \(v_1\), the orbital speed for the \(n\) = 1 Bohr orbit in hydrogen. (b) What is the largest value of Z for which the \(n\) = 1 orbital speed is less than 10\(\%\) of the speed of light in vacuum?

A certain atom has an energy level 2.58 eV above the ground level. Once excited to this level, the atom remains in this level for 1.64 \(\times\) 10\(^{-7}\) s (on average) before emitting a photon and returning to the ground level. (a) What is the energy of the photon (in electron volts)? What is its wavelength (in nanometers)? (b) What is the smallest possible uncertainty in energy of the photon? Give your answer in electron volts. (c) Show that \(\mid \Delta{E}/E \mid = \mid \Delta \lambda/\lambda \mid if \mid \Delta \lambda/\lambda \mid \ll\) 1. Use this to calculate the magnitude of the smallest possible uncertainty in the wavelength of the photon. Give your answer in nanometers.

What is the de Broglie wavelength for an electron with speed (a) \(v\)= 0.480\(c\) and (b) \(v\) = 0.960\(c\)? (\(Hint\): Use the correct relativistic expression for linear momentum if necessary.)

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