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Chapter 28: Problem 3

A laser used to weld detached retinas emits light with a wavelength of \(652 \mathrm{nm}\) in pulses that are \(20.0 \mathrm{~ms}\) in duration. The average power expended during each pulse is \(0.600 \mathrm{~W}\). (a) How much energy is in each pulse, in joules? In electronvolts? (b) What is the energy of one photon in joules? In electronvolts? (c) How many photons are in each pulse?

Short Answer

Expert verified
Pulse energy: 0.012 J (\(7.49 \times 10^{16}\) eV); Photon energy: \(3.05 \times 10^{-19}\) J (1.90 eV); Photons per pulse: \(3.93 \times 10^{16}\).

Step by step solution

01

Calculate Energy per Pulse in Joules

The energy of a pulse can be calculated by multiplying the power of the pulse by its duration. Use the formula: \[ E = P \times t \]where \( E \) is the energy, \( P = 0.600 \text{ W} \) is the average power, and \( t = 20.0 \times 10^{-3} \text{ s} \) is the duration of the pulse. Thus, \[ E = 0.600 \times 20.0 \times 10^{-3} = 0.012 \text{ J} \]
02

Convert Energy in Joules to Electronvolts for Each Pulse

To convert the energy from joules to electronvolts, use the conversion factor: \[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]Thus, the energy in electronvolts is: \[ E = \frac{0.012}{1.602 \times 10^{-19}} \approx 7.49 \times 10^{16} \text{ eV} \]
03

Calculate Energy of One Photon in Joules

The energy of a single photon is given by: \[ E = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \text{ Js} \) is Planck's constant, \( c = 3.00 \times 10^{8} \text{ m/s} \) is the speed of light, and \( \lambda = 652 \times 10^{-9} \text{ m} \) is the wavelength. Thus, \[ E = \frac{6.626 \times 10^{-34} \times 3.00 \times 10^{8}}{652 \times 10^{-9}} \approx 3.05 \times 10^{-19} \text{ J} \]
04

Convert Photon Energy in Joules to Electronvolts

To find energy of one photon in electronvolts, again use the conversion factor:\[ 1 \text{ eV} = 1.602 \times 10^{-19} \text{ J} \]Thus, the energy per photon in electronvolts is: \[ E = \frac{3.05 \times 10^{-19}}{1.602 \times 10^{-19}} \approx 1.90 \text{ eV} \]
05

Calculate Number of Photons per Pulse

The number of photons in each pulse can be found by dividing the total energy per pulse by the energy of one photon:\[ n = \frac{E_{ ext{pulse}}}{E_{ ext{photon}}} \]where \( E_{\text{pulse}} = 0.012 \text{ J} \) and \( E_{\text{photon}} = 3.05 \times 10^{-19} \text{ J} \). Thus, \[ n \approx \frac{0.012}{3.05 \times 10^{-19}} \approx 3.93 \times 10^{16} \text{ photons} \]

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Key Concepts

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

Laser Physics
Laser physics delves into the science of light amplification by stimulated emission of radiation. Lasers are devices that produce a coherent beam of photons. This beam is characterized by having all photons at the same frequency and phase. The laser used in this problem operates at a specific wavelength of 652 nm. It emits light in pulses, which is often required in precise applications like medical procedures.
  • Coherence: Means all photons are in phase.
  • Monochromatic: Light is of a single wavelength.
  • High intensity: Due to the large number of photons in a small area.
In laser physics, understanding how these beams can be manipulated provides the groundwork for applications in industries ranging from medical to communication.
Energy Conversion
Energy conversion in this context involves transforming energy from one form to another. In the problem, the average power of the laser pulse is given, and you are required to find the energy in joules and electronvolts. The fundamental formula used is \[E = P \times t\]where energy is derived by the product of power and time. This offers energy in joules, a common SI unit. Since laser applications often deal with microscopic scales, electronvolts, which are units of energy commonly used in the realm of atoms and particles, are introduced.
  • Joules: Standard unit of energy in physics.
  • Electronvolts: More suitable for atomic-scale and photonic energy levels. Conversion factor is \(1 \text{ eV} = 1.602 \times 10^{-19} \text{ J}\).
Converting between these units is crucial for making comparisons in photon energy contexts.
Photon Counting
Photon counting involves determining the number of photons present in a beam or pulse of light. It's an important concept in laser applications, where understanding the number of photons can inform the efficiency and effectiveness of the process. For the calculation:\[ n = \frac{E_{\text{pulse}}}{E_{\text{photon}}} \]where \(E_{\text{pulse}}\) represents the total energy of the pulse, and \(E_{\text{photon}}\) is the energy of a single photon.
  • Importance: Determining photon count is vital for tasks that require precise dosages of energy, such as in medical and scientific applications.
  • Precision: Ensures accurate delivery of energy in technological applications.
Being able to calculate this provides insight into how powerful or effective a laser pulse will be for its intended use.
Wavelength and Frequency
Wavelength and frequency are two intrinsic properties of photons that are inversely related. Wavelength is typically described as the distance between two consecutive peaks of a wave. It is often given in nanometers for light waves. Frequency, on the other hand, refers to how many waves pass a point per second and is measured in hertz.The relationship can be described by the equation:\[ c = \lambda \times f \]where \(\lambda\) is the wavelength, \(f\) is the frequency, and \(c\) is the speed of light.
  • Speed of Light: A constant value \(3.00 \times 10^{8} \text{ m/s}\).
  • Inverse Relationship: As wavelength increases, frequency decreases, and vice versa.
Understanding this relationship helps determine the energy of a photon, as it is directly linked to frequency.
Quantum Mechanics
Quantum mechanics is the foundation that explains the behavior of photons and other particles at the smallest scales. It introduces the concept that energy is not continuous but rather comes in discrete packets called quanta. For photons, this discrete energy can be described with Planck鈥檚 relation:\[ E = \frac{hc}{\lambda} \]where \(h\) is Planck鈥檚 constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength.
  • Planck's Constant: A fundamental notion in quantum mechanics at \(6.626 \times 10^{-34} \text{ Js}\).
  • Photon Energy: Directly related to the frequency and inversely related to the wavelength of light.
This framework is essential for understanding phenomena at the micro-scale, which are inexplicable by classical physics.

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Most popular questions from this chapter

Why is it easier to use helium ions rather than neutral helium atoms in an atomic microscope? A. Helium atoms are not electrically charged, and only electrically charged particles have wave properties. B. Helium atoms form molecules, which are too large to have wave properties. C. Neutral helium atoms are more difficult to focus with electric and magnetic fields. D. The much larger mass of a helium atom compared to a helium ion makes it more difficult to accelerate.

A proton and an electron are each given a kinetic energy of \(1000 \mathrm{eV}\) by accelerating them through a potential difference of \(1000 \mathrm{~V}\). Find the ratio of the proton's de Broglie wavelength to that of the electron.

PRK surgery. Photorefractive keratectomy (PRK) is a laserbased surgery process that corrects near- and farsightedness by removing part of the lens of the eye to change its curvature and hence focal length. This procedure can remove layers \(0.25 \mu \mathrm{m}\) thick in pulses lasting \(12.0 \mathrm{~ns}\) with a laser beam of wavelength \(193 \mathrm{nm} .\) Low-intensity beams can be used because each individual photon has enough energy to break the covalent bonds of the tissue. (a) In what part of the electromagnetic spectrum does this light lie? (b) What is the energy of a single photon? (c) If a \(1.50 \mathrm{~mW}\) beam is used, how many photons are delivered to the lens in each pulse?

If a photon of wavelength \(0.04250 \mathrm{nm}\) strikes a free electron and is scattered at an angle of \(35.0^{\circ}\) from its original direction, find (a) the change in the wavelength of this photon, (b) the wavelength of the scattered light, (c) the change in energy of the photon (is it a loss or a gain?), and (d) the energy gained by the electron.

Protons are accelerated from rest by a potential difference of \(4.00 \mathrm{kV}\) and strike a metal target. If a proton produces one photon on impact, what is the minimum wavelength of the resulting X-rays? How does your answer compare to the minimum wavelength if \(4.00 \mathrm{keV}\) electrons are used instead? Why do X-ray tubes use electrons rather than protons to produce X-rays?
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