/*! 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 32 Pulsed dye lasers emit light of ... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

Pulsed dye lasers emit light of wavelength 585 nm in 0.45-ms pulses to remove skin blemishes such as birthmarks. The beam is usually focused onto a circular spot 5.0 mm in diameter. Suppose that the output of one such laser is 20.0 W. (a) What is the energy of each photon, in eV? (b) How many photons per square millimeter are delivered to the blemish during each pulse?

Short Answer

Expert verified
(a) Energy of each photon is 2.12 eV. (b) 1.35 \times 10^{15} photons/mm^2 per pulse.

Step by step solution

01

Calculate the Energy of a Photon

The energy of a photon is given by the equation \[ E = \frac{hc}{\lambda} \] where \( h = 6.626 \times 10^{-34} \text{ J s} \) is Planck's constant, \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light, and \( \lambda = 585 \text{ nm} = 585 \times 10^{-9} \text{ m} \) is the wavelength. Substituting these values in, we find:\[ E = \frac{(6.626 \times 10^{-34} \, \text{J s}) \times (3.00 \times 10^8 \, \text{m/s})}{585 \times 10^{-9} \, \text{m}} = 3.39 \times 10^{-19} \, \text{J} \]Since we require the energy in electron volts (eV), we divide by the charge of an electron \( e = 1.602 \times 10^{-19} \text{ C} \):\[ E = \frac{3.39 \times 10^{-19} \, \text{J}}{1.602 \times 10^{-19} \text{ C}} = 2.12 \, \text{eV} \].
02

Calculate the Total Energy in One Pulse

The power output of the laser is given as \( 20.0 \, \text{W} \), which means it outputs 20 J every second. Given that the pulse duration is \( 0.45 \, \text{ms} = 0.45 \times 10^{-3} \, \text{s} \), the total energy per pulse is \[ E_{\text{pulse}} = \text{Power} \times \text{Time} = 20.0 \, \text{W} \times 0.45 \times 10^{-3} \, \text{s} = 9.0 \times 10^{-3} \, \text{J} \].
03

Calculate the Number of Photons per Pulse

Given the energy of each photon is \( 3.39 \times 10^{-19} \, \text{J} \), the number of photons per pulse \( N \) can be calculated as \[ N = \frac{E_{\text{pulse}}}{E_{\text{photon}}} = \frac{9.0 \times 10^{-3} \, \text{J}}{3.39 \times 10^{-19} \, \text{J/photon}} = 2.65 \times 10^{16} \text{ photons} \].
04

Calculate the Area of the Laser Spot

The diameter of the laser's focus spot is given as \( 5.0 \, \text{mm} \), and thus the radius \( r \) is \( 2.5 \, \text{mm} = 2.5 \times 10^{-3} \, \text{m} \). Using the formula for the area of a circle, \( A = \pi r^2 \), we get:\[ A = \pi \times (2.5 \times 10^{-3} \, \text{m})^2 = 19.635 \times 10^{-6} \, \text{m}^2 \].
05

Calculate Photons per Square Millimeter

To find the number of photons per square millimeter, convert the area from square meters to square millimeters (\[ 19.635 \times 10^{-6} \, \text{m}^2 = 19.635 \, \text{mm}^2 \]), and divide the total photons by this area:\[ \text{Photons/mm}^2 = \frac{2.65 \times 10^{16} \text{ photons}}{19.635 \text{mm}^2} = 1.35 \times 10^{15} \text{ photons/mm}^2 \].

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Photon Energy
Photon energy is a fundamental concept when dealing with light and lasers, like pulsed dye lasers. It refers to the amount of energy carried by a single photon, which is the smallest unit of light. To calculate this energy, we use the formula:\[ E = \frac{hc}{\lambda} \]where:
  • \( h = 6.626 \times 10^{-34} \text{ J s} \) is Planck's constant, a universal constant that plays a central role in quantum mechanics.
  • \( c = 3.00 \times 10^8 \text{ m/s} \) is the speed of light, another fundamental constant.
  • \( \lambda \) is the wavelength of the light. For the problem at hand, the wavelength is 585 nm, which we convert to meters (\( 585 \times 10^{-9} \text{ m} \)).
Plugging these values into our equation gives us the photon's energy in joules. This energy can then be converted into electron volts (eV) by dividing by the charge of an electron, \( 1.602 \times 10^{-19} \text{ C} \). Photons in pulsed dye lasers typically exhibit energies around 2.12 eV, allowing them to perform tasks like blemish removal.
Laser Power
Laser power is the rate at which a laser emits energy, measured in watts (W). For pulsed dye lasers, understanding the power helps determine the intensity of the laser pulse. The power of a laser is given by the equation:\[ \text{Power} = \frac{\text{Energy}}{\text{Time}} \].In our problem, the pulsed dye laser has an output power of 20.0 W. This means it releases 20 joules of energy every second. However, since the laser pulse duration is only 0.45 milliseconds, we multiply the power by the pulse duration to find the total energy per pulse:\[ E_{\text{pulse}} = 20.0 \text{ W} \times 0.45 \times 10^{-3} \text{ s} = 9.0 \times 10^{-3} \text{ J} \].This energy calculation is crucial for determining how effective a laser will be in practical applications like dermatology, where precise energy delivery is vital.
Wavelength Calculation
Wavelength calculation is vital in analyzing the properties of light emitted by a laser. The wavelength, denoted by \( \lambda \), is the distance over which the wave's shape repeats. It is directly related to a photon's energy and determines the type of light; in this case, visible light of 585 nm used by pulsed dye lasers.Calculating the wavelength in meters by converting from nanometers helps in determining the photon energy using the formula:\[ \lambda = \frac{c}{f} \]where \( f \) is the frequency of the wave. However, in our exercise, we primarily focus on using given wavelength information to calculate other properties, like energy. Understanding the wavelength allows us to pinpoint the specific color and application of the laser, which is crucial in fields like biomedical engineering where only certain wavelengths are effective.
Laser Beam Focus
Laser beam focus refers to the concentration of the laser light onto a specific spot. Focusing the laser beam is crucial for applications that require precision. The focused beam can be described by the spot size, which is circular in the case of the pulsed dye laser.To find the beam's focused area, we calculate the area of a circle using the radius, \( r = \frac{d}{2} \), where \( d \) is the diameter of the spot. For a 5.0 mm diameter, the radius is 2.5 mm or 2.5 \( \times 10^{-3} \) m. Calculate the area as:\[ A = \pi r^2 \]Translating this into square millimeters helps determine how many photons hit each square millimeter of the treated area. The precise focusing of lasers ensures targeted treatment in dermatology, reducing damage to the surrounding tissue. With pinpoint accuracy, pulsed dye lasers are highly efficient in various medical applications.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

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

Light from an ideal spherical blackbody 15.0 cm in diameter is analyzed by using a diffraction grating that has 3850 lines/cm. When you shine this light through the grating, you observe that the peak-intensity wavelength forms a first-order bright fringe at \(\pm\)14.4\(^\circ\) from the central bright fringe. (a) What is the temperature of the blackbody? (b) How long will it take this sphere to radiate 12.0 MJ of energy at constant temperature?

An electron is moving with a speed of 8.00 \(\times\) 10\(^6\) m/s. What is the speed of a proton that has the same de Broglie wavelength as this electron?

A beam of electrons is accelerated from rest through a potential difference of 0.100 kV and then passes through a thin slit. When viewed far from the slit, the diffracted beam shows its first diffraction minima at \(\pm\)14.6\(^\circ\) from the original direction of the beam. (a) Do we need to use relativity formulas? How do you know? (b) How wide is the slit?

The radii of atomic nuclei are of the order of 5.0 \(\times\) 10\(^{-15}\) m. (a) Estimate the minimum uncertainty in the momentum of an electron if it is confined within a nucleus. (b) Take this uncertainty in momentum to be an estimate of the magnitude of the momentum. Use the relativistic relationship between energy and momentum, Eq. (37.39), to obtain an estimate of the kinetic energy of an electron confined within a nucleus. (c) Compare the energy calculated in part (b) to the magnitude of the Coulomb potential energy of a proton and an electron separated by 5.0 \(\times\) 10\(^{-15}\) m. On the basis of your result, could there be electrons within the nucleus? (\(Note\): It is interesting to compare this result to that of Problem 39.72.)

See all solutions

Recommended explanations on Physics Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.