/*! 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 30 PRK Surgery. Photorefractive ker... [FREE SOLUTION] | 91Ó°ÊÓ

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PRK Surgery. Photorefractive keratectomy (PRK) is a laser-based surgical procedure that corrects near- and farsightedness by removing part on the lens of the eye to change its curvature and hence focal length. This procedure can remove layers 0.25\(\mu \mathrm{m}\) thick using pulses lasting 12.0 \(\mathrm{ns}\) from 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 he? (b) What is the energy of a single photon? (c) If 1.50-mW beam is used, how many photons are delivered to the lens in each pulse?

Short Answer

Expert verified
(a) Ultraviolet spectrum; (b) Energy = \(1.03 \times 10^{-18} \text{ J}\); (c) \(1.75 \times 10^{7}\) photons per pulse.

Step by step solution

01

Identify the Part of the Electromagnetic Spectrum

The wavelength of the laser is given as 193 nm. First, convert this to meters: \(193 \text{ nm} = 193 \times 10^{-9} \text{ m}\). Ultraviolet (UV) light typically covers the range of wavelengths from 10 nm to 400 nm. Since 193 nm falls within this range, the light used in the procedure is part of the ultraviolet region of the electromagnetic spectrum.
02

Calculate the Energy of a Single Photon

The energy of a photon \(E\) can be calculated using the equation \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant \(6.63 \times 10^{-34} \text{ Js}\), \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\), and \(\lambda\) is the wavelength in meters. Substitute the values:\[E = \frac{6.63 \times 10^{-34} \text{ Js} \times 3.00 \times 10^8 \text{ m/s}}{193 \times 10^{-9} \text{ m}} = 1.03 \times 10^{-18} \text{ J}\]
03

Determine the Number of Photons Delivered Per Pulse

First, calculate the energy delivered by the beam in each pulse. Power \(P\) is given as 1.50 mW or \(1.50 \times 10^{-3} \text{ W}\), and time \(t\) for each pulse is 12.0 ns or \(12.0 \times 10^{-9} \text{ s}\). Energy per pulse \(E_{\text{pulse}} = P \times t = 1.50 \times 10^{-3} \text{ W} \times 12.0 \times 10^{-9} \text{ s} = 1.80 \times 10^{-11} \text{ J}\). Then, the number of photons \(N\) is calculated by dividing the energy per pulse by the energy per photon: \(N = \frac{1.80 \times 10^{-11} \text{ J}}{1.03 \times 10^{-18} \text{ J/photon}} \approx 1.75 \times 10^{7}\).

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Key Concepts

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

Electromagnetic Spectrum
The electromagnetic spectrum is a vast range of all types of electromagnetic radiation, from radio waves to gamma rays. Radiation is energy that travels and spreads out as it moves. In this spectrum:
  • Longer wavelengths are associated with lower energy radiation like radio waves.
  • Shorter wavelengths mean higher energy, covering ultraviolet, X-rays, and gamma rays.
  • Visible light falls somewhere in the middle of this spectrum with moderate wavelengths and energy levels.
In the context of PRK surgery, we focus on the ultraviolet region. The laser used for this procedure emits light with a wavelength of 193 nm. This places it within the ultraviolet (UV) segment, which ranges between 10 nm and 400 nm in wavelength. UV light is higher in energy compared to visible light, making it very effective for medical applications where precise laser operations require breaking chemical bonds in tissues.
Ultraviolet Light
Ultraviolet light is a type of electromagnetic radiation with wavelengths shorter than visible light but longer than X-rays. Its wavelength range is between 10 nm to 400 nm. Here are some characteristics and uses:
  • Because of its shorter wavelength, UV light carries more energy compared to visible light.
  • This higher energy is sufficient to break molecular bonds, which is why it must be used with care.
  • In PRK surgery, the energy from UV photons enables precision in removing microscopic layers of tissue from the eye.
The laser at 193 nm falls well into the UV range. Beyond medical applications, UV light is also used in sterilization and fluorescent lighting. However, it's essential to protect your skin and eyes from excessive UV exposure as it can be harmful.
Photon Counting
Photon counting is all about determining the number of photons, the basic unit of light, interacting with a substance. When dealing with lasers, especially in medical surgeries like PRK, this concept is crucial for assessing the energy delivered:
  • A photon is a particle representing a quantum of light; its energy is given by the formula: \[E = \frac{hc}{\lambda}\]Here, \(h\) is Planck's constant \(6.63 \times 10^{-34} \text{ Js}\), \(c\) is the speed of light \(3.00 \times 10^8 \text{ m/s}\), and \(\lambda\) is the wavelength.
  • For the laser in PRK, each photon carries an energy of \(1.03 \times 10^{-18} \text{ J}\).
  • Knowing the energy per photon and the total energy delivered by the laser allows us to calculate the number of photons. This is done by dividing the total energy by the energy of one photon.
In PRK surgery, for a 1.50 mW beam and pulses lasting 12.0 ns, about \(1.75 \times 10^7\) photons are delivered per pulse. This precise calculation helps in achieving the exact surgical outcomes required.

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Most popular questions from this chapter

Two stars, both of which behave like ideal blackbodies, radiate the same total energy per second. The cooler one has a surface temperature \(T\) and 3.0 times the diameter of the hotter star. (a) What is the temperature of the hotter star in terms of \(T\) ? (b) What is the ratio of the peak-intensity wavelength of the hot star to the peak-intensity wavelength of the cool star?

Radiation has been detected from space that is characteristic of an ideal radiator at \(T=2.728 \mathrm{K}\) . (This radiation is a relic of the Big Bang at the beginning of the universe.) For this temperature, at what wavelength does the Planck distribution peak? In what part of the electromagnetic spectrum is this wavelength?

(a) What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x rays with a wavelength of 0.150 \(\mathrm{nm}\) ? (b) What is the shortest wavelength produced in an \(\mathrm{x}\) -ray tube operated at 30.0 \(\mathrm{kV}\) ?

A hydrogen atom initially in the ground level absorbs a photon, which excites it to the \(n=4\) level. Determine the wave- length and frequency of the photon.

Consider Compton scattering of a photon by a moving electron. Before the collision the photon has wavelength \(\lambda\) and is moving in the \(+x\) -direction, and the electron is moving in the - \(x\) -direction with total energy \(E\) (including its rest energy \(m c^{2} )\) . The photon and electron collide head on. After the collision, both are moving in the \(-x\) -direction (that is, the photon has been scattered by \(180^{\circ}\) ). (a) Derive an expression for the wavelength \(\lambda^{\prime}\) of the scattered photon. Show that if \(E \gg m c^{2}\) , where \(m\) is the rest mass of the electron, your result reduces to $$ \Lambda^{\prime}=\frac{h c}{E}\left(1+\frac{m^{2} c^{4} \lambda}{4 h c E}\right) $$ (b) A beam of infrared radiation from a \(\mathrm{CO}_{2}\) laser \((\lambda=10.6 \mu \mathrm{m})\) collides head-on with a beam of electrons, each of total energy \(E=10.0 \mathrm{GeV}\left(1 \mathrm{GeV}=10^{9} \mathrm{eV}\right) .\) Calculate the wavelength \(\lambda^{\prime}\) of the scattered photons, assuming a \(180^{\circ}\) scattering angle. (c) What kind of scattered photons are these (infrared, microwave, ultraviolet, etc.)? Can you think of an application of this effect?

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