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BIO 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 during each pulse is 0.600 \(\mathrm{W}\) (a) How much energy is in each pulse in joules? In electron volts? (b) What is the energy of one photon in joules' In electron volts? (b) How many photons are in each pulse?

Short Answer

Expert verified
Each pulse has 0.012 J or \( 7.4904 \times 10^{16} \; \text{eV} \) of energy. Each photon has \( 3.054 \times 10^{-19} \; \text{J} \) or 1.906 eV. There are about \( 3.93 \times 10^{16} \) photons per pulse.

Step by step solution

01

Calculate Energy per Pulse in Joules

To find the energy in each pulse in joules, use the formula: \[ E = P \times t \]where \( E \) is the energy, \( P \) is the power, and \( t \) is the time duration of the pulse. Here, the given power \( P = 0.600 \; \text{W} \) and the pulse duration \( t = 20.0 \; \text{ms} = 0.020 \; \text{s} \). Thus, substitute these values into the formula:\[E = 0.600 \; \text{W} \times 0.020 \; \text{s} = 0.012 \; \text{J}\]So, the energy per pulse is \( 0.012 \; \text{J} \).
02

Convert Energy per Pulse to Electron Volts

Convert the energy from joules to electron volts using the conversion factor: \[1 \; \text{J} = 6.242 \times 10^{18} \; \text{eV}\]For \( 0.012 \; \text{J} \),\[E = 0.012 \; \text{J} \times 6.242 \times 10^{18} \; \text{eV/J} = 7.4904 \times 10^{16} \; \text{eV}\]Thus, the energy per pulse in electron volts is \( 7.4904 \times 10^{16} \; \text{eV} \).
03

Calculate Energy of One Photon in Joules

To find the energy of a single photon, use the formula:\[E_{\text{photon}} = \frac{hc}{\lambda} \]where \( h = 6.626 \times 10^{-34} \; \text{J} \cdot \text{s} \) 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. Substituting these values gives:\[E_{\text{photon}} = \frac{6.626 \times 10^{-34} \; \text{J} \cdot \text{s} \times 3.00 \times 10^8 \; \text{m/s}}{652 \times 10^{-9} \; \text{m}} = 3.054 \times 10^{-19} \; \text{J}\]So, the energy of one photon is \( 3.054 \times 10^{-19} \; \text{J} \).
04

Convert Energy of One Photon to Electron Volts

Convert the energy of the photon from joules to electron volts:\[E_{\text{photon}} = 3.054 \times 10^{-19} \; \text{J} \times 6.242 \times 10^{18} \; \text{eV/J} = 1.906 \; \text{eV}\]Thus, the energy of one photon is \( 1.906 \; \text{eV} \).
05

Calculate Number of Photons per Pulse

To determine the number of photons in each pulse, divide the total energy of the pulse by the energy of one photon:\[N = \frac{E}{E_{\text{photon}}}\]Where \( E = 0.012 \; \text{J} \) is the energy of the pulse and \( E_{\text{photon}} = 3.054 \times 10^{-19} \; \text{J} \). Substituting these values:\[N = \frac{0.012 \; \text{J}}{3.054 \times 10^{-19} \; \text{J}} \approx 3.93 \times 10^{16}\]So, there are approximately \( 3.93 \times 10^{16} \) photons per pulse.

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Key Concepts

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

Photon Energy
Photon energy is a fundamental concept in laser physics. Each photon, a tiny packet of light energy, carries energy that can be calculated using Planck’s equation: \[ E_{\text{photon}} = \frac{hc}{\lambda} \]where:
  • \( h \) is Planck’s constant \( (6.626 \times 10^{-34} \ \text{J s}) \)
  • \( c \) is the speed of light \( (3.00 \times 10^8 \ \text{m/s}) \)
  • \( \lambda \) is the wavelength of light
The energy of a photon is inversely proportional to the wavelength. That means, shorter wavelengths (like blue light) have more energy, and longer wavelengths (like red light) have less. In the case of laser light used for retina welding, knowing the photon energy is crucial for precise medical procedures. Understanding this energy can help predict how the light will interact with biological tissues and the effects it will have.
Power and Energy Calculations
Energy and power calculations are an integral part of understanding laser functioning. Power is the rate of energy transfer and is usually measured in watts \((\text{W})\). To find out how much energy a laser pulse contains, you use the formula: \[ E = P \times t \]Here, \(E\) is the energy in joules, \(P\) is the power in watts, and \(t\) is the duration of the pulse in seconds. For example, if a pulse lasts 20 ms, and the power is 0.600 W, the energy is:\[ E = 0.600 \ \text{W} \times 0.020 \ \text{s} = 0.012 \ \text{J} \]This equation helps quantify how much energy is delivered in each pulse, which is crucial for applications such as medical laser treatments, where precise energy delivery is required to achieve desired outcomes without damaging surrounding tissue.
Wavelength and Frequency
In the world of physics, wavelength and frequency are two sides of the same coin. For a given wave, as wavelength increases, frequency decreases, and vice versa. The relationship between wavelength \((\lambda)\) and frequency \((f)\) is given by the equation:\[ c = \lambda \times f \]where \(c\) is the speed of light. Understanding this relationship is key when working with lasers, as it determines the color of the light emitted (visible to the human eye as different colors). In our exercise, the laser light’s wavelength is 652 nm, signifying red light. Each wavelength has distinct properties and interacts differently with materials. Thus, controlling wavelength is vital in applications like retina welding, where specific tissue absorption is desired.
Retina Welding
Retina welding is a medical procedure that uses laser technology to repair retinal detachments in the eye. The laser emits a focused light that creates precise and controlled burns. These burns help reattach the retina by causing the tissue to scar and bond. The choice of wavelength (here, 652 nm) is critical because it must pass through the eye without being absorbed prematurely. This ensures the energy reaches and precisely affects the target area of the retina. The power and duration of the laser pulses must be carefully calculated to deliver sufficient energy for this effect, without damaging other delicate structures in the eye. Understanding photon energy and laser parameters is crucial to the success and safety of retina welding procedures.

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Most popular questions from this chapter

An incident x-ray photon of wavelength 0.0900 nm is scattered in the backward direction from a free electron that is initially at rest. (a) What is the magnitude of the momentum of the scattered photon? (b) What is the kinetic energy of the electron after the photon is scattered?

(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 x-ray tube operated at 30.0 \(\mathrm{kV}\) ?

A horizontal beam of laser light of wavelength 585 \(\mathrm{nm}\) passes through a narrow slit that has width 0.0620 \(\mathrm{mm}\) . The intensity of the light is measured on a vertical screen that is 2.00 \(\mathrm{m}\) . from the slit. (a) What is the minimum uncertainty in the vertical component of the momentum of each photon in the beam after the photon has passed through the slit? (b) Use the result of part (a) to estimate the width of the central diffraction maximum that is observed on the screen.

(a) Calculate the maximum increase in photon wavelength that can occur during Compton scattering, (b) What is the energy (in electron volts) of the lowest- energy \(x\) -ray photon for which Compton scattering could result in doubling the original wavelength?

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 \(E>m c^{2},\) 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, ett.. \(.\) Can you think of an application of this effect?

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