/*! 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 25 A pulsed dye laser emits light o... [FREE SOLUTION] | 91影视

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Chapter 38: Problem 25

A pulsed dye laser emits light of wavelength \(585 \mathrm{nm}\) in \(450 \mu \mathrm{s}\) pulses. Because this wavelength is strongly absorbed by the hemoglobin in the blood, the method is especially effective for removing various types of blemishes due to blood, such as portwine-colored birthmarks. To get a reasonable estimate of the power required for such laser surgery, we can model the blood as having the same specific heat and heat of vaporization as water \(\left(4190 \mathrm{~J} / \mathrm{kg} \cdot \mathrm{K}, 2.256 \times 10^{6} \mathrm{~J} / \mathrm{kg}\right)\) Suppose that each pulse must remove \(2.0 \mu \mathrm{g}\) of blood by evaporating it, starting at \(33^{\circ} \mathrm{C}\). (a) How much energy must each pulse deliver to the blemish? (b) What must be the power output of this laser? (c) How many photons does each pulse deliver to the blemish?

Short Answer

Expert verified
The energy needed for each pulse is \(5.074 \times 10{-3} J\), the power output of the laser should be \(11.276 W\), and each pulse should contain \(1.492 \times 10^{16}\) photons.

Step by step solution

01

Calculate the Energy Needed to Heat the Blood

The process of heating involves two steps: First, the blood is heated from \(33^{\circ}\-C\) to \(100^{\circ}\). This requires us to utilize the specific heat formula, which is \[q=mc\Delta T\]. Here, \(m=2\mu g = 2 \times 10^{-9} kg\), \(c=4190 J/kg \cdot K, and \Delta T = 100-33 = 67^{\circ} C = 67 K\]. By substituting these values into the formula, we get \[q = 2 \times 10^{-9} \times 4190 \times 67 = 561.86 \times 10^{-6} J\]
02

Calculate the Energy Needed to Vaporise the Blood

In the second step, the blood is vaporized at \(100^{\circ}C\). We can use the formula for heat of vaporization which is \(q= mL\), where \(m=2\mu g = 2 \times 10^{-9} kg\) and \(L= 2.256 \times 10^6 J/kg\). By substituting these values, we obtain: \(q = 2 \times 10^{-9} \times 2.256 \times 10^6 = 4.512 \times 10^{-3} J\)
03

Total Energy Required to Remove the Blemish

Combine the results from step 1 and 2 to get the total energy required. So, it is \(561.86 \times 10^{-6} J + 4.512 \times 10^{-3} J = 5.074 \times 10^{-3} J\)
04

Find Out the Power Output of the Laser

The power output can be acquired by using the formula \(P = \frac{E}{t}\) where \(E = 5.074 \times 10^{-3} J\), and \(t = 450 \times 10^{-6}\) seconds. Thus, \(P = \frac{5.074 \times 10^{-3}}{450 \times 10^{-6}} = 11.276 W\)
05

Calculate the Number of Photons Delivered

The formula to find the number of photons is \(N = \frac{E}{E_{photon}}\), where the energy of a photon, \(E_{photon} = \frac{hc}{\lambda}\). Here, \(h = 6.626 \times 10^{-34} J \cdot s\) (Planck's constant), \(c = 3 \times 10^{8} m/s\) (speed of light), and \(\lambda = 585 \times 10^{-9} m\). By substituting these values in the formula for \(E_{photon}\), we get \(E_{photon} = \frac{6.626 \times 10^{-34} \times 3 \times 10^{8}}{585 \times 10^{-9}} = 3.400 \times 10^{-19} J\). Hence, the number of photons in each pulse, \(N = \frac{5.074 \times 10^{-3}}{3.400 \times 10^{-19}} = 1.492 \times 10^{16}\)

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Key Concepts

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

Specific Heat
Specific heat is a crucial concept when understanding the physics behind the pulsed dye laser used in medical procedures. It refers to the amount of heat required to raise the temperature of one kilogram of a substance by one Kelvin. Represented by the formula \( q = mc\Delta T \), where \(q\) is the heat absorbed or released, \(m\) is the mass of the substance, \(c\) is the specific heat capacity, and \(\Delta T\) is the change in temperature.

In our scenario, the blood's specific heat is similar to that of water, which is \(4190 \mathrm{J/kg \cdot K}\). To calculate the energy necessary to heat the blood from 33掳C to 100掳C, the change in temperature \(\Delta T\) is 67K, and the mass of blood is \(2 \mu g\), which is equivalent to \(2 \times 10^{-9} kg\). This understanding allows us to determine the thermal energy required in the first phase of the laser treatment. It is imperative for students to comprehend this primary thermal concept to apply it across diverse subjects such as physics, chemistry, and biology, as well as in practical technologies like laser surgery.
Heat of Vaporization
The heat of vaporization is the energy required to convert a given amount of a substance from a liquid into a vapor at its boiling point without changing its temperature. This is a crucial concept in the realm of thermodynamics and plays a pivotal role in the physiology handled by the pulsed dye laser.

The formula to calculate the heat of vaporization is \( q = mL \), where \(q\) is the heat energy, \(m\) is the mass, and \(L\) represents the heat of vaporization constant for the substance. In our example, \(L\) for water (and thus for blood, by approximation) is \(2.256 \times 10^6 \mathrm{J/kg}\).

When a pulsed dye laser is directed at a blemish, it needs to vaporize a certain amount of blood, meaning it must provide enough energy to overcome the blood's heat of vaporization. This phase of treatment requires significantly more energy compared to merely heating the blood to its boiling point. Understanding the heat of vaporization can help elucidate why certain processes require such high amounts of energy and the intricate balance needed for delicate procedures like laser surgery.
Photon Energy
Photon energy is fundamental to the physics of lasers and is the energy carried by a single photon. The energy of a photon is determined by the equation \( E_{photon} = \frac{hc}{\lambda} \), with \(h\) being Planck's constant (\(6.626 \times 10^{-34} J \cdot s\)), \(c\) the speed of light (\(3 \times 10^{8} m/s\)), and \(\lambda\) the wavelength of the light.

In the exercise, the wavelength of the pulsed dye laser is \(585 nm\). This wavelength is particularly absorbed by the hemoglobin in the blood, making the laser effective for treating vascular blemishes. The calculated photon energy helps us to understand the interaction between the laser light and the biological tissue.

The number of photons in each pulse is an impressive figure, showcasing not just the sheer quantity of these elementary particles involved but also the precise control over energy delivery required in medical lasers. Aspiring scientists and medical professionals need to appreciate the magnitude of photon interactions with matter, paving the way for advancements in laser technologies and therapeutic applications.
  • Photon Energy Calculation: Helps in understanding the power delivered to the target tissue.
  • Photon-Tissue Interaction: Crucial for safe and effective laser treatments.
  • Technological Application: Knowledge of photon energy informs the design and usage of lasers in various fields.

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Most popular questions from this chapter

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?); (d) the energy gained by the electron.

The photoelectric work function of potassium is \(2.3 \mathrm{eV}\). If light that has a wavelength of \(190 \mathrm{nm}\) falls on potassium, find (a) the stopping potential in volts; (b) the kinetic energy, in electron volts, of the most energetic electrons ejected; (c) the speed of these electrons.

To test the photon concept, you perform a Compton-scattering experiment in a research lab. Using photons of very short wavelength, you measure the wavelength \(\lambda^{\prime}\) of scattered photons as a function of the scattering angle \(\phi,\) the angle between the direction of a scattered photon and the incident photon. You obtain these results. $$ \begin{array}{l|ccccc} \boldsymbol{\phi}(\mathrm{deg}) & 30.6 & 58.7 & 90.2 & 119.2 & 151.3 \\ \hline \boldsymbol{\lambda}^{\prime}(\mathbf{p m}) & 5.52 & 6.40 & 7.60 & 8.84 & 9.69 \end{array} $$ Your analysis assumes that the target is a free electron at rest. (a) Graph your data as \(\lambda^{\prime}\) versus \(1-\cos \phi .\) What are the slope and \(y\) -intercept of the best-fit straight line to your data? (b) The Compton wavelength \(\lambda_{\mathrm{C}}\) is defined as \(\lambda_{\mathrm{C}}=h / m c,\) where \(m\) is the mass of an electron. Use the results of part (a) to calculate \(\lambda_{\mathrm{C}} .\) (c) Use the results of part (a) to calculate the wavelength \(\lambda\) of the incident light.

A positron with speed \(v\) impinges horizontally upon an electron at rest. These particles annihilate and produce two photons with the same wavelength \(\lambda,\) each of which travels at angle \(\phi\) with respect to the horizontal. (a) What is the condition for energy conservation in terms of \(m, \gamma, c,\) and \(h,\) where \(m\) is the electron mass and \(\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2} ?(\mathrm{~b}) \mathrm{What}\) is the condition for momentum conservation in terms of the same parameters and the angle \(\phi ?\) (c) Solve these equations and determine the energy of each photon \(E_{\text {photon }}\) and the angle \(\phi\), the latter as a function of only \(\gamma\). (d) If the positron had \(5.11 \mathrm{MeV}\) of kinetic energy, then what is the energy of each photon, and what is the angle \(\phi ?\) (e) What positron speed would result in the photons traveling off perpendicular to each other? (f) If the positron with speed \(v\) was traveling in the \(+x\) -direction, then with what speed \(u\) should we perform a Lorentz velocity transformation to bring us to the "center of momentum" frame, where the total momentum is zero?

A photon with wavelength \(\lambda\) is incident on a stationary particle with mass \(M,\) as shown in Fig. \(\mathbf{P 3 8 . 3 9 .}\) The photon is annihilated while an electron-positron pair is produced. The target particle moves off in the original direction of the photon with speed \(V_{M}\). The electron travels with speed \(v\) at angle \(\phi\) with respect to that direction. Owing to momentum conservation, the positron has the same speed as the electron. (a) Using the notation \(\gamma=\left(1-v^{2} / c^{2}\right)^{-1 / 2}\) and \(\gamma_{M}=\left(1-V_{M}^{2} / c^{2}\right)^{-1 / 2},\) write a relativistic expression for energy conservation in this process. Use the symbol \(m\) for the electron mass. (b) Write an analogous expression for momentum conservation. (c) Eliminate the ratio \(h / \lambda\) between your energy and momentum equations to derive a relationship between \(v / c, V_{M} / c,\) and \(\phi\) in terms of the ratio \(m / M\), keeping \(\gamma\) and \(\gamma_{M}\) as a useful shorthand notation. (d) Consider the case where \(V_{M} \ll c\), and use a binomial expansion to derive an expression for \(\gamma_{M}\) to the first order in \(V_{M}\). Use that result to rewrite your previous result as an expression for \(V_{M}\) in terms of \(c, v, m / M,\) and \(\phi .\) (e) Are there choices of \(v\) and \(\phi\) for which \(V_{M}=0 ?\) (f) Suppose the target particle is a proton. If the electron and positron remain stationary, so that \(v=0\), then with what speed does the proton move, in \(\mathrm{km} / \mathrm{s} ?\) (g) If the electron and positron each have total energy \(5.00 \mathrm{MeV}\) and move with \(\phi=60^{\circ}\), then what is the speed of the proton? (Hint: First solve for \(\gamma\) and \(v .)\) (h) What is the energy of the incident photon in this case?
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