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Chapter 6: Problem 86

The watt is the derived SI unit of power, the measure of energy per unit time: \(1 \mathrm{~W}=1 \mathrm{~J} / \mathrm{s}\). A semiconductor laser in a DVD player has an output wavelength of \(650 \mathrm{nm}\) and a power level of $5.0 \mathrm{~mW}$. How many photons strike the DVD surface during the playing of a DVD 90 minutes in length?

Short Answer

Expert verified
The number of photons that strike the DVD surface during the playing of a 90-minute DVD is approximately \(8.8 \cdot 10^{19}\) photons.

Step by step solution

01

Calculate the energy per photon

To calculate the energy per photon, we will use the formula: \[E = h \cdot f\] Where E is the energy, h is the Planck's constant (approx. \(6.63 \cdot 10^{-34} Js\)), and f is the frequency of the light. Since we are given the wavelength \(\lambda\), we can calculate the frequency using the speed of light (approx. \(3\cdot 10^8 m/s\)): \[f = \frac{c}{\lambda}\] Now plug in the values to find the frequency and then the energy per photon: \[ f = \frac{3\cdot 10^8 m/s}{650 \cdot 10^{-9} m} \Rightarrow f \approx 4.62 \cdot 10^{14} s^{-1} \] \[ E = h \cdot f = (6.63 \cdot 10^{-34} Js) \cdot (4.62 \cdot 10^{14} s^{-1}) \Rightarrow E \approx 3.06 \cdot 10^{-19} J \]
02

Calculate the total energy delivered by the laser

We have the power of the laser given as \(5.0 mW\). To find the total energy delivered by the laser, we will use the formula: \[E_\text{total} = P \cdot t\] Where \(E_\text{total}\) is the total energy, P is the power, and t is the time. First, we need to convert the power to watts and the time to seconds: \[ P = 5.0 mW = 5.0 \cdot 10^{-3} W \] \[ t = 90 \text{ minutes} = (90)(60) \text{ seconds} = 5400 \text{ seconds} \] Now, plug in the values to find the total energy delivered in the given time: \[ E_\text{total} = (5.0 \cdot 10^{-3} W) \cdot (5400 s) \Rightarrow E_\text{total} \approx 27 J \]
03

Calculate the number of photons striking the DVD surface

Now we can find the number of photons by dividing the total energy by the energy per photon: \[ \text{Number of photons} = \frac{E_\text{total}}{E} = \frac{27 J}{3.06 \cdot 10^{-19} J} \] \[ \text{Number of photons} \approx 8.8 \cdot 10^{19} \] So, approximately \(8.8 \cdot 10^{19}\) photons strike the DVD surface during the playing of a 90-minute DVD.

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Most popular questions from this chapter

Label each of the following statements as true or false. For those that are false, correct the statement. (a) Visible light is a form of electromagnetic radiation. (b) Ultraviolet light has longer wavelengths than visible light. (c) X rays travel faster than microwaves. (d) Electromagnetic radiation and sound waves travel at the same speed.

Consider a transition in which the electron of a hydrogen atom is excited from \(n=1\) to \(n=\infty\). (a) What is the end result of this transition? (b) What is the wavelength of light that must be absorbed to accomplish this process? (c) What will occur if light with a shorter wavelength than that in part (b) is used to excite the hydrogen atom? (d) How are the results of parts \((\mathrm{b})\) and \((\mathrm{c})\) related to the plot shown in Exercise \(6.88 ?\)

Consider a fictitious one-dimensional system with one electron. The wave function for the electron, drawn below, is \(\psi(x)=\sin x\) from \(x=0\) to $x=2 \pi .\( (a) Sketch the probability density, \)\psi^{2}(x),\( from \)x=0\( to \)x=2 \pi .(\mathbf{b})\( At what value or values of \)x$ will there be the greatest probability of finding the electron? (c) What is the probability that the electron will be found at \(x=\pi ?\) What is such a point in a wave function called? [Section 6.5\(]\)

The first 25 years of the twentieth century were momentous for the rapid pace of change in scientists' understanding of the nature of matter. (a) How did Rutherford's experiments on the scattering of \(\alpha\) particles by a gold foil set the stage for Bohr's theory of the hydrogen atom? (b) In what ways is de Broglie's hypothesis, as it applies to electrons, consistent with J. J. Thomson's conclusion that the electron has mass? In what sense is it consistent with proposals preceding Thomson's work that the cathode rays are a wave phenomenon?

(a) For an He+ ion, do the \(2 s\) and \(2 p\) orbitals have the same energy? If not, which orbital has a lower energy? (b) If we add one electron to form the He atom, would your answer to part (a) change?
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