Question

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 CD player has an output wavelength of \(780 \mathrm{~nm}\) and a power level of \(0.10 \mathrm{~mW}\). How many photons strike the CD surface during the playing of a CD 69 minutes in length?

Short answer

Approximately \(1.63 \times 10^{18}\) photons strike the CD surface during the playing of a CD 69 minutes in length.

Step-by-step solution

Convert the wavelength to energy per photon

To convert the wavelength to energy per photon, we can use the equation: \(E = \dfrac{hc}{\lambda}\) where \(E\) = energy of the photon, \(h\) = Planck's constant (\(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}\)), \(c\) = speed of light (\(3.00 \times 10^8 \mathrm{~m/s}\)), and, \(\lambda\) = wavelength (given as 780 nm). First, convert the given wavelength (780 nm) to meters: \(780 \mathrm{~nm} \times \dfrac{1 \mathrm{~m}}{10^9 \mathrm{~nm}} = 7.80 \times 10^{-7} \mathrm{~m}\) Now, calculate the energy per photon: \(E = \dfrac{(6.626 \times 10^{-34} \mathrm{~J} \cdot \mathrm{s}) (3.00 \times 10^8 \mathrm{~m/s})}{7.80 \times 10^{-7} \mathrm{~m}} \approx 2.54 \times 10^{-19} \mathrm{~J}\)

Convert power level to energy per second

Power is given as 0.10 mW. To convert the power level to energy per second, multiply it by \(10^{-3}\) to convert from milliwatts to watts: \(0.10 \mathrm{~mW} \times \dfrac{1 \mathrm{~W}}{10^3 \mathrm{~mW}} = 1.0 \times 10^{-4} \mathrm{~W}\) As \(1 \mathrm{~W} = 1 \mathrm{~J/s}\), the energy per second is: \(1.0 \times 10^{-4} \mathrm{~J/s}\)

Calculate the number of photons produced per second

We have the energy per second and energy per photon. Divide the energy per second by the energy per photon to find the number of photons produced per second: \(\dfrac{1.0 \times 10^{-4} \mathrm{~J/s}}{2.54 \times 10^{-19} \mathrm{~J}} \approx 3.94 \times 10^{14} \ \text{photons/s}\)

Calculate the total number of photons during the 69 minutes

Now, we have the number of photons produced per second. We need to find how many photons strike the CD surface during 69 minutes. First, convert the given time to seconds: \(69 \ \text{minutes} \times \dfrac{60 \ \text{seconds}}{1 \ \text{minute}} = 4140 \ \text{seconds}\) Next, multiply the number of photons per second by the total number of seconds: \(3.94 \times 10^{14} \ \text{photons/s} \times 4140 \ \text{seconds} \approx 1.63 \times 10^{18} \ \text{photons}\) So, approximately \(1.63 \times 10^{18}\) photons strike the CD surface during the playing of a CD 69 minutes in length.

Key concepts

SI Unit of Power

When discussing the rate at which energy is transformed or transferred, the SI (International System of Units) unit known as the watt comes into play. Essentially, it quantifies how swiftly work is done. One watt is equivalent to one joule of energy being used per second, expressed as \(1 W = 1 J/s\).

In practical terms, when a device is rated at, say, \(100 W\), it is consuming energy at a rate of 100 joules per second. This is crucial for understanding how much energy is emitted or consumed over time by electrical appliances and in this case, by a semiconductor laser in a CD player. To make the concept clear to students, imagine if a lightbulb has a power rating of \(60 W\); this means that it uses \(60 J\) of electrical energy every second to stay lit.

Energy per Photon

Diving deep into the particle nature of light, we encounter photons, the fundamental particles of light. Their energy is calculated by considering both Planck's constant (\(h\)) and the speed of light (\(c\)), in relation to their wavelength (\(\lambda\)). The formula \(E = \frac{hc}{\lambda}\) is pivotal, and it underscores the inverse relationship between energy and wavelength - shorter wavelengths mean higher energy photons and vice versa.

To contextualize, if a semiconductor laser emits light at a wavelength of \(780 nm\), converting this to meters and applying the formula gives us the energy for each photon. This calculation is integral for applications such as determining the intensity of light sources and understanding the interactions of light with matter, a common exercise in physics classes. By recognizing the energy per photon, students can unravel why different light sources have different effects, like how some lasers can be used for delicate surgeries or cutting through materials.

Semiconductor Laser

Semiconductor lasers, often found in CD players, are intricate devices that produce coherent light through the process of stimulated emission in a semiconductor. Known for their compact size and efficiency, they emit light at precise wavelengths, like the \(780 nm\) mentioned in the exercise, commonly used for reading compact discs.

The semiconductor laser in our exercise not only presents a real-world application of physics but also showcases their practical use in everyday technology. Their operation involves energizing electrons within a semiconductor material to create photons when these electrons return to a lower energy state. In fact, the power level of the laser emitting these photons, such as \(0.10 mW\), directly affects the number of photons striking the CD surface during playtime. By illustrating examples like the CD player, students can visualize how abstract scientific principles are employed in technologies they regularly interact with.