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A diode laser emits at a wavelength of \(987 \mathrm{~nm}\). (a) In what portion of the electromagnetic spectrum is this radiation found? (b) All of its output energy is absorbed in a detector that measures a total energy of \(0.52 \mathrm{~J}\) over a period of \(32 \mathrm{~s}\). How many photons per second are being emitted by the laser?

Short Answer

Expert verified
(a) The radiation is found in the Infrared portion of the electromagnetic spectrum. (b) The diode laser emits approximately \(8.042 \times 10^{16}\) photons per second.

Step by step solution

01

Electromagnetic Spectrum:

Given the wavelength of the emitted radiation is \(987 \mathrm{~nm}\). To identify the radiation's portion within the electromagnetic spectrum, let's check the ranges of the most common types of EM radiation in increasing order of wavelength: 1. X-Rays: \(10^{-8} - 10^{-11} \mathrm{~m}\) 2. Ultraviolet (UV): \(10^{-8} - 10^{-7} \mathrm{~m}\) 3. Visible light: \(390 - 700 \mathrm{~nm}\) 4. Infrared (IR): \(700 \mathrm{~nm} - 1 \mathrm{~mm}\) 5. Microwave: \(1 \mathrm{~mm} - 30 \mathrm{~cm}\) 6. Radio waves: \(30 \mathrm{~cm} - 1000 \mathrm{~m}\) Since the wavelength of the diode laser is \(987 \mathrm{~nm}\), it falls within the Infrared range of the electromagnetic spectrum. (a) The radiation is found in the Infrared portion of the electromagnetic spectrum.
02

Calculate Energy per Photon:

To calculate the energy of each photon produced by the diode laser, we will use the following formula: \[E = h \frac{c}{\lambda}\] Where: - \(E\) is the energy of a single photon; - \(h\) is the Planck's constant, \(6.626 \times 10^{-34} \mathrm{Js}\); - \(c\) is the speed of light in a vacuum, \(2.998 \times 10^8 \mathrm{m/s}\); - \(\lambda\) is the wavelength of the emitted radiation, \(987 \mathrm{~nm}\), which is equal to \(987 \times 10^{-9} \mathrm{m}\). Now, let's plug in the values and find \(E\): \(E = (6.626 \times 10^{-34} \mathrm{Js})\frac{2.998 \times 10^8 \mathrm{m/s}}{987 \times 10^{-9} \mathrm{m}} \) \(E = 2.020 \times 10^{-19} \mathrm{J}\)
03

Calculate Number of Photons per Second:

The total energy absorbed by the detector is given as \(0.52 \mathrm{~J}\), and this measurement took place over a period of \(32 \mathrm{~s}\). Therefore, the energy per second is \(\frac{0.52 \mathrm{J}}{32 \mathrm{s}} = 0.01625 \mathrm{J/s}\). Now, we need to find out how many photons per second are being emitted by the laser. To do this, we'll divide the energy per second by the energy per photon: \(Number\: of\: Photons/s = \frac{0.01625 \mathrm{J/s}}{2.020 \times 10^{-19} \mathrm{J}} \) \(Number\: of\: Photons/s = 8.042 \times 10^{16} \) (b) The diode laser emits approximately \(8.042 \times 10^{16}\) photons per second.

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Key Concepts

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

IR radiation
Infrared (IR) radiation is a type of electromagnetic energy that is prevalent in our daily lives, though it is invisible to the naked eye. It occupies a part of the electromagnetic spectrum that lies beyond the visible red light—hence 'infrared', which means 'below red'—with wavelengths ranging from about 700 nanometers (nm) to 1 millimeter (mm). This places IR radiation right between the visible spectrum and microwave radiation.
Electromagnetic spectrum refers to the entire range of electromagnetic radiation, which spans from short gamma rays to long radio waves. The part of the spectrum where a particular radiation lies is determined by its wavelength. In practical terms, IR radiation is commonly associated with heat, as objects emit this form of radiation dependent on their temperature. For instance, humans emit IR radiation, which is why we can be detected by thermal imaging cameras. IR radiation can be further subdivided into near-infrared (close to visible light), mid-infrared, and far-infrared (closer to microwaves). Applications of IR radiation include remote controls, night-vision equipment, and thermal insulation.
Photon energy calculation
The energy of a photon can be thought of as a packet of electromagnetic radiation energy, and it is proportional to the photon's frequency. Since energy and frequency are directly related, with a decrease in wavelength, there comes an increase in both the frequency and the energy of the photons.
To calculate the energy of a photon, we use a basic formula derived from quantum physics:

\(E = h \frac{c}{\lambda}\)

where \(E\) is the energy of the photon in joules (J), \(h\) is Planck's constant (approximately \(6.626 \times 10^{-34} \mathrm{Js}\)), \(c\) represents the speed of light in a vacuum (which is about \(2.998 \times 10^8 \mathrm{m/s}\)), and \(\lambda\) is the wavelength of the photon in meters (m). This relationship is key for understanding how energy is quantized on a microscopic scale, such as in light-emitting diodes (LEDs) and lasers, which are vital in various technologies including communication devices and medical instruments. Knowing how many joules of energy are in a single photon allows us to calculate things like the number of photons emitted per second by a light source, given its total energy output.
Planck's constant
Planck's constant is a fundamental quantity in quantum mechanics, symbolized by \(h\), and its value is approximately \(6.626 \times 10^{-34} \) joule-seconds (Js). This constant is remarkably important because it sets the scale of quantum effects, and connects the amount of energy a photon carries with the frequency of its electromagnetic wave.
Max Planck, a German physicist, introduced this constant in 1900 when explaining blackbody radiation, eventually leading to the birth of quantum mechanics. The constant has numerous applications in the field of physics, including the photoelectric effect, which provided evidence for the quantum nature of light and earned Albert Einstein the Nobel Prize in Physics in 1921. Planck's constant has such a fundamental role in quantum mechanics that the International System of Units (SI) has redefined the kilogram in terms of it, indicative of its well-established and precise value in physics. Understanding Planck's constant allows us to comprehend the discrete nature of energy levels in atoms and molecules, contrasting classical physics' interpretation of energy as a continuous spectrum.

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Most popular questions from this chapter

Identify the specific element that corresponds to each of the following electron configurations: (a) \(1 s^{2} 2 s^{2}\), (b) \(1 s^{2} 2 s^{2} 2 p^{4}\), (c) \([\mathrm{Ar}] 4 s^{1} 3 d^{5}\), (d) \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\), (e) \(1 s\) ?

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 that preceded Thomson's work, that the cathode rays are a wave phenomenon?

The "magic numbers" in the periodic table are the atomic numbers of elements with high stability (the noble gases): \(2,10,18,36,54\), and \(86 .\) In terms of allowed values of orbitals and spin quantum numbers, explain why these electron arrangements correspond to special stability.

Microwave ovens use microwave radiation to heat food. The energy of the microwaves is absorbed by water molecules in food, then transferred to other components of the food. (a) Suppose that the microwave radiation has a wavelength of \(11.2 \mathrm{~cm}\). How many photons are required to heat \(200 \mathrm{~mL}\) of coffee from \(23^{\circ} \mathrm{C}\) to \(60{ }^{\circ} \mathrm{C} ?\) (b) Suppose the microwave's power is \(900 \mathrm{~W}\) (1 Watt \(=1\) Joule/sec). How long would you have to heat the coffee in part (a)?

When the spectrum of light from the Sun is examined in high resolution in an experiment similar to that illustrated in Figure 6.11, dark lines are evident. These are called Fraunhofer lines, after the scientist who studied them extensively in the early nineteenth century. Altogether, about 25,000 lines have been identified in the solar spectrum between \(2950 \AA\) and \(10,000 \AA\). The Fraunhofer lines are attributed to absorption of certain wavelengths of the Sun's "white" light by gaseous elements in the Sun's atmosphere. (a) Describe the process that causes absorption of specific wavelengths of light from the solar spectrum. (b) If a scientist wanted to know which Fraunhofer lines belonged to a given element, say neon, what experiments could she conduct here on Earth to provide data?

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