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 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 \(8.06 \times 10^{16}\) photons per second.

Step by step solution

01

Identify the portion of the electromagnetic spectrum

To identify the portion of the electromagnetic spectrum, compare the given wavelength (987 nm) with the ranges of different wavelengths in the electromagnetic spectrum. The main portions are (in increasing order of wavelength): Gamma rays, X-rays, Ultraviolet, Visible light, Infrared, Microwaves, and Radio waves. Visible light has a wavelength range of about 400 nm to 700 nm, and Infrared light has a wavelength range of about 700 nm to 1 mm. As the given wavelength is 987 nm, it falls within the Infrared region. Answer to (a): The radiation is found in the Infrared portion of the electromagnetic spectrum.
02

Calculate the energy of one photon

To calculate the energy of one photon, use the Planck's equation: \(E = h\frac{c}{\lambda}\) where \(E\) is the energy of the photon, \(h\) is the Planck's constant \((h \approx 6.626 \times 10^{-34} \text{ Js})\), \(c\) is the speed of light \((c \approx 3 \times 10^8 \text{ m/s})\), and \(\lambda\) is the wavelength of the photon. First, convert the wavelength from nanometers to meters: \(\lambda = 987 \text{ nm} \times \frac{1 \text{ m}}{10^9 \text{ nm}} = 9.87 \times 10^{-7} \text{ m}\) Now, calculate the energy of one photon: \(E = (6.626 \times 10^{-34} \text{ Js}) \frac{3 \times 10^8 \text{ m/s}}{9.87 \times 10^{-7} \text{ m}} \approx 2.018 \times 10^{-19} \text{ J}\)
03

Calculate the total number of emitted photons

Since we know the total energy absorbed in the detector (0.52 J) and the energy of one photon, we can calculate the total number of emitted photons: \(n = \frac{\text{Total absorbed energy}}{\text{Energy of one photon}}\) \(n = \frac{0.52 \text{ J}}{2.018 \times 10^{-19} \text{ J}} \approx 2.58 \times 10^{18}\)
04

Compute the average number of photons emitted per second

Divide the total number of emitted photons by the given time period (32 s) to find the average number of photons emitted per second: \(\text{Photons per second} = \frac{\text{Total number of emitted photons}}{\text{Time}}\) \(\text{Photons per second} = \frac{2.58 \times 10^{18}}{32 \text{ s}} \approx 8.06 \times 10^{16} \text{ photons/s}\) Answer to (b): The diode laser emits approximately \(8.06 \times 10^{16}\) photons per second.

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Key Concepts

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

Electromagnetic spectrum
The electromagnetic (EM) spectrum encompasses all types of electromagnetic radiation, which varies in wavelength and frequency. This range includes, from shortest to longest wavelength: gamma rays, X-rays, ultraviolet rays, visible light, infrared radiation, microwaves, and radio waves. Wavelengths are typically measured in meters, with visible light wavelengths ranging from about 400 nm to 700 nm. As we move towards infrared, microwaves, and radio waves, the wavelengths get progressively longer, while gamma rays and X-rays have much shorter wavelengths.

This spectrum is fundamental to understanding how different types of electromagnetic waves interact with matter, as each category of wave has unique properties and applications. For instance, X-rays can penetrate soft tissues, making them invaluable in medical imaging, while radio waves are long enough to be used for broadcasting signals over large distances.
Infrared radiation
Infrared radiation is a type of electromagnetic radiation with wavelengths longer than visible light but shorter than microwaves, typically ranging from 700 nm to 1 mm. It is invisible to the human eye, but we can feel it as heat, which signifies how it transfers energy. Infrared is often divided into subcategories: near-infrared (closest to visible light), mid-infrared, and far-infrared (closest to microwaves).

Objects at room temperature emit infrared as thermal radiation. This property is harnessed in various technologies, such as night-vision equipment and thermal imaging, which detect infrared radiation from warm objects against cooler backgrounds. In the context of the exercise, infrared radiation is identified by the laser's wavelength of 987 nm, situating it in the infrared region of the spectrum.
Planck's equation
Planck's equation, devised by physicist Max Planck, is a fundamental relationship in quantum mechanics connecting the energy of a photon to its frequency (or through the speed of light, its wavelength). The equation is expressed as: \[ E = h\frac{c}{\lambda} \] where \(E\) denotes the energy of the photon, \(h\) is Planck’s constant (approximately \(6.626 \times 10^{-34} Js\)), \(c\) is the speed of light in a vacuum (roughly \(3 \times 10^8 m/s\)), and \(\lambda\) is the wavelength of the photon.

Understanding this equation is crucial in fields such as quantum physics, chemistry, and astronomy because it bridges the gap between the macroscopic classical world and the microscopic quantum world. It indicates how the quantized nature of light can lead to phenomena like the photoelectric effect and plays a pivotal role in determining the characteristics of photons emitted or absorbed by atoms.
Energy of a photon
The energy of a photon is intrinsic to its identity in the quantum world. According to Planck's equation, the energy is directly proportional to its frequency and inversely proportional to its wavelength. This means that photons of light with higher frequencies (and thus shorter wavelengths) like gamma rays pack more energy than those with lower frequencies such as radio waves.

To compute the energy of a single photon in practical applications like our exercise, you'd rearrange Planck's equation to substitute for wavelength if the frequency is unknown. Given that the energy of each photon is quantized, when lasers or other light sources emit photons, each one carries a discrete 'packet' of energy. By knowing the energy of one photon and the total energy emitted or absorbed, as shown in the exercise solution, one can determine the quantity of photons involved in the process which is crucial for scientific experiments and industrial applications like laser technology.

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Most popular questions from this chapter

Bohr's model can be used for hydrogen-like ions-ions that have only one electron, such as \(\mathrm{He}^{+}\) and \(\mathrm{Li}^{2+} .\) (a) Why is the Bohr model applicable to He \(^{+}\) ions but not to neutral He atoms? (b) The ground-state energies of \(\mathrm{H}, \mathrm{He}^{+},\) and \(\mathrm{Li}^{2+}\) are tabulated as follows: $$ \begin{array}{l}{\text { Atom or ion } \quad \quad\quad\quad\quad\quad \mathrm{H} \quad\quad\quad\quad\quad\quad \text { He }^{+} \quad\quad\quad\quad\quad\quad\quad \mathrm{Li}^{2+}} \\ {\text { Ground- state }\quad-2.18 \times 10^{-18} \mathrm{J}\quad-8.72 \times 10^{-18} \mathrm{J}\quad-1.96 \times 10^{-17} \mathrm{J}} \\ {\text { energy }}\end{array} $$ By examining these numbers, propose a relationship between the ground-state energy of hydrogen-like systems and the nuclear charge, \(Z .(\mathbf{c})\) Use the relationship you derive in part (b) to predict the ground-state energy of the \(\mathrm{C}^{5+}\) ion.

Place the following transitions of the hydrogen atom in order from shortest to longest wavelength of the photon emitted: \(n=5\) to \(n=3, n=4\) to \(n=2, n=7\) to \(n=4,\) and \(n=3\) to \(n=2\) .

In the television series Star Trek, the transporter beam is a device used to "beam down" people from the Starship Enterprise to another location, such as the surface of a planet. The writers of the show put a "Heisenberg compensator" into the transporter beam mechanism. Explain why such a compensator (which is entirely fictional) would be necessary to get around Heisenberg's uncertainty principle.

The rays of the Sun that cause tanning and burning are in the ultraviolet portion of the electromagnetic spectrum. These rays are categorized by wavelength. So-called UV-A radiation has wavelengths in the range of \(320-380 \mathrm{nm},\) whereas UV-B radiation has wavelengths in the range of \(290-320 \mathrm{nm} .\) (a) Calculate the frequency of light that has a wavelength of 320 \(\mathrm{nm}\) . (b) Calculate the energy of a mole of 320 -nm photons. (c) Which are more energetic, photons of UV-A radiation or photons of UV-B radiation? (d) The UV-B radiation from the Sun is considered a greater cause of sunburn in humans than is UV-A radiation. Is this observation consistent with your answer to part (c)?

(a) What are "valence electrons"? (b) What are "core electrons"? (c) What does each box in an orbital diagram represent? (d) What object is represented by the half arrows in an orbital diagram? What does the direction of the arrow signify?

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