You've acquired a laser for your dental practice. It produces \(400-\mathrm{mJ}\) pulses at 2.94 - \(\mu\) m wavelength. A patient wonders about the number of photons in each pulse, and where they lie in the EM spectrum. Your answer?

Short Answer

Expert verified
Each pulse contains approximately 1.25 x \(10^{16}\) photons and they reside in the Infrared region of the Electromagnetic Spectrum.

Step by step solution

01

Calculate Energy Per Photon

Firstly, convert wavelength from micrometers to meters and energy of each pulse to Joules. Wavelength λ is 2.94 × \(10^{-6}\) m, and energy E is 400 × \(10^{-3}\) J. Utilize Planck's equation to define energy per photon as \(E_{photon} = \frac{hc}{λ}\). Here, \(h\) is Planck's constant equal to 6.63 × \(10^{-34}\) J*s and \(c\) is the speed of light equal to \(3×10^8\) m/s. Plug the known quantities into the equation to find the energy per photon.
02

Calculate the Number of Photons

The total energy of the pulse is given as 400 mJ. Therefore, to find the number of photons, divide the total energy of the pulse by the energy per photon using the formula \(N = \frac{E}{E_{photon}}\). This gives the number of photons in each pulse.
03

Identify their Position on the EM Spectrum

Refer to the electromagnetic spectrum. The given wavelength is 2.94 µm, which falls under the category of Infrared.

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Key Concepts

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

Planck's Equation
Planck's equation is fundamental for understanding the energy of photons, which are light particles. This equation is elegantly simple yet powerful:
ewline ewline Planck's equation is expressed as ewline ewline \( E_{photon} = \frac{hc}{ewline ewline ewline ewline } \).ewline ewline Here, \( E_{photon} \) represents the energy of a single photon, \( h \) is Planck's constant (6.63 × \(10^{-34}\) Joule seconds), \( c \) is the speed of light in a vacuum (approximately \(3×10^{8}\) meters per second), and \( ewline ewline \) is the wavelength of the photon. This fundamental relationship allows scientists and professionals, including those in dental practices with advanced laser equipments, to compute the energy carried by a single photon if its wavelength is known.
ewline ewline Planck's equation is a cornerstone in quantum mechanics, tying together the discreet particle aspects of light with its wave nature. Understanding this equation enables practitioners to calculate the energy used in various laser applications, like those in medical treatments.
Energy of a Photon
The energy of a photon is directly related to its frequency and inversely related to its wavelength. Photons, which are quantum of light, embody energy that can be calculated using Planck's equation. The higher the frequency (which corresponds to a shorter wavelength), the greater the energy of the photon.ewline ewline In practical use, like a laser for dental procedures, the energy per photon determines how much energy is transmitted in each pulse of laser light. This is vital for ensuring the safety and effectiveness of the treatment. To clarify the concept, knowing that energy is measured in Joules (J) and considering our example where the laser emits at a wavelength of 2.94 µm, you can calculate the specific energy per photon. It plays a crucial role in determining the appropriate settings for medical equipment and ensures that the treatments are within the safe and therapeutic thresholds.
Electromagnetic Spectrum
The electromagnetic (EM) spectrum encompasses all types of electromagnetic radiation - from the very short wavelengths, like gamma rays and X-rays, to the very long wavelengths like radio waves. The visible spectrum, which is what our eyes can detect, is only a small portion of the entire EM spectrum.ewline ewline Wavelengths are typically measured in meters (m), but they can range significantly in size. The EM spectrum is divided into categories based on wavelength: gamma rays have the shortest wavelength, then X-rays, ultraviolet light, visible light, infrared light, microwaves, and finally, radio waves with the longest wavelength.ewline ewline Infrared waves, which include wavelengths from about 700 nanometers (nm) to 1 millimeter (mm), are just below visible red light in the EM spectrum and are not visible to the human eye. They are frequently used in various applications, including remote controls, night-vision equipment, and medical procedures involving lasers, such as the one mentioned in our dental laser example. Understanding where these wavelengths fall on the EM spectrum helps in choosing the correct type of laser and ensuring it is fit for the intended use.
Wavelength to Energy Conversion
The conversion of wavelength to energy is an integral part of understanding how photonic devices work, like lasers used in medicine and telecommunications. Through Planck's equation, it's clear that there is an inverse relationship between the energy of a photon and its wavelength. This means that as the wavelength decreases, the energy of the photon increases, and vice versa.ewline ewline Converting the wavelength into energy is critical when calibrating devices that release energy in the form of light, as each wavelength can have different effects on various materials, such as human tissue in medical applications. By using the formula \( E_{photon} = \frac{hc}{ewline ewline \} \), where \( h \) and \( c \) are constants, one can substitute the wavelength value to receive an accurate energy value for each photon emitted in a light pulse. This is particularly beneficial when specifying laser settings for patient treatments or any technical application where precision is paramount.
ewline ewline For example, if a dental laser emits light with a wavelength of 2.94 µm, it's essential for patient safety and treatment effectiveness to convert this wavelength into energy to know exactly how much energy is delivered per pulse.ewline ewline In conclusion, mastering wavelength to energy conversion can optimize the application of photonic devices and ensure predictable outcomes in their use.

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Most popular questions from this chapter

Show that the wavelength \(\lambda\) in \(\mathrm{nm}\) of a photon with energy \(E\) in eV is \(\lambda=1240 / E\)

With sufficient energy, it's possible to eject an electron from an inner atomic orbital. A higher-energy electron will then drop into the unoccupied state, emitting a photon with energy equal to the difference between the two levels. For inner-shell electrons, photon energies are in the keV range, putting them in the X-ray region of the spectrum. These characteristic X rays are labeled with the letter indicating the shell to which the electron drops, followed by a Greek letter indicating the higher level from which it drops; thus \(K \alpha\) designates a transition from the \(L\) shell to the \(K\) shell. Characteristic X rays provide scientists and physicians with an important diagnostic tool. Environmental scientists bombard pollution samples with high- energy electrons, knocking out inner-shell electrons and thus producing X-ray spectra that help identify contaminants (Fig. \(36.20 a\) ). Geologists do the same with rocks. Medical radiologists reverse the process, exploiting the fact that X rays cause inner-shell transitions as well as complete ejection of inner-shell electrons. In particular, radiologists use the element barium in this way to produce high-contrast X-ray images of the intestinal tract \((\text { Fig. } 36.20 b)\)(GRAPH CANNOT COPY) (a) An \(\mathrm{X}\) -ray spectrum from air pollutants trapped on a filter. The labeled peaks show the presence of lead (Pb) and arsenic (As), as evidenced by \(K \alpha, K \beta, L \alpha,\) and \(L \beta\) characteristic X rays. (b) \(\mathrm{X}\) -ray of an intestinal tract, made by coating the intestinal wall with X-ray-opaque barium Elements \(A\) and \(B\) have atomic numbers \(Z_{A}\) and \(Z_{B}=2 Z_{A} .\) How do you expect element \(B^{\prime}\) s \(K \alpha\) X-ray energy to compare with that of element \(A\) ? a. \(B^{\prime}\) s \(K \alpha\) energy should be about one-fourth that of \(A\) b. \(B\) 's \(K \alpha\) energy should be about half that of \(A\) c. \(B^{\prime}\) s \(K \alpha\) energy should be about twice that of \(A\) d. \(B^{\prime}\) s \(K \alpha\) energy should be about four times that of \(A\)

A selection rule for the infinite square well allows only those transitions in which \(n\) changes by an odd number. Suppose an infinite square well of width 0.200 nm contains an electron in the \(n=4\) state. (a) Draw an energy-level diagram showing all allowed transitions that could occur as this electron drops toward the ground state, including transitions from lower levels that could be reached from \(n=4 .\) (b) Find all the possible photon energies emitted in these transitions.

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