/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 33 How many photons per second are ... [FREE SOLUTION] | 91Ó°ÊÓ

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How many photons per second are emitted by a \(7.50 \mathrm{~mW}\) \(\mathrm{CO}_{2}\) laser that has a wavelength of \(10.6 \mu \mathrm{m} ?\)

Short Answer

Expert verified
Around \(3.99 \times 10^{16}\) photons per second are emitted by the laser.

Step by step solution

01

Convert Laser Power to Watts

Firstly, convert the power from milliwatts to watts. As there are 1000 milliwatts in a watt, this equates to \(7.50~mW = 7.50 \times 10^{-3}\) watts.
02

Calculate Frequency of Laser

Secondly, you will compute the frequency of the laser. Light frequency can be calculated by the equation \(f=c/\lambda\), where \(c\) is the speed of light in a vacuum and \(\lambda\) is the wavelength of the light. Substituting the given values \(c = 3 \times 10^{8} \, m/s\) and \(\lambda = 10.6 \, \mu m = 10.6 \times 10^{-6}\) meters gives you \(f = \frac {3 \times 10^{8}}{10.6 \times 10^{-6}}\), which approximately equals \(2.83 \times 10^{14}\) hertz.
03

Calculate Energy Per Photon Emitted

In the third step, you have to calculate the energy of a single photon. This can be given using the Planck-Einstein equation \(E=hf\), where \(h\) is Planck's constant with a value of \(6.63 \times 10^{-34}\) joules/second. Substituting the value of \(f\) from step 2, the energy \(E\) of a photon will be \(E = 6.63 \times 10^{-34} \times 2.83 \times 10^{14}\), which gives approximately \(1.8779 \times 10^{-19}\) joules.
04

Calculate the Number of Photons Emitted Per Second

Finally, calculate the number of photons emitted per second by dividing the power of the laser by the energy of a single photon using the formula \(N=P/E\). From step 1, power \(P = 7.50 \times 10^{-3}\) watts, and from step 3, the energy per photon is \(E = 1.8779 \times 10^{-19}\) joules. Substituting these values into the formula results in \(N = 7.50 \times 10^{-3} / 1.8779 \times 10^{-19}\), or approximately \(3.99 \times 10^{16}\) photons are emitted per second.

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Key Concepts

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

Laser Power
Laser power is a crucial element when determining photon emission. In our problem, we begin with a 7.50 milliwatt ( 7.50 mW) CO eutrons per second are at play for every ounce of power. Calculating laser power involves converting the power from milliwatts to watts for easier understanding and computations. Given 1 watt equals 1000 milliwatts, 7.50 mW can be expressed as 7.50 × 10^{-3} watts. This conversion provides a manageable number when later determining photon emission through energy calculations. Remember, lasers translate electrical energy into light, and their power indicates how much energy passes through per unit of time. For engineers and scientists working with lasers, understanding their power is essential to controlling and predicting the outcomes of laser-based activities.
Wavelength
The wavelength is a defining characteristic of a wave, including light, which tells us the distance over which the wave's shape repeats. It's often measured in micrometers ( µm). For our particular CO 2 laser, the wavelength is given as 10.6 µm. Wavelength influences many aspects of light's behavior, including its energy and frequency. Relating wavelength to energy, shorter wavelengths correspond with higher energies and vice versa. This relationship is pivotal when deciphering the energy of each photon in laser calculations. Wavelengths help also define different areas of the electromagnetic spectrum, distinguishing between infrared, visible, ultraviolet, and so on. In a practical sense, lasers like the CO 2, with its infrared wavelength, have applications in fields like medicine and manufacturing, emphasizing the importance of understanding wavelength properties.
Frequency
Frequency refers to how often wave peaks pass a given point per second and is measured in hertz (Hz).It is deeply connected with both wavelength and the speed of light.For light traveling in a vacuum, the frequency can be calculated using the formula\(f = \frac{c}{\lambda}\).For our exercise, substituting the speed of light (\(3 \times 10^{8}\) meters per second) and the given wavelength (10.6 micrometers or 10.6 × 10^{-6} meters), frequencies are about \(2.83 \times 10^{14}\) Hz.Frequency is essential in defining a light wave's energy and behavior, and it determines where the photon lands in the electromagnetic spectrum.In lasers, controlling frequency influences properties like color and energy output, making it a fundamental aspect of design and functionality.
Planck-Einstein Equation
The Planck-Einstein equation bridges the gap between a photon's frequency and its energy.Formulated as \(E = hf\), it uses Planck's constant \(h\) (6.63 \times 10^{-34} joules/second) and the frequency \(f\) of the photon to find its energy \(E\).Importantly, it supports the understanding that photons with higher frequencies carry more energy.In our solution, inserting the laser's frequency (as calculated earlier at \(2.83 \times 10^{14}\) Hz) gives each photon's energy as approximately \(1.8779 \times 10^{-19}\) joules.The Planck-Einstein equation is a foundational physics tool for unraveling light properties, enabling precise calculations that are fundamental for numerous applications,from understanding cosmic phenomena to developing technologies like lasers, solar panels, and even computing devices.

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Most popular questions from this chapter

A hydrogen atom initially in its ground level absorbs a photon, which excites the atom to the \(n=3\) level. Determine the wavelength and frequency of the photon.

High-speed electrons are used to probe the interior structure of the atomic nucleus. For such electrons the expression \(\lambda=h / p\) still holds, but we must use the relativistic expression for momentum, \(p=m v / \sqrt{1-v^{2} / c^{2}}\). (a) Show that the speed of an electron that has de Broglie wavelength \(\lambda\) is \(v=\frac{c}{\sqrt{1+(m c \lambda / h)^{2}}}\) (b) The quantity \(h / m c\) equals \(2.426 \times 10^{-12} \mathrm{~m}\). (As we saw in Section 38.3 , this same quantity appears in Eq. \((38.7),\) the expression for Compton scattering of photons by electrons.) If \(\lambda\) is small compared to \(h / m c,\) the denominator in the expression found in part (a) is close to unity and the speed \(v\) is very close to \(c .\) In this case it is convenient to write \(v=(1-\Delta) c\) and express the speed of the electron in terms of \(\Delta\) rather than \(v .\) Find an expression for \(\Delta\) valid when \(\lambda \ll h / m c .[\)Hint: Use the binomial expansion \((1+z)^{n}=1+n z+\left[n(n-1) z^{2} / 2\right]+\cdots,\) valid for the case \(|z|<1 .]\) (c) How fast must an electron move for its de Broglie wavelength to be \(1.00 \times 10^{-15} \mathrm{~m}\), comparable to the size of a proton? Express your answer in the form \(v=(1-\Delta) c\), and state the value of \(\Delta\).

A triply ionized beryllium ion, \(\mathrm{Be}^{3+}\) (a beryllium atom with three electrons removed), behaves very much like a hydrogen atom except that the nuclear charge is four times as great. (a) What is the ground-level energy of \(\mathrm{Be}^{3+}\) ? How does this compare to the groundlevel energy of the hydrogen atom? (b) What is the ionization energy of \(\mathrm{Be}^{3+} ?\) How does this compare to the ionization energy of the hydrogen atom? (c) For the hydrogen atom, the wavelength of the photon emitted in the \(n=2\) to \(n=1\) transition is \(122 \mathrm{nm}\) (see Example 39.6 ). What is the wavelength of the photon emitted when a \(\mathrm{Be}^{3+}\) ion undergoes this transition? (d) For a given value of \(n\), how does the radius of an orbit in \(\mathrm{Be}^{3+}\) compare to that for hydrogen?

For crystal diffraction experiments (discussed in Section 39.1 ) wavelengths on the order of \(0.20 \mathrm{nm}\) are often appropriate. Find the energy in electron volts for a particle with this wavelength if the particle is (a) a photon; (b) an electron; (c) an alpha particle \(\left(m=6.64 \times 10^{-27} \mathrm{~kg}\right)\)

A \(10.0 \mathrm{~g}\) marble is gently placed on a horizontal tabletop that is \(1,75 \mathrm{~m}\) wide. (a) What is the maximum uncertainty in the horizontal position of the marble? (b) According to the Heisenberg uncertainty principle, what is the minimum uncertainty in the horizontal velocity of the marble? (c) In light of your answer to part (b), what is the longest time the marble could remain on the table? Compare this time to the age of the universe, which is approximately 14 billion years. (Hint: Can you know that the horizontal velocity of the marble is exactly zero?)

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