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Chapter 6: Problem 26

(a) A green laser pointer emits light with a wavelength of \(532 \mathrm{~nm}\). What is the frequency of this light? (b) What is the energy of one of these photons? (c) The laser pointer emits light because electrons in the material are excited (by a battery) from their ground state to an upper excited state. When the electrons return to the ground state, they lose the excess energy in the form of 532 -nm photons. What is the energy gap between the ground state and excited state in the laser material?

Short Answer

Expert verified
(a) The frequency of the light is \(5.639 \times 10^{14} \mathrm{~Hz}\). (b) The energy of one photon is \(3.734 \times 10^{-19} \mathrm{~J}\). (c) The energy gap between the ground state and the excited state in the laser material is \(3.734 \times 10^{-19} \mathrm{~J}\).

Step by step solution

01

Part (a): Finding the frequency of light

To find the frequency of the light, we can use the formula: \(c = \lambda\nu\) where \(c\) is the speed of light in a vacuum (\(3.00 \times 10^8 \mathrm{~m/s}\)), \(\lambda\) is the wavelength, and \(\nu\) is the frequency. We are given the wavelength (\(532 \mathrm{~nm}\)) and need to find the frequency. First, we need to convert the wavelength from nanometers to meters: \(532 \mathrm{~nm} = 532 \times 10^{-9} \mathrm{~m}\) Now we can plug the values into the formula and solve for the frequency: \(\nu = \frac{c}{\lambda} = \frac{3.00 \times 10^8 \mathrm{~m/s}}{532 \times 10^{-9} \mathrm{~m}} \) \(\nu = 5.639 \times 10^{14} \mathrm{~Hz}\) So, the frequency of the light is \(5.639 \times 10^{14} \mathrm{~Hz}\).
02

Part (b): Finding the energy of one photon

To find the energy of one photon, we can use the formula: \(E = h\nu\) where \(E\) is the energy of one photon, \(h\) is Planck's constant (\(6.63 \times 10^{-34} \mathrm{~Js}\)), and \(\nu\) is the frequency. We found the frequency in part (a): \(5.639 \times 10^{14} \mathrm{~Hz}\). Now, we can plug in the values and calculate the energy: \(E = (6.63 \times 10^{-34} \mathrm{~Js}) \times (5.639 \times 10^{14} \mathrm{~Hz})\) \(E = 3.734 \times 10^{-19} \mathrm{~J}\) The energy of one photon is \(3.734 \times 10^{-19} \mathrm{~J}\).
03

Part (c): Finding the energy gap between ground state and excited state

The energy gap between the ground state and the excited state in the laser material is equal to the energy of one photon emitted by the laser. This is because electrons in the material are excited from their ground state to an upper excited state, and when they return to the ground state, they lose the excess energy in the form of photons. We found the energy of one photon in part (b): \(3.734 \times 10^{-19} \mathrm{~J}\). Therefore, the energy gap between the ground state and the excited state in the laser material is \(3.734 \times 10^{-19} \mathrm{~J}\).

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Key Concepts

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

Laser Wavelength
Understanding the wavelength of a laser light is crucial, as it determines many of its properties. Wavelength, denoted by \( \lambda \), is the distance between two consecutive peaks of a wave. In the case of the green laser pointer from the exercise, the wavelength is given as \(532 \, \text{nm}\). It鈥檚 important to convert this to meters when performing calculations, as standard scientific measurements are often in the SI unit. Thus, \(532 \, \text{nm}\) equals \(532 \times 10^{-9} \, \text{m}\).

Understanding wavelength helps in applications like determining the type of material or its suitability for tasks like cutting or medical procedures. Different wavelengths in laser light correspond to different energies and can, therefore, penetrate or affect materials differently.
Frequency Calculation
Frequency, symbolized as \(u\), tells us how often the peaks of a wave pass a point per second, measured in Hertz (Hz). It鈥檚 what dictates the behavior and properties of light, much like the vibrating strings of a guitar producing sound.

The formula \(c = \lambda u\) relates light's speed in a vacuum \(c\), the wavelength \(\lambda\), and the frequency \(u\). Since we know both the speed of light, \(3.00 \times 10^8 \, \text{m/s}\), and the wavelength \(532 \times 10^{-9} \, \text{m}\), we can solve for frequency:
  • Firstly, solve for \(u\) with \(u = \frac{c}{\lambda}\)
  • Then substitute the known values into the formula to calculate: \(u = \frac{3.00 \times 10^8 \, \text{m/s}}{532 \times 10^{-9} \, \text{m}}\)
  • The calculated frequency becomes \(5.639 \times 10^{14} \, \text{Hz}\)
This frequency is crucial, as it鈥檚 used to calculate other properties like energy, which shows how energetic a photon is.
Energy Gap
The energy gap in a laser material is linked to how electrons move between energy states. When a material is excited by an input like a battery, electrons get energized and jump to a higher energy state. Once the energy source is removed, electrons fall back to their original lower energy, releasing energy as photons.

In the context of the laser pointer, the energy emitted equals the energy gap. We determined that the energy per photon (which equates to the gap) using Planck鈥檚 equation \(E = hu\), where \(h\) is Planck鈥檚 constant \(6.63 \times 10^{-34} \, \text{Js}\) and the calculated frequency is \(5.639 \times 10^{14} \, \text{Hz}\).
  • Using these values, the energy gap is \(3.734 \times 10^{-19} \, \text{J}\).
This energy gap effectively governs the color and energy output of the laser, reflecting the specific needs and design of the laser for purposes, from point measurements to complex surgeries.

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Most popular questions from this chapter

For orbitals that are symmetric but not spherical, the contour representations (as in Figures \(6.23\) and 6.24) suggest where nodal planes exist (that is, where the electron density is zero). For example, the \(p_{x}\) orbital has a node wherever \(x=0\). This equation is satisfied by all points on the \(y z\) plane, so this plane is called a nodal plane of the \(p_{x}\) orbital. (a) Determine the nodal plane of the \(p_{z}\) orbital. (b) What are the two nodal planes of the \(d_{x y}\) orbital? (c) What are the two nodal planes of the \(d_{x^{2}-y^{2}}\) orbital?

The visible emission lines observed by Balmer all involved \(n_{\mathrm{f}}=2\). (a) Which of the following is the best explanation of why the lines with \(n_{\mathrm{f}}=3\) are not observed in the visible portion of the spectrum: (i) Transitions to \(n_{\mathrm{f}}=3\) are not allowed to happen, (ii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the infrared portion of the spectrum, (iii) transitions to \(n_{\mathrm{f}}=3\) emit photons in the ultraviolet portion of the spectrum, or (iv) transitions to \(n_{\mathrm{f}}=3\) emit photons that are at exactly the same wavelengths as those to \(n_{\mathrm{f}}=2\). (b) Calculate the wavelengths of the first three lines in the Balmer series-those for which \(n_{\mathrm{i}}=3,4\), and 5 -and identify these lines in the emission spectrum shown in Figure 6.11.

(a) Consider the following three statements: (i) A hydrogen atom in the \(n=3\) state can emit light at only two specific wavelengths, (ii) a hydrogen atom in the \(n=2\) state is at a lower energy than the \(n=1\) state, and (iii) the energy of an emitted photon equals the energy difference of the two states involved in the emission. Which of these statements is or are true? (b) Does a hydrogen atom "expand" or "contract" as it moves from its ground state to an excited state?

(a) State the Pauli exclusion principle in your own words. (b) The Pauli exclusion principle is, in an important sense, the key to understanding the periodic table. (c) Explain.

(a) Calculate the energy of a photon of electromagnetic radiation whose frequency is \(2.94 \times 10^{14} \mathrm{~s}^{-1}\). (b) Calculate the energy of a photon of radiation whose wavelength is \(413 \mathrm{~nm}\). (c) What wavelength of radiation has photons of energy \(6.06 \times 10^{-19} \mathrm{~J}\) ?
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