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(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 \(532-\mathrm{nm}\) photons. What is the energy gap between the ground state and excited state in the laser material?

Short Answer

Expert verified
(a) The frequency is approximately \(5.64 \times 10^{14} \, \text{Hz}\). (b) Photon energy is approximately \(3.73 \times 10^{-19} \, \text{J}\). (c) The energy gap is approximately \(3.73 \times 10^{-19} \, \text{J}\).

Step by step solution

01

Understanding the Relationship Between Wavelength and Frequency

To find the frequency of light, we can use the equation \(c = \lambda u\), where \(c\) is the speed of light \( (3.00 \times 10^8 \, \text{m/s}) \), \(\lambda\) is the wavelength, and \(u\) is the frequency. Rearrange the equation to solve for frequency: \(u = \frac{c}{\lambda}\).
02

Convert Wavelength to Meters

The wavelength given is \(532 \mathrm{nm}\). To use the equation, convert this to meters. Thus, \(532 \, \text{nm} = 532 \times 10^{-9} \, \text{m}\).
03

Calculate the Frequency of the Green Laser Light

Substitute \(c = 3.00 \times 10^8 \, \text{m/s}\) and \(\lambda = 532 \times 10^{-9} \, \text{m}\) into the frequency equation: \(u = \frac{3.00 \times 10^8}{532 \times 10^{-9}}\).
04

Compute the Frequency

Perform the division: \(u \approx 5.64 \times 10^{14} \, \text{Hz}\). This is the frequency of the green laser light.
05

Understanding Photon Energy

The energy of a photon can be calculated using the equation \(E = hu\), where \(h\) is Planck's constant \( (6.626 \times 10^{-34} \, \text{J} \cdot \text{s}) \) and \(u\) is the frequency.
06

Calculate the Energy of One Photon

Substitute \(u = 5.64 \times 10^{14} \, \text{Hz}\) into the energy equation: \(E = 6.626 \times 10^{-34} \times 5.64 \times 10^{14}\).
07

Compute the Energy

Calculate: \(E \approx 3.73 \times 10^{-19} \, \text{J}\). This is the energy of one photon emitted by the laser pointer.
08

Determine the Energy Gap Between States

The energy gap between the ground state and the excited state in the laser material is equal to the energy of the photons emitted, since returning electrons emit photons of the same energy they absorbed. Therefore, the energy gap is \(3.73 \times 10^{-19} \, \text{J}\).

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Key Concepts

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

Wavelength to Frequency Conversion
The wavelength of light is intrinsically linked to its frequency. Understanding this relationship is fundamental when working with light emitted by lasers. Since light behaves as both a wave and a particle, it becomes important to explore its wave characteristics first.

Let's consider a green laser pointer, which emits light at a wavelength of \(532\ \text{nm}\). Wavelength is the distance between two consecutive peaks of a wave. However, to find the frequency of the light wave, we need to know how often these peaks pass a point each second. This is where frequency comes in, measured in hertz (Hz).

To convert wavelength to frequency, we use the equation:
  • \(c = \lambda u\)
where \(c\) is the speed of light \((3.00 \times 10^8\ \text{m/s})\), \(\lambda\) is the wavelength, and \(u\) is the frequency.

To solve for the frequency \((u)\), rearrange the formula:
  • \(u = \frac{c}{\lambda}\)
Since wavelengths are often given in nanometers, remember to convert them to meters first; here \(532\ \text{nm} = 532 \times 10^{-9}\ \text{m}\). By substituting the values, you get the frequency of the green laser light as approximately \(5.64 \times 10^{14}\ \text{Hz}\). This frequency indicates how many cycles the light wave completes in one second.
Photon Energy Calculation
Light behaves not only as a wave but also as a stream of particles called photons. To find the energy of these photons, we turn to the relationship between energy and frequency.

The energy of a photon is dictated by Planck's equation:
  • \(E = hu\)
In this equation, \(E\) represents the energy of a photon, \(h\) is Planck's constant \((6.626 \times 10^{-34} \text{J} \cdot \text{s})\), and \(u\) is the frequency.

By substituting the frequency we calculated earlier (\(u = 5.64 \times 10^{14} \ \text{Hz}\)) into the equation, we can determine the energy of one photon emitted by the green laser pointer:
  • \(E = 6.626 \times 10^{-34} \times 5.64 \times 10^{14}\)
Doing the multiplication gives us approximately \(3.73 \times 10^{-19} \text{J}\). This tells us how much energy each emitted photon carries, illustrating the quantum nature of light.
Energy Gap in Laser Materials
Lasers operate on the principle of stimulated emission, where electrons transition between energy states or levels. In the case of a green laser pointer, electrons in the material are initially excited to a higher energy state by an external energy source, usually a battery.

Once the electrons return to their original or ground state, they release energy in the form of light, specifically photons with a wavelength of \(532\ \text{nm}\). The energy gap between these two states corresponds directly to the energy of the emitted photons. This principle forms the basis of calculating the energy gap in laser materials.

Since each photon has an energy of \(3.73 \times 10^{-19} \text{J}\), this also represents the energy difference between the excited state and the ground state of the electrons. These calculations are crucial in understanding and designing laser technology.

Overall, the energy gap indicates how much energy is needed to excite an electron from the ground state to an excited state, providing insights into the efficiency and color of the laser light emitted.

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Most popular questions from this chapter

The discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71)\) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 .\) Write a balanced chemical equation for the reaction. Starting with a 55.4 -g sample of \(\mathrm{ZrO}_{2},\) calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and \(\mathrm{Hf}\) form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

Write the condensed electron configurations for the following atoms, using the appropriate noble-gas core abbreviations: \((\mathbf{a}) \mathrm{Cs},(\mathbf{b}) \mathrm{Ni},(\mathbf{c}) \mathrm{Se},(\mathbf{d}) \mathrm{Cd},(\mathbf{e}) \mathrm{U},(\mathbf{f}) \mathrm{Pb} .\)

What is the maximum number of electrons in an atom that can have the following quantum numbers? (a) \(n=3, m_{l}=-1 ;\) (b) \(n=4, l=2\); (c) \(n=4, l=3, m_{l}=-2 ;\) (d) \(n=5, l=2\), \(m_{l}=0\)

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

(a) For \(n=4,\) what are the possible values of \(l ?(\mathbf{b})\) For \(l=2\), what are the possible values of \(m_{l} ?\) (c) If \(m_{l}\) is \(2,\) what are the possible values for \(l ?\)

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