Problem 21 - Chemistry: The Central Science

Question

(a) A red laser pointer emits light with a wavelength of \(650 \mathrm{~nm}\). What is the frequency of this light? (b) What is the energy of 1 mole 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 \(650 \mathrm{~nm}\) photons. What is the energy gap between the ground state and excited state in the laser material?

Step-by-Step Solution

Answer

The frequency of the red laser light is \(4.62 \times 10^{14} \mathrm{~Hz}\). The energy of 1 mole of photons is \(1.84 \times 10^5 \mathrm{~J/mol}\). The energy gap between the ground state and excited state in the laser material is \(3.06 \times 10^{-19} \mathrm{~J}\).

Step 1: (a.1) Rearrange the speed of light equation

To find the frequency, we'll use the speed of light equation, \(c = λν\), where \(c\) is the speed of light in a vacuum, \(λ\) is the wavelength, and \(ν\) is the frequency. We'll rearrange this equation to solve for the frequency: \[ν = \frac{c}{λ}\]

Step 2: (a.2) Calculate the frequency

Given that the wavelength (\(λ\)) of the red laser pointer is \(650 \mathrm{~nm}\) or \(6.50 \times 10^{-7} \mathrm{~m}\) and the speed of light (\(c\)) in a vacuum is approximately \(3.00 \times 10^8 \mathrm{~m/s}\), we can find the frequency (\(ν\)) using the rearranged equation: \[ν = \frac{3.00 \times 10^8 \mathrm{~m/s}}{6.50 \times 10^{-7} \mathrm{~m}} = 4.62 \times 10^{14} \mathrm{~Hz}\] Thus, the frequency of the red laser light is \(4.62 \times 10^{14} \mathrm{~Hz}\). #b) Finding the energy of 1 mole of photons#

Step 3: (b.1) Calculate the energy of a single photon

Using the equation for the energy of a photon, \(E = hν\), where \(h\) is Planck's constant (approximately \(6.63 \times 10^{-34} \mathrm{~Js}\)) and \(ν\) is the frequency we found in part (a), we can find the energy of a single photon: \[E = (6.63 \times 10^{-34} \mathrm{~Js}) (4.62 \times 10^{14} \mathrm{~Hz}) = 3.06 \times 10^{-19} \mathrm{~J}\] Therefore, the energy of a single photon is \(3.06 \times 10^{-19} \mathrm{~J}\).

Step 4: (b.2) Calculate the energy of 1 mole of photons

Now, we'll multiply the energy of a single photon by Avogadro's number (\(N_A\)) to find the energy of 1 mole of photons: \[(3.06 \times 10^{-19} \mathrm{~J})(6.022 \times 10^{23} \text{ photons/mol}) = 1.84 \times 10^5 \mathrm{~J/mol}\] Thus, the energy of 1 mole of photons is \(1.84 \times 10^5 \mathrm{~J/mol}\). #c) Finding the energy gap between the ground state and excited state in the laser material#

Step 5: (c) Determine the energy gap

The energy gap between the ground state and excited state in the laser material can be found by calculating the energy of a single photon using the energy of a photon equation \(E = hν\). We already found the energy of a single photon in part (b.1): \(3.06 \times 10^{-19} \mathrm{~J}\). Therefore, the energy gap between the ground state and excited state in the laser material is \(3.06 \times 10^{-19} \mathrm{~J}\).

Key Concepts

Photon Energy

Photon energy is the energy carried by a single photon. A photon is a particle of light, and its energy is determined by the equation \(E = h u\), where \(E\) is the energy, \(h\) is Planck's constant, and \(u\) is the frequency of the photon. This equation shows a direct relationship between the energy and frequency; higher frequency means higher energy.

In practice, calculating the energy of a photon helps us understand various phenomena in laser physics and quantum mechanics. For instance, in our exercise, we found that each photon of red laser light with a frequency of \(4.62 \times 10^{14} \text{ Hz}\) has an energy of \(3.06 \times 10^{-19} \text{ J}\).

• This fundamental concept applies to everything from medical lasers to everyday electronics.
• The ability to calculate photon energy enables engineers and scientists to design efficient light-based technology.

Frequency Calculation

Frequency calculation is an essential step in determining various properties of a wave, including light. The frequency of a wave is the number of complete wave cycles that pass a specific point in one second, measured in Hertz (Hz).

To calculate the frequency of a light wave, we use the formula: \(c = \lambda u\), where \(c\) is the speed of light, \(\lambda\) is the wavelength, and \(u\) is the frequency. By rearranging this formula to \(u = \frac{c}{\lambda}\), we can find the frequency of any given wavelength.

• In the exercise above, the wavelength of the red laser is \(650 \text{ nm}\) or \(6.50 \times 10^{-7} \text{ m}\).
• Given the speed of light as \(3.00 \times 10^8 \text{ m/s}\), we calculate the frequency to be \(4.62 \times 10^{14} \text{ Hz}\).

This calculation is crucial in understanding the properties and behavior of light. For example, frequency is directly related to energy, which is important in studies of light-matter interaction.

Wavelength

Wavelength is an important concept when discussing light and various forms of electromagnetic radiation. It is the distance between two consecutive peaks of a wave. Measured in meters (m) or nanometers (nm), the wavelength determines the wave's characteristics, such as its color in the visible spectrum.

In laser physics, the wavelength is a critical parameter because it defines the color of the laser and its applications. In our scenario, the red laser has a wavelength of \(650 \mathrm{~nm}\), which falls in the red region of the visible spectrum. This is important because different wavelengths can have different effects and uses in applications:

• Red lasers are commonly used in teaching and presentations as pointers due to their visibility.
• They may also be used in various scientific and industrial contexts.

Understanding wavelengths helps in designing and utilizing lasers effectively for specific purposes.

Planck's Constant

Planck's constant, denoted as \(h\), is a fundamental constant in quantum mechanics. With a value of approximately \(6.63 \times 10^{-34} \text{ Js}\), it plays a critical role in the quantization of energy levels in quantum systems.

Planck's constant is significant in calculations involving the energy of photons. Used in the formula \(E = h u\), it links the frequency of a photon with its energy:

• This constant enables us to relate classical wave concepts with quantum phenomena.
• It provides a bridge between macroscopic wave characteristics and atomic-level energy interactions.

In applications like laser physics, understanding Planck's constant allows scientists and engineers to develop technology that changes electronic states in materials precisely, like what happens when a laser emits light. As evident from our exercise, Planck's constant was key in calculating the photon energy and hence the energy gap within the laser material.