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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 -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’s 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’s 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’s 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’s equation \(E = hu\), where \(h\) is Planck’s 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

Use the de Broglie relationship to determine the wavelengths of the following objects: (a) an \(85-\mathrm{kg}\) person skiing at \(50 \mathrm{~km} / \mathrm{hr}\), (b) a \(10.0-\mathrm{g}\) bullet fired at \(250 \mathrm{~m} / \mathrm{s}\), (c) a lithium atom moving at \(2.5 \times 10^{5} \mathrm{~m} / \mathrm{s}\), (d) an ozone \(\left(\mathrm{O}_{3}\right)\) molecule in the upper atmosphere moving at \(550 \mathrm{~m} / \mathrm{s}\).

Microwave ovens use microwave radiation to heat food. The energy of the microwaves is absorbed by water molecules in food and then transferred to other components of the food. (a) Suppose that the microwave radiation has a wavelength of \(11.2 \mathrm{~cm}\). How many photons are required to heat \(200 \mathrm{~mL}\) of coffee from 23 to \(60^{\circ} \mathrm{C}\) ? (b) Suppose the microwave's power is \(900 \mathrm{~W}\) ( 1 watt \(=1\) joule-second). How long would you have to heat the coffee in part (a)?

(a) Account for formation of the following series of oxides in terms of the electron configurations of the elements and the discussion of ionic compounds in Section \(2.7 :\) $\mathrm{K}_{2} \mathrm{O}, \mathrm{CaO}, \mathrm{Sc}_{2} \mathrm{O}_{3}, \mathrm{Ti} \mathrm{O}_{2}, \mathrm{V}_{2} \mathrm{O}_{5}, \mathrm{CrO}_{3} .$ (b) Name these oxides. (c) Consider the metal oxides whose enthalpies of formation (in kJ mol \(^{-1}\) ) are listed here. Calculate the enthalpy changes in the following general reaction for each case: $$ \mathrm{M}_{n} \mathrm{O}_{m}(s)+\mathrm{H}_{2}(g) \longrightarrow n \mathrm{M}(s)+m \mathrm{H}_{2} \mathrm{O}(g) $$ (You will need to write the balanced equation for each case and then compute \(\Delta H^{\circ} . )(\mathbf{d})\) Based on the data given, estimate a value of \(\Delta H_{f}^{\circ}\) for \(S c_{2} \mathrm{O}_{3}(s) .\)

The familiar phenomenon of a rainbow results from the diffraction of sunlight through raindrops. (a) Does the wavelength of light increase or decrease as we proceed outward from the innermost band of the rainbow? (b) Does the frequency of light increase or decrease as we proceed outward? (c) Suppose that instead of sunlight, the visible light from a hydrogen discharge tube (Figure 6.10) was used as the light source. What do you think the resulting "hydrogen discharge rainbow" would look like? [Section 6.3]

One of the emission lines of the hydrogen atom has a wavelength of \(93.07 \mathrm{~nm}\). (a) In what region of the electromagnetic spectrum is this emission found? (b) Determine the initial and final values of \(n\) associated with this emission.

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