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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
The frequency of the light emitted by the green laser pointer is \(f = \frac{3.00 \times 10^8\, m/s}{532 \times 10^{-9} m} \approx 5.63 \times 10^{14} Hz\). The energy of one photon is \(E = (6.63 \times 10^{-34}\, Js) * f \approx 3.73 \times 10^{-19} J\). The energy gap between the ground state and the excited state in the laser material is \(\Delta E = E \approx 3.73 \times 10^{-19} J\).

Step by step solution

01

Calculate the frequency of the light

To find the frequency (f) of the light, we can use the formula relating the speed of light (c), wavelength (λ), and frequency (f): \[c = λ * f\] Rearrange the formula to solve for frequency: \[f = \frac{c}{λ}\] The speed of light (c) is approximately \(3.00 \times 10^8 \, m/s\), and the wavelength (λ) of the light is \(532\, nm\), which can be converted to meters: \(\lambda = 532 * 10^{-9} m\). Now, substitute the values of c and λ into the frequency formula: \[f = \frac{3.00 \times 10^8\, m/s}{532 \times 10^{-9} m}\]
02

Calculate the energy of one photon

To find the energy (E) of one photon, we can use the formula relating Planck's constant (h) and frequency (f): \[E = h * f\] Planck's constant (h) is approximately \(6.63 \times 10^{-34}\, Js\). Use the frequency calculated in Step 1 to find the energy of one photon: \[E = (6.63 \times 10^{-34}\, Js) * f\]
03

Calculate the energy gap between ground state and excited state

The energy gap (ΔE) between the ground state and the excited state in the laser material is equal to the energy of one photon, as calculated in Step 2. So, the energy gap is: \[\Delta E = E\] Calculate the values for frequency (f), energy of one photon (E), and the energy gap between ground state and excited state (ΔE) using the formulas and constants provided.

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Key Concepts

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

energy of photons
In the world of physics, understanding how light energy comes in packets rather than a continuous stream is crucial. These packets of light energy are what we call photons. The energy of a single photon is directly related to its frequency by using Planck’s constant, symbolized as \( h \).
The relationship can be expressed through the formula:
  • \( E = h \times f \)
Where \( E \) is the photon energy, and \( f \) is the frequency of the light.
This implies that higher frequency light carries more energy in its photons. Conversely, lower frequency light carries less energy.
One can see this directly in visible light – blue light, with a higher frequency, has more photon energy than red light, which has a lower frequency.
laser material energy states
Lasers operate through a fascinating process involving the energy states of electrons in the material used. When a laser pointer emits light, electrons within the material are excited to a higher energy level by some external energy source, usually a battery.

Once these electrons gain energy, they move from a ground state to a higher, excited state. This transition is temporary. The electron cannot stay in the excited state indefinitely and eventually falls back to its ground state.
When it does, it must release the excess energy it gained. This is accomplished through the emission of a photon - in this case, the green light photons we observe from the laser.
  • The energy of this emitted photon equals the gap between the two energy states.
Understanding this process helps us gain insight into the fundamental operation of laser technology and the critical role played by electronic energy states in materials.
Planck's constant
Max Planck introduced a revolutionary idea when he presented the concept of quantization, introducing Planck's constant \( h \). This constant is fundamental in describing the behavior of particles at a microscopic level.

Planck’s constant is a tiny number, approximately \( 6.63 \times 10^{-34} \) Joule seconds \((Js)\), which might seem insignificant in size. Yet, it underpins much of quantum mechanics and modern physics.
  • It provides the bridge between different world scales in physics: the macroscopic and the quantum.
  • It’s used in calculating the energy of photons through the formula \( E = h \times f \).
The role of Planck's constant is critical in the realms of quantum phenomena, explaining why, for instance, energy exchanges are not continuous but come in discrete packets or quanta.

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Most popular questions from this chapter

Is energy emitted or absorbed when the following electronic transitions occur in hydrogen? (a) from \(n=3\) to \(n=2\), (b) from an orbit of radius \(0.846 \mathrm{nm}\) to one of radius 0.212 \(\mathrm{nm},(\mathbf{c})\) an electron adds to the \(\mathrm{H}^{+}\) ion and ends up in the \(n=2\) shell?

If a sample of calcium chloride is introduced into a nonluminous flame, the color of the flame turns to orange ("flame test"). The light is emitted because calcium atoms become excited; their return to the ground state results in light emission. (a) The wavelength of this emitted light is \(422.7 \mathrm{nm} .\) Calculate its frequency. (b) What is the energy of \(1.00 \mathrm{~mol}\) of these photons (a mole of photons is called an Einstein)? (c) Calculate the energy gap between the excited and ground states for the calcium atom.

The electron microscope has been widely used to obtain highly magnified images of biological and other types of materials. When an electron is accelerated through a particular potential field, it attains a speed of \(9.47 \times 10^{6} \mathrm{~m} / \mathrm{s}\) What is the characteristic wavelength of this electron? Is the wavelength comparable to the size of atoms?

The series of emission lines of the hydrogen atom for which \(n_{f}=4\) is called the Brackett series. (a) Determine the region of the electromagnetic spectrum in which the lines of the Brackett series are observed. (b) Calculate the wavelengths of the first three lines in the Brackett series - those for which \(n_{i}=5,6,\) and 7.

One type of sunburn occurs on exposure to UV light of wavelength in the vicinity of \(325 \mathrm{nm}\) (a) What is the energy of a photon of this wavelength? (b) What is the energy of a mole of these photons? (c) How many photons are in a \(1.00 \mathrm{~mJ}\) burst of this radiation? \((\mathbf{d})\) These \(\mathrm{UV}\) photons can break chemical bonds in your skin to cause sunburn-a form of radiation damage. If the \(325-\mathrm{nm}\) radiation provides exactly the energy to break an average chemical bond in the skin, estimate the average energy of these bonds in \(\mathrm{kJ} / \mathrm{mol}\).

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