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Photon in a Dye Laser. An electron in a long, organic molecule used in a dye laser behaves approximately like a particle in a box with width 4.18 nm. What is the wavelength of the photon emitted when the electron undergoes a transition (a) from the first excited level to the ground level and (b) from the second excited level to the first excited level?

Short Answer

Expert verified
The wavelength for transition (a) is approximately 516 nm, and for transition (b) is approximately 310 nm.

Step by step solution

01

Understand the Particle in a Box Model

In this problem, we consider an electron in a box of length \( L = 4.18 \text{ nm} \). The energy levels for a particle in a 1D box are given by \( E_n = \frac{n^2 h^2}{8mL^2} \), where \( n \) is the quantum number, \( h \) is Planck's constant, and \( m \) is the mass of the electron.
02

Calculate Energy Difference for Transition (a)

To find the energy difference \( \Delta E \) between the first excited state \( (n=2) \) and the ground state \( (n=1) \), calculate: \( \Delta E = E_2 - E_1 = \frac{(2^2 - 1^2)h^2}{8mL^2} = \frac{3h^2}{8mL^2} \).
03

Convert Energy Difference to Wavelength for Transition (a)

The energy difference corresponds to the energy of the photon emitted, given by \( \Delta E = \frac{hc}{\lambda} \). Solve for \( \lambda \) to get \( \lambda = \frac{hc}{\Delta E} = \frac{8mL^2c}{3h} \).
04

Calculate Energy Difference for Transition (b)

For the transition from the second excited state to the first excited state \((n=3 \text{ to } n=2)\), the energy difference is \( \Delta E = E_3 - E_2 = \frac{(3^2 - 2^2)h^2}{8mL^2} = \frac{5h^2}{8mL^2} \).
05

Convert Energy Difference to Wavelength for Transition (b)

Again, using the relationship \( \Delta E = \frac{hc}{\lambda} \), solve for \( \lambda \): \( \lambda = \frac{hc}{\Delta E} = \frac{8mL^2c}{5h} \).
06

Final Calculations

Substitute numerical values: \( h = 6.626 \times 10^{-34} \text{ J s} \), \( m = 9.109 \times 10^{-31} \text{ kg} \), \( c = 3 \times 10^8 \text{ m/s} \), and \( L = 4.18 \times 10^{-9} \text{ m} \) into the equations for each transition. Perform the calculations to find the wavelengths.

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Key Concepts

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

Photon Emission
Photon emission occurs when an electron transitions between energy levels in an atom or molecule, releasing energy in the form of a photon. This concept is fundamental to understanding phenomena in quantum mechanics, like the operation of lasers, including dye lasers. In the context of the exercise provided, as the electron in the molecule transitions from one energy level to another, it emits a photon. The energy of this photon is directly correlated to the difference in energy between the two levels.
Photon emission happens as discrete packets of energy, based on the principle that energy levels in atoms and molecules are quantized. This means that electrons can only exist in specific energy states and not in between these states. When an electron moves from a higher energy state to a lower one, the excess energy is emitted as a photon. The specific wavelength of this photon can be calculated using the energy difference, providing valuable insights into the electronic transitions occurring within the molecule.
Quantum Mechanics
Quantum mechanics underlies the framework for "Particle in a Box" problems. In this approach, particles like electrons are described not as classical points but by wavefunctions. These wavefunctions represent the probability of finding a particle in a particular location.
In quantum mechanics, the "Particle in a Box" model serves as a simplified illustration where a particle is confined to move freely within a perfectly rigid and impenetrable boundary. It is a fundamental problem that provides insight into the quantized nature of energy levels. The electron behaves like a wave, and only certain wavelengths \( \lambda \) that fit the box exactly (called standing waves) are allowed. This quantization results in discrete energy levels calculated by the formula:
  • \( E_n = \frac{n^2 h^2}{8mL^2} \)
where \( E_n \) is the energy of the level \( n \), \( h \) is Planck's constant, \( m \) is the mass of the electron, and \( L \) is the length of the box. Hence, the understanding of quantum mechanics is crucial to predict the behavior of particles in such constrained systems.
Energy Levels
Energy levels refer to the discrete set of energies that electrons can possess in an atom or molecule. These states are often depicted as a ladder, with each rung representing one quantized energy level. For the "Particle in a Box" model, energy levels arise from the restrictions on the electronic wavefunctions within a confined space.
The specific energy for each level \( n \) is determined by:
  • \( E_n = \frac{n^2 h^2}{8mL^2} \)
In this context, energy levels are crucial for understanding photon emission because the difference in these levels gives rise to the particular wavelength of the photon emitted during electronic transitions. Only when electrons transition from higher to lower levels do they emit energy as photons, and this emitted energy corresponds to the difference between the energy levels involved. This model helps visualize what happens as electrons shift between quantized states.
Wavelength Calculation
Calculating the wavelength of a photon involved in electronic transitions is essential for understanding the photon's energy and characteristics. Based on the exercise solution, the calculation commences by establishing the difference in energy levels (\( \Delta E \)). Once \( \Delta E \) is determined, the wavelength \( \lambda \) can be calculated using the formula:
  • \( \Delta E = \frac{hc}{\lambda} \)
Here, \( h \) is Planck's constant, and \( c \) is the speed of light. Rearranging the equation gives:
  • \( \lambda = \frac{hc}{\Delta E} \)
This relationship shows how the energy of the emitted photon (related to the energy difference) can be directly converted into a measurable wavelength.
Understanding wavelength calculation is pivotal in fields like spectroscopy, where the analysis of wavelengths helps determine the structure and properties of molecules. By calculating the exact wavelengths emitted (as done in the exercise), scientists can gather precise information about the energy levels and behavior of electrons within atoms or molecules.

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Most popular questions from this chapter

A particle is described by a wave function \(\psi(x)=A e^{-\alpha x^{2}}\) where \(A\) and \(\alpha\) are real, positive constants. If the value of \(\alpha\) is increased, what effect does this have on (a) the particle's uncertainty in position and (b) the particle's uncertainty in momentum? Explain your answers.

Consider a particle in a box with rigid walls at \(x=0\) and \(x=L .\) Let the particle be in the ground level. Calculate the probability \(|\psi|^{2} d x\) that the particle will be found in the interval \(x\) to \(x+d x\) for \((\) a \() x=L / 4 ;(\) b) \(x=L / 2 ;(\) c) \(x=3 L / 4\)

A particle moving in one dimension (the \(x\) -axis) is described by the wave function \(\psi(x)=\left\\{\begin{array}{ll}{A e^{-b x},} & {\text { for } x \geq 0} \\ {A e^{b x},} & {\text { for } x<0}\end{array}\right.\) where \(b=2.00 \mathrm{m}^{-1}, A>0,\) and the \(+x\) -axis points toward the right. (a) Determine \(A\) so that the wave function is normalized. (b) Sketch the graph of the wave function. (c) Find the probability of finding this particle in each of the following regions: (i) within 50.0 \(\mathrm{cm}\) of the origin, (ii) on the left side of the origin (can you first guess the answer by looking at the graph of the wave function? (iii) between \(x=0.500 \mathrm{m}\) and \(x=1.00 \mathrm{m} .\)

A proton is in a box of width \(L .\) What must the width of the box be for the ground-level energy to be \(5.0 \mathrm{MeV},\) a typical value for the energy with which the particles in a nucleus are bound? Compare your result to the size of a nucleus- that is, on the order of \(10^{-14} \mathrm{m} .\)

A fellow student proposes that a possible wave function for a free particle with mass \(m\) (one for which the potential-energy function \(U(x)\) is zero) is $$\psi(x)=\left\\{\begin{array}{ll}{e^{+\kappa x},} & {x<0} \\ {e^{-\kappa x},} & {x \geq 0}\end{array}\right.$$ where \(\kappa\) is a positive constant. (a) Graph this proposed wave function. (b) Show that the proposed wave function satisfies the Schrodinger equation for \(x < 0\) if the energy is \(E=-\hbar^{2} \kappa^{2} / 2 m-\) that is, if the energy of the particle is negative. (c) Show that the proposed wave function also satisfies the Schrodinger equation for \(x \geq 0\) with the same energy as in part (b). (d) Explain why the proposed wave function is nonetheless not an acceptable solution of the Schrodinger equation for a free particle. (Hint: What is the behavior of the function at \(x=0 ?\) ) It is in fact impossible for a free particle (one for which \(U(x)=0 )\) to have an energy less than zero.

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