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Chapter 35: Problem 29

An electron is confined to a cubical box. For what box width will a transition from the first excited state to the ground state result in emission of a 950 -nm infrared photon?

Short Answer

Expert verified
Calculating the energy of the photon and then equating it with the difference between the first excited state and the ground state in a particle in a cubic box allows us to solve for the box width, \(L\).

Step by step solution

01

Determine the Energy of the Photon

Using the given wavelength of 950nm, we first find the energy of the photon using Einstein's equation \(E=hf\), and the fact that the speed of light, \(c\), is equal to the frequency, \(f\), times the wavelength, \(\lambda\). Substituting \(f\) with \( \frac{c}{\lambda} \), we get \(E = \frac {hc}{\lambda}\). Here, \(h\) is Planck’s constant \(6.626 \times 10^{-34} Js\) and \(c\) 3\( \times 10^{8} ms^{-1}\). Replace \(\lambda\) with \(950 \times 10^{-9} m\) to calculate \(E\)
02

Determine the Energy Difference

The energy for an electron confined in a three dimensional box is given by \(E = \frac{{h^2(n_x^2+n_y^2+n_z^2)}}{{8mL^2}}\), where \(n_x, n_y, n_z\) are quantum numbers that hold positive integer values corresponding to the energy state, \(m\) is the mass of the electron and \(L\) is the size (width) of the box. Here, the electron makes a transition from the first excited state (represented by quantum numbers \(2,1,1\)) to the ground state (represented by quantum numbers \(1,1,1\)). So, calculate the energy difference \(\Delta E = E_{211} - E_{111}.\)
03

Calculate the dimension of the box

Given \(\Delta E = \frac {hc}{\lambda}\), the energy difference equals the energy of the photon. Substitute \(\Delta E\) with the expression, solve for \(l\) to obtain the width of the box.

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Most popular questions from this chapter

Is quantization significant for macro molecules confined to biological cells? To find out, consider a protein of mass 250,000 u confined to a \(10 \mu \mathrm{m}\) -diameter cell. Treating this as a particle in a one-dimensional square well, find the energy difference between the ground state and the first excited state. Given that biochemical reactions typically involve energies on the order of \(1 \mathrm{eV}\), what do you conclude about the role of quantization?

The ground-state energy for an electron in infinite square well \(A\) is equal to the energy of the first excited state for an electron in well B. How do the wells' widths compare?

In terms of de Broglie's matter-wave hypothesis, how does making the sides of a box different lengths remove the degeneracy associated with a particle confined to that box?

(a) Using the potential energy \(U=\frac{1}{2} m \omega^{2} x^{2}\) discussed on page 635 develop the Schrodinger equation for the harmonic oscillator. (b) Show by substitution that \(\psi_{0}(x)=A_{0} e^{-\alpha^{2} x^{2} / 2}\) satisfies your equation, where \(\alpha^{2}=m \omega / \hbar\) and the energy is given by Equation 35.7 with \(n=0 .\) (c) Find the normalization constant \(A_{0} .\) You then have the ground-state wave function for the harmonic oscillator.

A particle's wave function is \(\psi=A e^{-x / a^{2}},\) where \(A\) and \(a\) are constants. (a) Where is the particle most likely to be found? (b) Where is the probability per unit length half its maximum value?
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