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Chapter 35: Problem 35

An electron drops from the \(n=7\) to the \(n=6\) level of an infinite square well 1.5 nm wide. Find (a) the energy and (b) the wavelength of the photon emitted.

Short Answer

Expert verified
The energy of the emitted photon is obtained by subtracting the energy at the \(n=6\) level from the energy at the \(n=7\) level, using the formula for well's energy levels. Once this energy is known, the wavelength is obtained using the energy-wavelength relation for a photon.

Step by step solution

01

Understanding the energy levels inside an infinite square well

In an infinite square well, the energy levels are given by the formula: \(E_n = \frac{n^2 h^2}{8mL^2}\) where \(n\) is the energy level, \(h\) is Planck's constant, \(m\) is the mass of the electron, and \(L\) is the width of the well. It is necessary to calculate the energy levels \(E_7\) and \(E_6\) to find the energy difference, which equals the energy of the emitted photon.
02

Calculate the energy difference between the two levels

First, let's calculate the energy for the level \(n=7\), \(E_7 = \frac{(7^2)h^2}{8m*(1.5*10^{-9})^2}\). Next, we calculate the energy for the level \(n=6\), \(E_6 = \frac{(6^2)h^2}{8m*(1.5*10^{-9})^2}\). We subtract these to get the energy difference which equals the energy of the photon: \(\Delta E = E_7 - E_6\). We can now insert the known values \(h = 6.63 * 10^{-34} Js\) and \(m = 9.11*10^{-31}kg\) which yields the energy of the photon.
03

Calculate the wavelength of the emitted photon

The energy of a photon is related to its wavelength through the equation \(E = \frac{hc}{\lambda}\), where \(h\) is Planck's constant, \(c\) is the speed of light, and \(\lambda\) is the wavelength. Solving for \(\lambda\), we get \(\lambda=\frac{hc}{E}\). With \(h\), \(c\) and the previously calculated \(E\), compute \(\lambda\), which will be the wavelength of the photon.

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Most popular questions from this chapter

Find an expression for the normalization constant \(A\) for the wave function given by \(\psi=0\) for \(|x|>b\) and \(\psi=A\left(b^{2}-x^{2}\right)\) for \(-b \leq x \leq b\)

One reason we don't notice quantum effects in everyday life is that Planck's constant \(h\) is so small. Treating yourself as a particle (mass \(60 \mathrm{kg}\) ) in a room-sized one-dimensional infinite square well (width \(2.6 \mathrm{m}\) ), how big would \(h\) have to be if your minimum possible energy corresponded to a speed of \(1.0 \mathrm{m} / \mathrm{s} ?\)

(a) Using the potential energy \(U=\frac{1}{2} m \omega^{2} x^{2}\) discussed on page \(675,\) develop the Schr枚dinger 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.

The ground-state energy of a harmonic oscillator is 4.0 eV. Find the energy separation between adjacent quantum states.

The ground-state wave function for a quantum harmonic oscillator has a single central peak. Why is this at odds with classical physics?
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