Extra Credit 15

https://phys.libretexts.org/Bookshel...cs_(Exercises)

Solve (i.e., write reasonable step-by-step tutorials for the solutions) of these exercises from the text)

#52, #65, #82

Problem 52

To excite an electron in a one-dimensional box from its first excited state to its third excited state requires 20.0 eV. What is the width of the box?

Solution

An electron in a 1D box will have an energy function

$$
E_{n}=\frac{h^{2}}{8 m L^{2}} n^{2}
$$

Consequently, any energy transition between levels will simply result in a difference of 

$$
\Delta E=\frac{h^{2}}{8 m L^{2}}\left(n_{1}\strut^{2}-n_{2}\strut^{2}\right)
$$

Since we know the energy levels, the transition energy, and the mass of the electron, we can simply rearrange Equation 2 to solve for the unknown length

$$
L=\sqrt{\frac{h^{2}\left(n_{1}\strut^{2}-n_{2}\strut^{2}\right)}{8 m_{e} \Delta E}}=h \sqrt{\frac{\left(n_{1}\strut^{2}-n_{2}\strut^{2}\right)}{8 m_{e} \Delta E}}
$$

Recall that the ground state for a particle in a 1D box is n = 1. Thus, the first excited state will be  n = 2, and the third excited state will be n = 4. For convenience, we will use units of eV for Planck's constant and \(\frac{\mathrm{MeV}}{c^2}\) for the mass of the electron where \(c\) is just the speed of light. Plugging values into Equation 3 we get

$$
L=\left(4.14 \times 10^{-15} \mathrm{~eV} \cdot \mathrm{s}\right) \sqrt{\frac{\left(4^{2}-2^{2}\right)}{8\left(0.511 ~\frac{\mathrm{MeV}}{c^{2}}\right)(20.0 ~\mathrm{eV})}}=4.76 \times 10^{-10} \mathrm{~m}
$$

 

 

Problem 65

Determine the expectation value of the potential energy for a quantum harmonic oscillator in the ground state. Use this to calculate the expectation value of the kinetic energy.

Solution

A quantum harmonic oscillator has the potential energy function

$$
\hat{U}=\frac{1}{2} k x^{2}
$$

A harmonic oscillator in the ground state can be represented by the wavefunction

$$
\Psi_{v=0}=\left(\frac{\alpha}{\pi}\right)^{\frac{1}{4}} e\strut^{-\frac{x^{2}}{2}}
$$
where \(\alpha = \dfrac{\mu \omega}{\hbar}\). The expectation value of the potential energy for a ground state harmonic oscillator can be represented in Dirac notation as

$$
\left\langle\Psi_{0}|\hat{U}| \Psi_{0}\right\rangle
$$

The operator commutes and the wavefunction has no complex parts, so the integral for the expectation value becomes

$$
\int_{\text {all space }} \frac{1}{2} k x^{2} \Psi_{0}\strut^{2} ~dx
$$

Therefore, 

$$
\langle\hat{U}\rangle_{\Psi}=\frac{1}{2} k\left(\frac{\alpha}{\pi}\right)^{\frac{1}{2}} \int_{-\infty}^{\infty} x^{2} e^{-x^{2}} ~dx
$$

Note that the integrand is an even function, so we can instead integrate from 0 to infinity and double the value of the integral. 

$$
\begin{gathered}
=\frac{1}{2} k\left(\frac{\alpha}{\pi}\right)^{\frac{1}{2}} \cdot 2 \int_{0}^{\infty} x^{2} e^{-x^{2}} ~dx \\ \\
=k\left(\frac{\alpha}{\pi}\right)^{\frac{1}{2}}\left[\frac{1}{2^{2} \alpha} \times\left(\frac{\pi}{\alpha}\right)^{\frac{1}{2}}\right] \\ \\
=\frac{k}{4 \alpha} \\ \\
=\frac{k}{4\left(\frac{\mu \omega}{\hbar}\right)} \\ \\
=\frac{\hbar}{4\omega} \cdot \frac{k}{\mu} \\ \\
=\frac{\hbar}{4\omega} \omega^2 \\ \\
=\frac{1}{2} \hbar \omega \times \frac{1}{2} \\ \\
=\frac{1}{2} E_{0}
\end{gathered}
$$

This value makes sense due to the particle's periodic oscillations between the turning points of the potential energy function. Since the kinetic energy function is also periodic and it shares the same range as the potential energy function, this must mean that the expectation value is also the same. Thus,

$$
\langle\hat{K}\rangle_{\Psi}=\frac{1}{2} E_{0}
$$

 

 

Problem 82

Find the expectation value of the position squared when the particle in the box is in its third excited state and the length of the box is L.

Solution

The normalized wavefunction for a particle in a box is

$$
\Psi(x)=\sqrt{\frac{2}{L}} \sin \left(\frac{n \pi x}{L}\right)
$$

The expectation value of the position operator squared can be expressed in Dirac notation as

$$
\left\langle\Psi\left|x^{2}\right| \Psi\right\rangle
$$

This operator commutes and the wavefunction has no complex parts, so the integral for the expectation value becomes

$$
\int_{\text {all space }} x^{2} \Psi^{2} d x
$$

The ground state for a particle in a box is n=1, so the third excited state corresponds to n=4. The particle in a box can only be found between x=0 and x=L, so these are the limits of integration. By plugging these values into Equation 13, we can evaluate the expectation value of the position squared.

$$
\begin{gathered}
\left\langle x^{2}\right\rangle_{\Psi}
=\frac{2}{L} \int_{0}^{L} x^{2} \sin \left(\frac{4 \pi x}{L}\right) ~dx \\ \\
=\frac{2}{L}\left[\frac{x^{3}}{6} - \left(\frac{x^{2}}{4\left(\frac{4 \pi}{L}\right)} - \frac{1}{8\left(\frac{4 \pi}{L}\right)^{3}}\right) \sin \left(2 \frac{4 \pi x}{L}\right) -\frac{x}{4\left(\frac{4 \pi}{L}\right)^{2}} \cos \left(2 \frac{4 \pi x}{L}\right)\right]_{0}^{L}
\end{gathered}
$$

$$
\begin{aligned}
=\frac{2}{L}\left[\frac{L^{3}}{6} - \left(\frac{L^{3}}{16 \pi}\right.\right.&\left.\left. - \frac{L^{3}}{512 \pi^{3}}\right) \sin (8 \pi) - \frac{L^{3}}{64 \pi^{2}} \cos (8 \pi)\right] \\ \\
&=2 L^{2}\left[\frac{1}{6}-\frac{1}{64 \pi^{2}}\right] \\ \\
&=2 L^{2}\left[\frac{32 \pi^{2}-3}{192 \pi^{2}}\right] \\ \\
&=\mathrm{L}^{2}\left[\frac{32 \pi^{2}-3}{96 \pi^{2}}\right]
\end{aligned}
$$

 

 

 

Answer (i.e., write the answer to these problems, but no need to show work. You can if you prefer)

 https://phys.libretexts.org/Bookshel...ves_(Exercise)

#6, #14, #29

Problem 6 

Discuss why some bodies appear black, other bodies appear red, and still other bodies appear white.

Solution

According to Kirchhoff’s law, a body having some finite temperature will emit all wavelengths of radiation, but the wavelengths themselves can be of different intensities like red wavelengths being emitted more intensely than the wavelengths. A blackbody radiator will emit a continuous distribution of wavelengths. The emission wavelength of greatest intensity (i.e. the most photons per second) is called \(\lambda_{max}\), and it will determine the color of the blackbody. According to Wien's Displacement Law, the absolute temperature of the blackbody radiator is inversely proportional to the \(\lambda_{max}\) such that higher temperatures will produce distributions centered about shorter wavelengths (higher energy), and lower temperatures will produce distributions centered about longer wavelengths (lower energy). If \(\lambda_{max}\) falls within the visible spectrum, then the blackbody will appear to be the corresponding color. If \(\lambda_{max}\) falls outside of the visible range such as with UV or microwave radiation, then the blackbody will appear black. If there is a distribution of relatively equal large intensities across the visible spectrum, the blackbody will appear white. Therefore, some bodies appear black, other bodies appear red, and still other bodies appear white.

 

 

Problem 14

Which aspects of the photoelectric effect cannot be explained by classical physics?

Solution

Incident radiation on metal will cause electrons to eject from the surface of the metal in a phenomenon known as the photoelectric effect. The ejected electrons can be directed through an electrical circuit such that the measured current is a measure of the rate of ejection. If light were just a plane wave, then the following should hold true:

1. The average kinetic energy of the electrons would need to increase over a certain time interval until enough energy had been transferred to produce a measurable current. 

2. The intensity of the light would be proportional to its energy and increasing the energy of the light should increase the current (i.e. the rate of electron ejection)

3. Increasing the intensity should also increase the average kinetic energy of the electrons.

But instead,

1. There was no time lag between currents produced by light of low intensity and high intensity. Low intensity light could produce a current just as quickly as high intensity light.

2. Regardless of intensity, only specific wavelengths of light could actually produce a current.

3. The average kinetic energy of the electrons was independent of the light intensity altogether.

 

 

Problem 29

Discuss the similarities and differences between Thomson’s model of the hydrogen atom and Bohr’s model of the hydrogen atom.

Solution

Both models of the atom attempt to explain the net neutral composition of the atom by proposing a positively charged core surrounded by negatively charged electrons. In J.J. Thompson’s model, the atom is a singular mass with a diffuse positive charge balanced by electrons embedded into the surface, much like pieces of fruit embedded into a plum pudding. In Bohr’s model, the atom is mostly air with a small, dense nucleus of positive charge at the center. Electrons then orbit around the nucleus in a planetary model.