Extra Credit 16

Last updated
Save as PDF

Page ID: 195741

\( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

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)

#53, #66, #83

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)

#7, #15, #30

Extra Credit Solutions

#53) An electron confined to a box of width 0.15 nm by infinite potential barriers emits a photon when it makes a transition from the first excited state to the ground state. Find the wavelength of the emitted photon.

Approach:

We want the wavelength, \( \lambda\), of the emitted photon from the n = 2 to n = 1 transition. We should begin by finding the corresponding energies for each state. The difference in energy between the two states will then give us the energy of the emitted photon. Using an equation that relates energy to wavelength, we can then solve for the wavelength.

Equations to use:

General expression for the eigenvalue of a particle in a box that follows the potential stated in the problem is:

\[E_n = \dfrac{hn^2}{8m_eL^2} \label{eigenvalue} \]

where:

\(n\) is the quantum number corresponding to the \( n^{th} \) energy state.

\(L\) is the width of the box.

\(h\) is Planks constant.

\( m_e\) is mass of the electron.

Expression for difference in energy states:

\[E_{n_{inital}} - E_{n_{final}} = \Delta E \label{energydifference} \]

Expression relating energy and wavelength:

\[E = \dfrac{hc}{\lambda} \label{energy_wavelength} \]

We can equate equation \ref{energydifference} to equation \ref{energy_wavelength} and solve for \( \lambda \) to get:

\[ \lambda = \dfrac{hc}{E_{n_{initial}} - E_{n_{final}}} \label{lambda} \]

Solution:

Lets first solve for the ground state energy, \(n=1\) using equation \ref{eigenvalue}:

\[ E_1 = \dfrac{h(1)^2}{8m_e(1.5*10^{-31} m)^2} \nonumber\]

\[ E_1 = 2.6776300*10^{-18} J \nonumber\]

Next, lets solve for the first excited state energy, \(n=2\):

\[ E_2 = \dfrac{h(2)^2}{8m_e(1.5*10^{-31} m)^2} \nonumber\]

\[ E_2 = 1.071052*10^{-17} J \nonumber\]

We then solve for the difference in energy states using equation \ref{energydifference}:

\[ \Delta E = (1.071052*10^{-17} - 2.6776300*10^{-18}) J = 8.03289*10^{-18} J \nonumber\]

Finally, using equation \ref{lambda} lets solve for, \(\lambda\), the wavelength of the emitted photon:

\[ \lambda = \dfrac{hc}{8.03289*10^{-18} J} = 24.7289 nm \nonumber\]

End of problem #53.

#83) Consider an infinite square well with wall boundaries x = 0 and x = L. Show that the function \(\Psi(x) = A\sin(kx)\) is the solution to the stationary Scrodinger equation for the particle in a box only if \(k = \dfrac{\sqrt{2mE}}{\hbar}\). Explain why this is an acceptable wave function only if k is an integer multiple of \(\dfrac{\pi}{L}\).

Approach:

We want to consider the boundary conditions for the particle in the box:

\[ \Psi(x = 0) = 0 \nonumber\]

and

\[ \Psi(x = L) = 0 \nonumber\]

We want to show that the function \( \Psi(x) = A\sin(kx) \) is a solution for a given value of k:

\[ k = \dfrac{\sqrt{2mE}}{\hbar} \label{kvalue} \]

So lets apply the boundary conditions to the function \( \Psi(x) = A\sin(kx) \) and see what value of, k, makes the equation true.

Solution:

\[ \Psi(x = 0) = A\sin(k*0) = 0 \nonumber\]

\( \sin(0) = 0 \) but that tells us nothing about the value of k so lets try the next boundary condition:

\[ \Psi(x = L) = A\sin(kL) = 0 \nonumber\]

to make this equation true:

\[ kL = n\pi \label{boundk} \]

and we know the value of k from equation \ref{kvalue} so we plug it into the equation above and get:

\[ \dfrac{\sqrt{2mE}}{\hbar}L = n\pi \nonumber\]

simplifying a bit and solving for E, we get:

\[ E = \dfrac{n^2\hbar^2\pi^2}{2mL^2} \nonumber \]

which matches the general expression for the eigenvalues of particle in a box.

If we return to equation \( \ref{boundk}\), and solve for k, we get:

\[ k = \dfrac{n\pi}{L} \]

which tells us \( \Psi(x) = A\sin(kx) \) is an acceptable wave function only if k is an integer multiple, where n is the integer, of \( \dfrac{\pi}{L}. \)

End of Problem #83.