- #1
TheSodesa
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Homework Statement
An electron is enclosed in a potential well, whose walls are ##V_0 = 8.0eV## high. If the energy of the ground state is ##E = 0.50eV##, approximate the width of the well.
Answer: ##0.72nm##
Homework Equations
For an electron in a potential well, whose energy is less than the height of the potential walls, outside the well the time-independent Schrödinger equation becomes:
\begin{equation}
\Psi ''(x) = \frac{2m}{\hbar^2}(V_0 - E) \Psi (x) = \alpha^2 \Psi (x),
\end{equation}
where
\begin{equation}
\alpha^2 = \frac{2m}{\hbar} (V_0 - E) > 0
\end{equation}
Inside the well, where ##V(x) = 0##, it is the familiar:
\begin{equation}
\Psi ''(x) = -k^2 \Psi (x), k^2 = \frac{2mE}{\hbar^2}
\end{equation}
By doing a whole bunch of math (by requiring that ##\Psi## be continuously differentiable at the potential walls and solving the resulting linear system with the assumption that it has more than one solution, meaning ##det() = 0##), we and up with the result
\begin{equation}
\frac{\sin (kL)}{\cos (kL)} = \tan (kL) = \frac{\alpha}{k}.
\end{equation}
Substituting ##\alpha## from ##(2)## and ##k## from ##(3)##, we get
\begin{equation}
\tan (kL) = \sqrt{\frac{V_0 - E}{E}}
\end{equation}
The Attempt at a Solution
Now my idea was to use ##(5)## to solve for ##L## as follows:
[tex]
\tan (kL) = \sqrt{\frac{V_0 - E}{E}}\\
\iff\\
kL = \arctan (\sqrt{\frac{V_0 - E}{E}})\\
\iff\\
L = k^{-1}\arctan (\sqrt{\frac{V_0 - E}{E}})
= \sqrt{\frac{2mE}{\hbar^2}}^{-1} \cdot \arctan (\sqrt{\frac{V_0 - E}{E}})\\
= \sqrt{\frac{2(9.109 \cdot 10^{-31}kg)(0.5 \cdot 1.6022 \cdot 10^{-19}J)}{(\frac{6.626\cdot 10^{-34}Js}{2\pi})^2}}^{-1} \cdot\\
\arctan (\sqrt{\frac{8eV - 0.5eV}{0.5eV}})\\
= 2.084 751 \cdot 10^{-8}m,
[/tex]
which is a bit off. Any idea what I'm doing wrong?
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