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**1. 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##

**2. 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}

**3. 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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