# The width of a finite potential well

#### TheSodesa

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:
$$\tan (kL) = \sqrt{\frac{V_0 - E}{E}}\\ \iff\\ kL = \arctan (\sqrt{\frac{V_0 - E}{E}})\\ \iff\\ L = k^{-1}\arctan (\sqrt{\frac{V_0 - E}{E}}) = \sqrt{\frac{2mE}{\hbar^2}}^{-1} \cdot \arctan (\sqrt{\frac{V_0 - E}{E}})\\ = \sqrt{\frac{2(9.109 \cdot 10^{-31}kg)(0.5 \cdot 1.6022 \cdot 10^{-19}J)}{(\frac{6.626\cdot 10^{-34}Js}{2\pi})^2}}^{-1} \cdot\\ \arctan (\sqrt{\frac{8eV - 0.5eV}{0.5eV}})\\ = 2.084 751 \cdot 10^{-8}m,$$
which is a bit off. Any idea what I'm doing wrong?

Last edited:
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#### TSny

Homework Helper
Gold Member
I grabbed a QM text off the shelf to check your equation (5). It appears to me that this equation is for a well that extends from x = -L to x = +L. So, L in (5) would be half the width of the well.

Also, make sure your calculator is in the radian mode for taking the arctan.

• TheSodesa

#### TheSodesa

I grabbed a QM text off the shelf to check your equation (5). It appears to me that this equation is for a well that extends from x = -L to x = +L. So, L in (5) would be half the width of the well.

Also, make sure your calculator is in the radian mode for taking the arctan.
Yeah, it was the radian thing, and the fact that the well goes from -L to L. The online material for our book (Tipler, Modern Physics, itself glosses over this part, the mathematics in particular) actually uses the letter $a$ in place of $L$, but I figured I could just change the place of the origin without doing anything to make up for it, so to speak.

Thanks again.

Last edited:

#### Simon Bridge

Science Advisor
Homework Helper
Apparently there are even and odd solutions. The online material for our book states, that the $-\cot(ka)$ actually corresponds to the odd case.
Yes?

#### TheSodesa

I as just noting that the link (the part of the page) you provided used $-\cot$ instead of $\tan$. At the time I didn't notice I could scroll up on that page, where the even solution could be found as well.

#### Simon Bridge

Science Advisor
Homework Helper
I as just noting that the link (the part of the page) you provided used $-\cot$ instead of $\tan$. At the time I didn't notice I could scroll up on that page, where the even solution could be found as well.
Yes? Where are you up to in your solution?

#### TheSodesa

Yes? Where are you up to in your solution?
Gah, sorry. I just fell of the map for a week, didn't I? I basically forgot to use radians in my calculation and also forgot, that the formula applies to a potential well from $-a$ to $a$, not $0$ to $L$.

Should I have posted my final numerical answer in this thread? I guess that would have been in good housekeeping...

The answer I got was $\approx 0.73nm$.

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