# Electron in a potential well (numerical problem)

• abstracted6
In summary, the width of the potential well of an electron is 2 angstrom and the energy of the electron can be calculated using the equation \nabla^2\psi+\frac{2m}{\hbar^2}(E-V)\psi=0 for various values of n. However, there seems to be a problem with the units of measure, as the calculated energy is 2.34675(10^20)eV instead of 9.39eV, as stated in the book. It is recommended to work with h bar in joules and convert to eV at the end of the calculation. Also, memorizing common combinations of constants such as \hbar c and mc^2 can simplify
abstracted6

## Homework Statement

The width of the potential well of an electron can be assumed to be 2angstrom. Calculate the energy of an electron (in joules and eV) from this information for various values of n.

I believe I have messed up my units somewhere in the final calculation, but I have included everything before that anyways.

For eV where n=1 the book says the answer is 9.39eV, however I'm getting 2.34675(10^20).

## Homework Equations

$$\nabla^2\psi+\frac{2m}{\hbar^2}(E-V)\psi=0$$

## The Attempt at a Solution

Started with
$$\nabla^2\psi+\frac{2m}{\hbar^2}(E-V)\psi=0$$ Only one direction, and no potential:
$$\frac{d^2\psi}{dx^2}+\frac{2m}{\hbar^2}E\psi=0$$
Letting $$\alpha=\sqrt{\frac{2m}{\hbar^2}E}$$
$$\therefore\psi(x)=Ae^{i\alpha x}+Be^{-i\alpha x}$$
Boundary Conditions:
$$\psi(0)=0=A+B\rightarrow A=-B$$
$$\psi(L)=0=Ae^{i\alpha L}+Be^{-i\alpha L}=A(e^{i\alpha L}-e^{-i\alpha L})$$From the identity:$$sin(\theta)=\frac{1}{2i}(e^{i\theta}-e^{-i\theta})$$
$$\therefore 0=A2isin(\alpha L)\rightarrow \alpha L=n\pi\rightarrow \alpha=\frac{n\pi}{L}$$
Putting alphas together
$$\frac{n\pi}{L}=\sqrt{\frac{2m}{\hbar^2}E}$$
$$\therefore E(L)=\frac{n^2\pi^2\hbar^2}{2mL^2}$$$$n=1,2,3...$$

These are the constants I am using for the final calculation:$$n=1$$$$\hbar=6.582(10^{-16})eVs$$$$m=9.11(10^{-31})Kg$$$$L=2(10^{-10})m$$

I get 2.34675(10^20)eV when it should be 9.39eV.

You have a problem with the units of measure. Thinking only of units what you have is

$\left[\frac{n^2\pi^2\hbar^2}{2mL^2}\right]=\frac{(eV)^2s^2}{kg\cdot m^2}=\frac{eV^2}{J}$

which is obvioulsy not in eV. You should work with h bar in joules and, at the end of your calculation, go to eV.

abstracted6 said:
These are the constants I am using for the final calculation:$$n=1$$$$\hbar=6.582(10^{-16})eVs$$$$m=9.11(10^{-31})Kg$$$$L=2(10^{-10})m$$
You can make your life a lot easier if you memorize some common combinations of the constants:
\begin{align*}
\hbar c &= 197\text{ eV nm} \\
mc^2 &= 511000\text{ eV}
\end{align*} If you multiply both the numerator and denominator of the expression for the energy by ##c^2##, you get
$$E = \frac{n^2\pi^2 (\hbar c)^2}{2 (mc^2) L^2}.$$ If you specify ##L## in nm, you can see that the units of ##E## will be eV.

## 1. What is an electron in a potential well?

An electron in a potential well refers to a hypothetical scenario in which an electron is confined to a specific region of space, typically by a potential energy barrier or "well". This confinement can lead to interesting behaviors and properties of the electron, which are often studied in physics and chemistry research.

## 2. How is the energy of an electron in a potential well calculated?

The energy of an electron in a potential well can be calculated using the Schrödinger equation, which is a fundamental equation in quantum mechanics. This equation takes into account the shape of the potential well, the mass of the electron, and other factors to determine the allowed energy levels for the electron.

## 3. What is the significance of the "quantum numbers" associated with an electron in a potential well?

The quantum numbers associated with an electron in a potential well are used to describe the properties and behavior of the electron within the well. These numbers include the energy level, the angular momentum, and the magnetic moment of the electron, among others. By changing these numbers, we can observe how the electron's behavior and energy level change within the potential well.

## 4. What is the relationship between the potential well depth and the energy of the electron?

The potential well depth is directly related to the energy of the electron. A deeper potential well corresponds to a lower energy for the electron, while a shallower potential well corresponds to a higher energy. This relationship is a fundamental aspect of the quantum mechanical behavior of electrons in confined spaces.

## 5. How is the wave function of an electron in a potential well represented?

The wave function of an electron in a potential well is typically represented graphically as a probability distribution, which shows the likelihood of finding the electron at different positions within the well. This representation allows us to visualize the behavior of the electron and understand its properties and interactions with the potential well.

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