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Chapter 20: Problem 4

Calculate the expectation value for the potential energy of the \(\mathrm{H}\) atom with the electron in the 1 s orbital. Compare your result with the total energy.

Short Answer

Expert verified
The expectation value of the potential energy for the hydrogen atom's electron in the 1s orbital can be calculated as: 鉄ㄏ堚倎鈧鈧|V|蠄鈧佲個鈧鉄 = -k * e^2 / (2 * a鈧) Comparing this to the total energy of the hydrogen atom in the 1s state, E = -k * e^2 / (2 * a鈧), we find that they are equal. Therefore, the expectation value of the potential energy of the electron in the 1s orbital of the hydrogen atom is equal to the total energy of the atom in this state.

Step by step solution

01

Write down the 1 s wavefunction

For the hydrogen atom, the wavefunction of the 1s orbital is given by: 蠄鈧佲個鈧(r) = (1 / 鈭毾) * (1 / a鈧)^(3/2) * e^(-r / a鈧) where a鈧 is the Bohr radius and r is the distance between the nucleus and the electron.
02

Write down the potential energy function

The potential energy function V(r) is given by the Coulomb's potential: V(r) = -k * e^2 / r
03

Calculate the expectation value of the potential energy

To find the expectation value of the potential energy, we integrate over all space using spherical coordinates (r, 胃, 蠁): 鉄ㄏ堚倎鈧鈧|V|蠄鈧佲個鈧鉄 = 鈭個^鈭炩埆鈧^蟺鈭個^2蟺 [蠄鈧佲個鈧*(r)V(r)蠄鈧佲個鈧(r)] r^2 * sin(胃) dr d胃 d蠁 Simplify the expression by plugging in the 1s wavefunction and the potential energy function: 鉄ㄏ堚倎鈧鈧|V|蠄鈧佲個鈧鉄 = (-k * e^2) * (1 / 蟺 * a鈧捖) * 鈭個^鈭 鈭個^蟺 鈭個^2蟺 r * e^(-2r / a鈧) * r^2 * sin(胃) dr d胃 d蠁 We can solve this triple integral by dividing it into three single integrals: 鉄ㄏ堚倎鈧鈧|V|蠄鈧佲個鈧鉄 = (-k * e^2) * (1 / 蟺 * a鈧捖) * (鈭個^2蟺 d蠁) * (鈭個^蟺 sin(胃) d胃) * (鈭個^鈭 r^3 * e^(-2r / a鈧) dr) These integrals can be solved individually as follows: 鈭個^2蟺 d蠁 = 2蟺 鈭個^蟺 sin(胃) d胃 = 2 鈭個^鈭 r^3 * e^(-2r / a鈧) dr = a鈧抆4 / 8 (using integration by parts or looking up the integral of r^n * e^(-ar) dr) Now multiply all terms together to get the expectation value: 鉄ㄏ堚倎鈧鈧|V|蠄鈧佲個鈧鉄 = (-k * e^2) * (1 / 蟺 * a鈧捖) * (2蟺) * (2) * (a鈧抆4 / 8) 鉄ㄏ堚倎鈧鈧|V|蠄鈧佲個鈧鉄 = -k * e^2 / (2 * a鈧)
04

Compare the expectation value of potential energy with the total energy

The total energy of the hydrogen atom in the 1 s state is: E = -k * e^2 / (2 * a鈧) Comparing this to the expectation value of the potential energy, we see that they are equal: 鉄ㄏ堚倎鈧鈧|V|蠄鈧佲個鈧鉄 = E Thus, the expectation value of the potential energy of the electron in the 1s orbital of the hydrogen atom is equal to the total energy of the atom in this state.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Hydrogen Atom
The hydrogen atom is the most basic and fundamental atom, consisting of only one proton and one electron. This simplicity makes it an important model in quantum mechanics. The behavior of the hydrogen atom can be explained using the Schr枚dinger equation, which provides insight into the probabilities of finding the electron in a particular region around the nucleus.
The solution to the Schr枚dinger equation for a hydrogen atom reveals that electrons can occupy certain stable orbits, called orbitals. Each orbital corresponds to a specific energy level. These orbits are named as 's', 'p', 'd', and 'f' orbitals.
The hydrogen atom is central to the study of quantum mechanics because it can be exactly solved, unlike more complex atoms. This makes the hydrogen atom useful for understanding general principles about atomic structure and spectra.
Potential Energy
Potential energy in the context of a hydrogen atom refers to the energy stored due to the electron's position relative to the nucleus. This is calculated using Coulomb's potential, which arises due to the electrostatic interaction between the negatively charged electron and the positively charged proton.
The potential energy function, expressed by the formula \( V(r) = -\frac{k \cdot e^2}{r} \), describes how the potential energy varies with the distance \( r \) from the nucleus.
  • \( k \) is the Coulomb's constant.
  • \( e \) is the elementary charge.
As \( r \) increases, the potential energy becomes less negative, indicating that more energy is "stored" in the system when the electron and nucleus are farther apart.
For the hydrogen atom, this potential energy helps dictate the electron's behavior, affecting both its speed and the nature of its orbits.
1s Orbital
The 1s orbital represents the simplest atomic orbital, found at the lowest energy level of a hydrogen atom. It is spherical in shape and is the most probable area where one finds the electron when it is in its ground state.
To mathematically describe the 1s orbital, we use the wave function \( \psi_{100}(r) = \frac{1}{\sqrt{\pi}} \left( \frac{1}{a_0} \right)^{3/2} e^{-r/a_0} \). This function illustrates how the probability density of finding an electron changes with distance \( r \) from the nucleus.
  • \( a_0 \) is the Bohr radius, a fundamental distance in atomic physics.
The 1s orbital is crucial when calculating the expectation values of various properties like energy, since it allows us to incorporate quantum mechanical principles in these computations.
Understanding the 1s orbital provides a clear insight into more complicated behaviors of electrons in other elements, setting the stage for more advanced quantum chemistry.

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Most popular questions from this chapter

Is the total energy wave function \\[ \psi_{310}(r, \theta, \phi)=\frac{1}{81}\left(\frac{2}{\pi}\right)^{1 / 2}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(6 \frac{r}{a_{0}}-\frac{r^{2}}{a_{0}^{2}}\right) e^{-r / 3 a_{0}} \cos \theta \\] an eigenfunction of any other operators? If so, which ones? What are the eigenvalues?

In spherical coordinates, \(z=r \cos \theta\). Calculate \(\langle z\rangle\) and \(\left\langle z^{2}\right\rangle\) for the \(H\) atom in its ground state. Without doing the calculation, what would you expect for \(\langle x\rangle\) and \(\langle y\rangle,\) and \(\left\langle x^{2}\right\rangle\) and \(\left(y^{2}\right\rangle ?\) Why?

Show that \(\psi_{2 p_{x}}(r, \theta, \phi)\) and \(\psi_{2 p_{y}}(r, \theta, \phi)\) can be written in the form \(N x e^{-r / 2 a_{0}}\) and \(N^{\prime} y e^{-r / 2 a_{0}}\) where \(N\) and \(N^{\prime}\) are normalization constants.

Show that the total energy eigenfunctions \(\psi_{100}(r)\) and \(\psi_{200}(r)\) are orthogonal.

Show by substitution that \(\psi_{100}(r, \theta, \phi)=\) \(1 / \sqrt{\pi}\left(1 / a_{0}\right)^{3 / 2} e^{-r / a_{0}}\) is a solution of \\[ \begin{array}{r} -\frac{\hbar^{2}}{2 m_{e}}\left[\begin{array}{c} \frac{1}{r^{2}} \frac{\partial}{\partial r}\left(r^{2} \frac{\partial \psi(r, \theta, \phi)}{\partial r}\right)+\frac{1}{r^{2} \sin \theta} \frac{\partial}{\partial \theta}\left(\sin \theta \frac{\partial \phi(r, \theta, \phi)}{\partial \theta}\right) \\ +\frac{1}{r^{2} \sin ^{2} \theta} \frac{\partial^{2} \psi(r, \theta, \phi)}{\partial \phi^{2}} \end{array}\right. \\ -\frac{e^{2}}{4 \pi \varepsilon_{0} r} \psi(r, \theta, \phi)=E \psi(r, \theta, \phi) \end{array} \\] What is the eigenvalue for the total energy? Use the relation \(a_{0}=\varepsilon_{0} h^{2} /\left(\pi m_{c} e^{2}\right)\) to simplify your answer.
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