For hydrogen atoms, the wave function for the state \(n=3\) \(\ell=0, m_{\ell}=0\) is $$ \psi_{300}=\frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27-18 \sigma+2 \sigma^{2}\right) e^{-\sigma \beta} $$ where \(\sigma=r / a_{0}\) and \(a_{0}\) is the Bohr radius \(\left(5.29 \times 10^{-11} \mathrm{m}\right)\) Calculate the position of the nodes for this wave function.

Short Answer

Expert verified
The positions of the nodes for the wave function are: \(r_1 \approx \frac{18 + \sqrt{108}}{4} \cdot 5.29 \times 10^{-11} \,\text{m}\) \(r_2 \approx \frac{18 - \sqrt{108}}{4} \cdot 5.29 \times 10^{-11} \,\text{m}\)

Step by step solution

01

Identify the relevant parts of the wave function

The wave function for the given hydrogen atom state is: \[\psi_{300} = \frac{1}{81 \sqrt{3 \pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2}\left(27 - 18 \sigma + 2 \sigma^{2}\right) e^{-\sigma \beta}\] where \(\sigma = \frac{r}{a_{0}}\). Our goal is to find the positions of the nodes corresponding to \(r\), so we focus on solving the equation: \[\psi_{300}(r) = 0\] It's important to note that the radial part of the wave function is multiplied by the exponential term, so we know that the radial part must equal zero to have a node: \(27 - 18\sigma + 2\sigma^2 = 0\)
02

Solve the quadratic equation for \(\sigma\)

The equation to solve is quadratic in \(\sigma\): \(2\sigma^2 - 18\sigma + 27 = 0\) We use the quadratic formula: \(\sigma = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) where \(a = 2\), \(b = -18\), and \(c = 27\). Plugging in the values: \(\sigma = \frac{18 \pm \sqrt{(-18)^2 - 4(2)(27)}}{2(2)}\)
03

Calculate \(\sigma\) and the corresponding \(r\)

Now, calculate the values of \(\sigma\): \(\sigma_1 = \frac{18 + \sqrt{324 - 216}}{4} = \frac{18 + \sqrt{108}}{4}\) \(\sigma_2 = \frac{18 - \sqrt{324 - 216}}{4} = \frac{18 - \sqrt{108}}{4}\) Notice that \(\sigma = \frac{r}{a_0}\), so to find the positions of the nodes with respect to \(r\), we need to multiply \(\sigma\) by \(a_0\): \(r_1 = \frac{18 + \sqrt{108}}{4} \cdot a_0 = \frac{18 + \sqrt{108}}{4} \cdot 5.29 \times 10^{-11} \,\text{m}\) \(r_2 = \frac{18 - \sqrt{108}}{4} \cdot a_0 = \frac{18 - \sqrt{108}}{4} \cdot 5.29 \times 10^{-11} \,\text{m}\) So, the positions of the nodes in meters are: \(r_1 \approx \frac{18 + \sqrt{108}}{4} \cdot 5.29 \times 10^{-11} \,\text{m}\) \(r_2 \approx \frac{18 - \sqrt{108}}{4} \cdot 5.29 \times 10^{-11} \,\text{m}\)

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Key Concepts

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

Quantum Mechanics
Quantum mechanics is a fundamental theory in physics that describes the physical properties of nature at the scale of atoms and subatomic particles. It introduces a completely different framework from classical mechanics, where energy, momentum, angular momentum, and other quantities of a bound system are not continuous, but quantized.

In the microscopic world of quantum mechanics, particles exhibit wave-like behavior, leading to the concept of the wave function, which describes the quantum state of a particle. The wave function provides all information about the probability of where a particle such as an electron can be found. Understanding the wave function is crucial as it helps predict where you might detect particles like electrons in an atom.
Wave Function Nodes
Wave function nodes are specific points where the probability density of finding an electron is zero. In other words, a node is where the wave function \(\psi\) is equal to zero. The presence of nodes in an atomic orbital can tell us a lot about the electron's energy level and its spatial distribution within the atom.

For the hydrogen atom specified in the exercise, the wave function nodes are calculated by finding the roots of the radial part of the wave function, ignoring the exponential decay factor. Identifying these nodes helps us visualize the areas within the atom where there is zero probability of finding an electron. This concept is essential in quantum chemistry and plays a significant role in understanding chemical bonding and the electron configuration of atoms.
Bohr Radius
The Bohr radius, denoted as \(a_0\), is a physical constant that represents the most probable distance between the nucleus and the electron in a hydrogen atom in its ground state. It has a value of approximately \(5.29 \times 10^{-11} \text{m}\).

In the context of the wave function for the hydrogen atom, the Bohr radius is used to normalize the radial distance \(r\) in terms of \(\sigma = r / a_0\). This non-dimensionalization helps in simplifying the mathematical treatment of the problem. Moreover, the Bohr radius is pivotal in the Bohr model of the atom, which, while less accurate than quantum mechanical descriptions, provides a stepping stone towards understanding atomic structure.
Quantum Numbers
Quantum numbers are values that describe the energy levels of electrons in an atom. These numbers can be thought of as the 'address' of an electron, indicating its position and movement in three-dimensional space around the nucleus. There are four quantum numbers: the principal quantum number (\(n\)), angular momentum quantum number (\(l\)), magnetic quantum number (\(m_l\)), and spin quantum number (\(m_s\)).

For this exercise, the wave function \(\psi_{300}\) is affected by the quantum numbers \(n=3\), \(l=0\), and \(m_l=0\), indicating an electron in the third energy level with no angular momentum. These quantum numbers are significant in determining the shape, size, and orientation of the electron's orbital, and in defining the probability distribution of finding an electron at a given position around the nucleus.

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Most popular questions from this chapter

Are the following statements true for the hydrogen atom only, true for all atoms, or not true for any atoms? a. The principal quantum number completely determines the energy of a given electron. b. The angular momentum quantum number, \(\ell,\) determines the shapes of the atomic orbitals. c. The magnetic quantum number, \(m_{\ell},\) determines the direction that the atomic orbitals point in space.

One bit of evidence that the quantum mechanical model is "correct" lies in the magnetic properties of matter. Atoms with unpaired electrons are attracted by magnetic fields and thus are said to exhibit paramagnetism. The degree to which this effect is observed is directly related to the number of unpaired electrons present in the atom. Consider the ground-state electron configurations for Li, N, Ni, Te, Ba, and Hg. Which of these atoms would be expected to be paramagnetic, and how many unpaired electrons are present in each paramagnetic atom?

An excited hydrogen atom with an electron in the \(n=5\) state emits light having a frequency of \(6.90 \times 10^{14} \mathrm{s}^{-1}\). Determine the principal quantum level for the final state in this electronic transition.

Rank the elements Be, B, C, N, and O in order of increasing first ionization energy. Explain your reasoning.

Carbon absorbs energy at a wavelength of \(150 . \mathrm{nm.}\) The total amount of energy emitted by a carbon sample is \(1.98 \times 10^{5} \mathrm{J}\) Calculate the number of carbon atoms present in the sample, assuming that each atom emits one photon.

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