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The wave function for the 2\(p_{z}\) orbital in the hydrogen atom is $$\psi_{2 p_{z}}=\frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\sigma / 2} \cos \theta$$ where \(a_{0}\) is the value for the radius of the first Bohr orbit in meters \(\left(5.29 \times 10^{-11}\right), \sigma\) is \(Z\left(r / a_{0}\right), r\) is the value for the distance from the nucleus in meters, and \(\theta\) is an angle. Calculate the value of \(\psi_{2 p_{z}}^{2}\) at \(r=a_{0}\) for \(\theta=0^{\circ}\left(z \text { axis ) and for } \theta=90^{\circ}\right.\) (xy plane).

Short Answer

Expert verified
For \(\theta = 0^\circ\), the square of the wave function is \(\psi_{2 p_{z}}^2 = \frac{1}{32\pi}\mathrm{e}^{-1}\). For \(\theta = 90^\circ\), the square of the wave function is \(\psi_{2 p_{z}}^2 = 0\).

Step by step solution

01

Plug in the values for \(r\) and \(a_0\) into the wave function formula

We are given the wave function for the 2\(p_{z}\) orbital as: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} \sigma \mathrm{e}^{-\sigma / 2} \cos \theta \] It is also given that \(r = a_0\), and since \(\sigma = Z\left(\frac{r}{a_0}\right)\), we can plug in these values: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}}\left(\frac{Z}{a_{0}}\right)^{3 / 2} Z \mathrm{e}^{-Z / 2} \cos \theta \] Note that for a hydrogen atom, \(Z = 1\), so the equation becomes: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2} \cos \theta \]
02

Calculate the square of the wave function for \(\theta = 0^\circ\)

At \(\theta = 0^\circ\), we have: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2} \cos 0^\circ \] Since \(\cos 0^\circ = 1\), the wave function is: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2} \] Now, we can find the square of the wave function: \[ \psi_{2 p_{z}}^2 = \left(\frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2}\right)^2 = \frac{1}{32\pi}\mathrm{e}^{-1} \]
03

Calculate the square of the wave function for \(\theta = 90^\circ\)

At \(\theta = 90^\circ\), we have: \[ \psi_{2 p_{z}} = \frac{1}{4 \sqrt{2 \pi}} \mathrm{e}^{-1 / 2} \cos 90^\circ \] Since \(\cos 90^\circ = 0\), the wave function is: \[ \psi_{2 p_{z}} = 0 \] Thus, the square of the wave function is: \[ \psi_{2 p_{z}}^2 = (0)^2 = 0 \]
04

Results

For \(\theta = 0^\circ\), the square of the wave function is \(\psi_{2 p_{z}}^2 = \frac{1}{32\pi}\mathrm{e}^{-1}\). For \(\theta = 90^\circ\), the square of the wave function is \(\psi_{2 p_{z}}^2 = 0\).

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Key Concepts

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

Wave Function
In quantum mechanics, a wave function is fundamental in describing the quantum state of a particle or a system. It provides complete information about the probability of finding a particle in different regions of space. The wave function is generally denoted as \( \psi \) and can have both real and imaginary components.
A wave function's absolute square, \( |\psi|^2 \), gives the probability density, indicating where the particle is likely to be found.
In this case, we explore the wave function associated with the 2\(p_{z}\) orbital of a hydrogen atom. The wave function describes an electron's behavior in this orbital and is significantly influential in predicting the electron's behavior in different chemical and physical processes.
When you calculate the square of the wave function, it becomes especially useful in visualizing electron density. This calculation can aid in understanding where an electron is most likely to be located within an atom.
Hydrogen Atom
The hydrogen atom is the simplest atom consisting of one proton in the nucleus and a single electron orbiting it. It serves as an ideal model for studying atomic structure due to its simplicity.
For hydrogen, the atomic number \( Z \) is 1, which influences the wave function of its electrons. The Bohr model and quantum mechanics provide insights into atomic interactions and electron distributions.
Hydrogen's electron energy levels are quantized, leading to discrete energy levels. These levels are depicted using quantum numbers. For instance, the principal quantum number \( n \) indicates the energy level, while \( l \) represents the orbital shape.
In our study, \( n = 2 \) and \( l = 1 \) correspond to the 2\(p_{z}\) orbital. Understanding this simple system provides a foundation for exploring more complex elements in quantum mechanics.
2p Orbital
In quantum mechanics, orbitals represent regions in an atom where electrons are likely to be found. The 2p orbitals are part of the second energy level of an atom and include three different types: 2p\(_x\), 2p\(_y\), and 2p\(_z\).
Each of these orbitals has a specific orientation in space. In this case, the 2p\(_{z}\) orbital refers to a dumbbell-shaped region oriented along the z-axis. This configuration affects how the electron cloud is spatially distributed around the nucleus.
The shape of the orbital is defined by angular wave functions, which include parameters such as \( \theta \) and \( \phi \). These angles affect the probability distribution of the electron's position.
  • When \( \theta = 0° \), the wave function has a maximum value.
  • When \( \theta = 90° \), the wave function is zero - indicating the node or region where the probability of finding an electron is zero.
Understanding these orbitals is critical for predicting chemical bonding and reactions.
Bohr Model
The Bohr model of the atom was one of the first early models to introduce quantum ideas to explain atomic structure. Proposed by Niels Bohr in 1913, it described electrons orbiting the nucleus at fixed distances, much like planets around the sun.
  • In Bohr's model, electrons occupy specific orbits with quantized energy levels, having defined radii and energies.
  • These levels correspond to the principal quantum numbers \( n \).
  • For hydrogen, the first orbit's radius, \( a_0 \), is known as the Bohr radius, approximately \( 5.29 \times 10^{-11} \) meters.
The model successfully explained the spectral lines of hydrogen but struggled with multi-electron systems. Despite its limitations, the Bohr model laid the groundwork for quantum mechanics and helps in visualizing atomic structure.

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Most popular questions from this chapter

In the ground state of element 115, Uup, a. how many electrons have \(n=5\) as one of their quantum numbers? b. how many electrons have \(\ell=3\) as one of their quantum numbers? c. how many electrons have \(m_{\ell}=1\) as one of their quantum numbers? d. how many electrons have \(m_{s}=-\frac{1}{2}\) as one of their quantum numbers?

An ion having a \(4+\) charge and a mass of 49.9 u has 2 electrons with principal quantum number \(n=1,8\) electrons with \(n=2\) and 10 electrons with \(n=3 .\) Supply as many of the properties for the ion as possible from the information given. (Hint: In forming ions for this species, the 4\(s\) electrons are lost before the 3\(d\) electrons.) a. the atomic number b. total number of \(s\) electrons c. total number of \(p\) electrons d. total number of \(d\) electrons e. the number of neutrons in the nucleus f. the ground-state electron configuration of the neutral atom

For each of the following pairs of elements $$(\mathrm{C} \text { and } \mathrm{N}) \quad(\text { Ar and Br })$$ pick the atom with a. more favorable (exothermic) electron affinity. b. higher ionization energy. c. larger size.

Which of the following sets of quantum numbers are not allowed in the hydrogen atom? For the sets of quantum numbers that are incorrect, state what is wrong in each set. a. \(n=3, \ell=2, m_{\ell}=2\) b. \(n=4, \ell=3, m_{\ell}=4\) c. \(n=0, \ell=0, m_{\ell}=0\) d. \(n=2, \ell=-1, m_{\ell}=1\)

In the second row of the periodic table, Be, N, and Ne all have endothermic (unfavorable) electron affinities, whereas the other second-row elements have exothermic (favorable) electron affinities. Rationalize why Be, N, and Ne have unfavorable electron affinities.

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