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What is the physical significance of the value of \(\psi^{2}\) at a particular point in an atomic orbital?

Short Answer

Expert verified
The physical significance of the value of \(\psi^2\) at a particular point in an atomic orbital is that it represents the probability density of finding an electron at that location. A higher value of \(\psi^2\) indicates a greater likelihood of the electron being present in that specific region, while a lower value denotes a lower probability. This helps in visualizing electron distribution in atomic orbitals and understanding their shapes and properties.

Step by step solution

01

Understanding the wave function

The wave function, denoted by \(\psi\), is a mathematical representation of the electron's quantum state within an atom. It provides information about the probability of the electron's position, momentum, and other properties.
02

Squaring the wave function

By squaring the wave function (\(\psi^2\)), we obtain the probability density. Probability density is used to determine the likelihood of finding a particle in a given region of space. In other words, the higher the value of \(\psi^2\) at a point, the more probable it is to find an electron in that region.
03

Probability density in atomic orbitals

In atomic orbitals, \(\psi^2\) represents the probability density of finding an electron in a specific location around the nucleus. It is essential to emphasize that \(\psi^2\) is not an exact position but rather an indication of the electron's probability distribution across space.
04

Physical significance of \(\psi^2\) in an atomic orbital

The physical significance of the value of \(\psi^2\) in an atomic orbital relates to the likelihood of finding an electron at that particular point. A higher value of \(\psi^2\) signifies a higher probability that the electron will be found in that location, while a lower value indicates a lower probability. This information allows us to visualize the electron distribution in atomic orbitals and understand their shapes and properties.

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Assume that we are in another universe with different physical laws. Electrons in this universe are described by four quantum numbers with meanings similar to those we use. We will call these quantum numbers \(p, q, r\), and \(s .\) The rules for these quantum numbers are as follows: \(p=1,2,3,4,5, \ldots\) \(q\) takes on positive odd integers and \(q \leq p\) \(r\) takes on all even integer values from \(-q\) to \(+q\). (Zero is considered an even number.) \(s=+\frac{1}{2}\) or \(-\frac{1}{2}\) a. Sketch what the first four periods of the periodic table will look like in this universe. b. Wh?t are the atomic numbers of the first four elements you would expect to be least reactive? c. Give an example, using elements in the forst four rows, of ionic compounds with the formulas \(\mathrm{XY}, \mathrm{XY}_{2}, \mathrm{X}_{2} \mathrm{Y}, \mathrm{XY}_{3}\), and \(\mathrm{X}_{2} \mathrm{Y}_{3}\) d. How many electrons can have \(p=4, q=3 ?\) e. How many electrons can have \(p=3, q=0, r=0 ?\) f. How many electrons can have \(p=6\) ?

Which of the following sets of quantum numbers are not allowed? For each incorrect set, state why it is incorrect. a. \(n=3, \ell=3, m_{\ell}=0, m_{s}=-\frac{1}{2}\) b. \(n=4, \ell=3, m_{\ell}=2, m_{s}=-\frac{1}{2}\) c. \(n=4, \ell=1, m_{\ell}=1, m_{s}=+\frac{1}{2}\) d. \(n=2, \ell=1, m_{\ell}=-1, m_{s}=-1\) e. \(n=5, \ell=-4, m_{\ell}=2, m_{s}=+\frac{1}{2}\) f. \(n=3, \ell=1, m_{\ell}=2, m_{s}=-\frac{1}{2}\)

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