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Consider a fictitious one-dimensional system with one electron. The wave function for the electron, drawn below, is \(\psi(x)=\sin x\) from \(x=0\) to \(x=2 \pi .\) (a) Sketch the probability density, \(\psi^{2}(x),\) from \(x=0\) to \(x=2 \pi .(\mathbf{b})\) At what value or values of \(x\) will there be the greatest probability of finding the electron? (c) What is the probability that the electron will be found at \(x=\pi ?\) What is such a point in a wave function called? [Section 6.5\(]\)

Short Answer

Expert verified
(a) The probability density function is \(\psi^2(x) = \sin^2 x\). (b) The greatest probability of finding the electron is at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\). (c) The probability of finding the electron at \(x=\pi\) is 0, and such a point is called a "node."

Step by step solution

01

(a) Sketching the Probability Density Function

To sketch the probability density function \(\psi^2(x)\), simply square the given wave function \(\psi(x)\) and plot it on the range from \(0\) to \(2\pi\). Given \(\psi(x)=\sin x\), the probability density function is: \[\psi^2(x)=(\sin x)^2=\sin^2 x\] Now, plot \(\sin^2 x\) on the range \([0, 2\pi]\).
02

(b) Finding Maximum Probability

To find the values of \(x\) with the greatest probability of finding the electron, we must look for the maximum points in the probability density function \(\psi^2(x)=\sin^2 x\). To find maxima, take the derivative of the function and set it to 0: \[\frac{d(\sin^2 x)}{dx}=0\] Using the chain rule, we get: \[\frac{d(\sin^2 x)}{dx}=2\sin x\cdot \cos x=0\] Solve for \(x\) and make sure to verify that they are maxima: \[2\sin x\cdot \cos x=0 \implies x=\{0, \frac{\pi}{2}, \pi, \frac{3\pi}{2}, 2\pi\}\] After checking for maxima, the greatest probability of finding the electron is at \(x=\frac{\pi}{2}\) and \(x=\frac{3\pi}{2}\).
03

(c) Probability at \(x=\pi\) and Terminology

To find the probability of finding the electron at \(x=\pi\), we need to evaluate the probability density function at this point: \[\psi^2(\pi)=\sin^2 \pi =0\] A point in a wave function where the probability of finding the electron is 0 is called a "node." Thus, at \(x=\pi\), the wave function \(\psi(x)=\sin x\) has a node.

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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, the wave function \(\psi(x)\) is a fundamental concept that describes the quantum state of a particle, like an electron, in a given system. For example, in the exercise provided, the wave function given is \(\psi(x) = \sin x\) over the interval from \(x = 0\) to \(x = 2\pi\). This function can be visualized as a wave, and it can be plotted as a sinusoidal curve within the specified interval.
Understanding the wave function is crucial because it contains all the information about the system. The most important part of the wave function is that it must be normalized. This means the total probability of finding the particle in the entire space should be equal to 1.
Key takeaways about wave functions include:
  • The wave function \(\psi(x)\) is typically a complex-valued function.
  • It is used to calculate probability densities, which help determine where a particle is likely to be found.
  • The wave function's shape and nature can vary depending on the system being studied, whether it's electrons in an atom or particles in a different quantum system.

This essentially provides the groundwork for calculating the probability densities and defining characteristics such as nodes.
Electron Probability
Electron probability is determined by the square of the wave function, \(\psi^2(x)\), also known as the probability density function. This concept helps us understand where an electron is likely to be found when measured. In the given exercise, the wave function \(\psi(x) = \sin x\) is squared to find the probability density: \(\sin^2 x\).
The probability density \(\psi^2(x)\) is always a non-negative function, indicating that probabilities cannot be negative. This squared function essentially indicates the regions where the electron is most likely to be found. For sections of the wave function where \(\psi^2(x)\) is larger, the probability of locating the electron is higher.
For instance, the probability is greatest at the midpoint between the troughs and peaks of \(\sin^2 x\), which occurs at \(x = \frac{\pi}{2}\) and \(x = \frac{3\pi}{2}\) where the function reaches its maximum. These points represent where the electron is statistically most likely to be encountered.
Nodes in Wave Functions
Nodes are specific points within the wave function where the probability density becomes zero. In other words, at a node, there is zero probability of finding an electron. This concept is crucial for the interpretation of many quantum mechanical systems.
In the exercise example with a wave function of \(\psi(x) = \sin x\), nodes occur where \(\sin x = 0\), which means the probability density \(\sin^2 x = 0\). Notably, at \(x = \pi\), \(\sin \pi = 0\) indicating a node.
The existence of nodes is significant as they reflect the energy states of particles within the system. The number of nodes is related to the quantum number of the system and provides insights into energy levels and configurations of an electron within atoms.
Understanding nodes is important because:
  • Nodes indicate where the wave function changes sign.
  • They define regions where electrons cannot be found.
  • The presence of nodes can affect the probability distribution and energy levels within quantum systems.
Nodes, therefore, not only tell us about the electron's position but also the characteristics of the wave functions that describe states in quantum systems.

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Most popular questions from this chapter

The discovery of hafnium, element number \(72,\) provided a controversial episode in chemistry. G. Urbain, a French chemist, claimed in 1911 to have isolated an element number 72 from a sample of rare earth (elements \(58-71\) ) compounds. However, Niels Bohr believed that hafnium was more likely to be found along with zirconium than with the rare earths. D. Coster and G. von Hevesy, working in Bohr's laboratory in Copenhagen, showed in 1922 that element 72 was present in a sample of Norwegian zircon, an ore of zirconium. (The name hafnium comes from the Latin name for Copenhagen, Hafnia). (a) How would you use electron configuration arguments to justify Bohr's prediction? (b) Zirconium, hafnium's neighbor in group \(4 \mathrm{~B}\), can be produced as a metal by reduction of solid \(\mathrm{ZrCl}_{4}\) with molten sodium metal. Write a balanced chemical equation for the reaction. Is this an oxidation- reduction reaction? If yes, what is reduced and what is oxidized? (c) Solid zirconium dioxide, \(\mathrm{ZrO}_{2}\), reacts with chlorine gas in the presence of carbon. The products of the reaction are \(\mathrm{ZrCl}_{4}\) and two gases, \(\mathrm{CO}_{2}\) and \(\mathrm{CO}\) in the ratio \(1: 2 .\) Write a balanced chemical equation for the reaction. Starting with a \(55.4-\mathrm{g}\) sample of \(\mathrm{ZrO}_{2}\), calculate the mass of \(\mathrm{ZrCl}_{4}\) formed, assuming that \(\mathrm{ZrO}_{2}\) is the limiting reagent and assuming \(100 \%\) yield. (d) Using their electron configurations, account for the fact that \(\mathrm{Zr}\) and Hf form chlorides \(\mathrm{MCl}_{4}\) and oxides \(\mathrm{MO}_{2}\).

(a) The average distance from the nucleus of a 3 s electron in a chlorine atom is smaller than that for a \(3 p\) electron. In light of this fact, which orbital is higher in energy? (b) Would you expect it to require more or less energy to remove a 3 s electron from the chlorine atom, as compared with a \(2 p\) electron?

Using Heisenberg's uncertainty principle, calculate the uncertainty in the position of (a) a 1.50 -mg mosquito moving at a speed of \(1.40 \mathrm{~m} / \mathrm{s}\) if the speed is known to within \(\pm 0.01 \mathrm{~m} / \mathrm{s} ;\) (b) a proton moving at a speed of \((5.00 \pm 0.01) \times 10^{4} \mathrm{~m} / \mathrm{s}\) (The mass of a proton is given in the table of fundamental constants in the inside cover of the text.)

Molybdenum metal must absorb radiation with an energy higher than \(7.22 \times 10^{-19} \mathrm{~J}\) ( "energy threshold") before it can eject an electron from its surface via the photoelectric effect. (a) What is the frequency threshold for emission of electrons? (b) What wavelength of radiation will provide a photon of this energy? (c) If molybdenum is irradiated with light of wavelength of \(240 \mathrm{nm}\), what is the maximum possible velocity of the emitted electrons?

Identify the specific element that corresponds to each of the following electron configurations and indicate the number of unpaired electrons for each: (a) \(1 s^{2} 2 s^{2}\), (b) \(1 s^{2} 2 s^{2} 2 p^{4}\) (d) \([\mathrm{Kr}] 5 s^{2} 4 d^{10} 5 p^{4}\) (c) \([\mathrm{Ar}] 4 s^{1} 3 d^{5}\)

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