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A wave function with a noninfinite wavelength- however approximate it might be - has nonzero momentum and thus non/ero kinetic energy. Even a single "bump" has kmetic energy. In either case. we can say that the function has kinetic energy because it has curvature - a second derivative. Indeed. the kinetic energy operator in any coondinute system involves a second derivative. The only function without kinefic energy would be a strajght line. As a special case, lhis includes a constant. which may be thought of as a function with an infinite wavelength. By looking at the cunature in the appropriate dimension(s). answer the following: For a given \(n\), is the kinetic energy solely (a) radial in the stinte of lowest \(\ell\) - that is, \(\ell=0\); and (b) rotational in the state of highest \(\ell\) - that is, \(\ell=n-1 ?\)

Short Answer

Expert verified
Yes, in simple terms, the kinetic energy is radial for 'l=0' state and rotational for 'l=n-1' state.

Step by step solution

01

Concept of Atomic Orbitals

Atomic orbitals are described by wave functions, with each wave function characterized by a unique set of quantum numbers 'n' and 'l'. The kinetic energy of the wave function is related to its curvature or second derivative, and this curvature can vary based on the quantum numbers. Will start by understanding 'l=0' state.
02

Understanding the 'l=0' State

For 'l=0', the orbital shape is spherical, and it is in the state of the lowest angular momentum. It is known as the s-orbital, and its wave function is symmetrical about the nucleus with no nodal surfaces. Hence, the kinetic energy is, indeed, radial in this state.
03

Understanding 'l=n-1' State

Now, for 'l=n-1', the orbital is in the state of highest angular momentum. This means that it has a complex shape with multiple nodal surfaces. This shape is more suggestive of a rotational motion rather than a radial motion. Hence, the kinetic energy is, indeed, mostly rotational in this state.

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Key Concepts

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

Wave Function
A wave function in quantum mechanics is a mathematical description of the quantum state of a particle or system. In simpler terms, think of it as a recipe that allows us to calculate the probability of finding a particle at any specific location. It's essential in determining how particles like electrons move and interact with other particles.

The wave function is often represented by the symbol \(\psi\) and it's crucial because it carries all the information needed to describe a particle's behavior. Furthermore, the wave function's shape and form can tell us a lot about the particle's properties, such as its energy. As mentioned in the exercise, kinetic energy in quantum mechanics is associated with the movement of the particle, which is revealed through the curvature or the second derivative of the wave function.
Quantum Numbers
Quantum numbers are like the address of an electron within an atom, specifying its location and behavior. There are several types of quantum numbers, but the most important are the principal quantum number (\(n\)), which indicates the energy level of an electron; the azimuthal quantum number (\(l\)), which determines the shape of the atomic orbital; and the magnetic quantum number, which gives the orientation of that orbital.

These numbers aren't just random figures; they are governed by strict rules and principles. For example, the quantum numbers for any electron are restricted by rules, such as \(l\) must be less than \(n\), and this segregation determines the distribution of electrons within an atom, affecting properties like bonding and reactivity.
Atomic Orbitals
Atomic orbitals are regions around an atom where there is a high probability of finding an electron. These orbitals are the 3D pictures of the wave functions, and each has a characteristic shape that can be spherical, dumbbell-shaped, and so on. The shape of an orbital is intricately linked with its quantum numbers, particularly the azimuthal quantum number (\(l\)).

For instance, when \(l=0\), we get what's known as an s-orbital, which is spherical. This shape means that the electron is likely to be found at equal probabilities at any point around the nucleus. The kinetic energy in this state is due to the movement radially away from or towards the nucleus.
Second Derivative
In mathematics, the second derivative is often seen as a measure of the curvature of a function. In quantum mechanics, this concept translates to the measure of the curvature of the wave function. The more curved the wave function is at a point, the higher the kinetic energy of the particle at that point.

The connection between kinetic energy and the second derivative is more than just metaphorical; it's encoded in the Schrödinger equation. Kinetic energy is actually represented by a second derivative term multiplied by a constant. This term indicates that particles with more curvature in their wave functions have higher kinetic energy, and therefore are moving faster or more vigorously.
Angular Momentum
Angular momentum in quantum mechanics is the rotational equivalent of linear momentum and is connected to the azimuthal quantum number (\(l\)). Just like an ice skater pulls in her arms to spin faster, electrons with higher angular momentum move in patterns that are less spherical and more complex.

In the context of the exercise, when discussing the 'state of the highest \(\ell\)', we are talking about the maximum amount of angular momentum for a given energy level, which results in a more complex, less symmetrical orbital. These are often p, d, or f orbitals, which display a rotational characteristic in their kinetic energy, indicating how the electron 'spins' or moves in more complex patterns rather than just 'in and out' of the nucleus like in s orbitals.

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Most popular questions from this chapter

Consider an electron in the ground state of a hydrogen atom. (a) Calculate the expectation value of its potential energy. (b) What is the expectation value of its kinetic energy? (Hint: What is the expectation value of the total energy?

Consider an electron in the ground state of a hydrogen atom. (a) Sketch plots of \(E\) and \(U(r)\) on the same axes (b) Show that. classically, an electron with this energy should not be able to get farther than \(2 a_{0}\) from the proton. (c) What is the probability of the electron being found in the classically forbidden region?

A hydrogen atom electron is in a \(2 p\) state. If no experiment have been done to establish a \(z\) -component of angular momentum, the atom is equally likely to be found with any allowed value of \(L_{i}\). Show that if the probability densities for these different possible states are added (with equal weighting). the result is independent of both \(\phi\) and \(\theta\).

Show that a transition where \(\mathrm{Am},=\pm 1\) conesponds to a dipole moment in the \(x y\) -plane. while \(1 m_{t}=0\) corresponds to a momentalong the z-avis. (You need consider only the \(\phi\) -parts of \(\psi_{1}\) and \(\psi_{\rho}\) which are of the form \(e^{-m}\).)

Classically, an orbiting charged particle radiates elecIromagnetic energy, and for an electron in atomic dimensions, if would lead to collupse in considerably less than the wink of an eye. (a) By equating the centripetal and Coulomb forces, show that for a classical charge \(-e\) of mass \(m\) held in circular orbit by its attraction to a fixed charge \(+e\), the following relationship holds: \(\omega=e r^{-1 / 2} / \sqrt{4 \pi \varepsilon_{0} m}\). (b) Electromagnetism tells us that a charge whose acceleration is a radiates power \(P=e^{2} a^{2} / 6 e_{f} f^{\prime}\). Shuw that thiv can also be expressed in terms of the orbit radius, as \(P=e^{6} /\left(96 m^{2} \varepsilon_{0}^{3} m c^{3} r^{\lambda}\right)\). Then calculate the energy lowt per orbit in terms of \(r\) by multiplying this power by the period \(T=2 \pi / \omega\) and using the formula from part (a) to eliminate \(\omega(\mathrm{c})\) in such a classical orbit. the total mechanical energy is half the potential energy (see Section 4.4), or \(E_{\text {orbu }}=-e^{2 / 8} \pi \varepsilon_{0} r .\) Calculate the change in energy per change in \(r, d E_{\text {obi }} l d r\). From this and the energy lost per orbit from part (b). determine the change in \(r\) per orbit and evaluate it for a typical orbit radius of \(10^{-10} \mathrm{~m}\). Would the electron's radius change much in a single orbit? (d) Argue that dividing \(d E_{\text {arbiu }} / d r\) by \(P\) and multiplying by dr gives the time required for \(r\) to change by \(d r\). Then. sum these times for all radii from \(r_{\text {initial }}\) to a final radius of \(0 .\) Evaluate your result for \(r_{\text {hiniul }}=10^{-10} \mathrm{~m}\). (One limitation of this extimate is that the electron would eventually be moving relativistically.)

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