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Chapter 8: Problem 78

In its ground state, carbon's \(2 p\) electrons interact to pro. duce \(j_{T}=0 .\) Given Hund's rule. what does this say about the total orbital angular momentum of these electrons?

Short Answer

Expert verified
The total orbital angular momentum of carbon's 2p electrons in their ground state, according to Hund's rule, is 0.

Step by step solution

01

Understanding Hund's Rule

According to Hund's Rule, for any set of orbitals (in this case, the 2p orbitals of carbon), electrons will fill each of these orbitals singly (with their spin pointing in the same direction) before they start pairing up in the same orbital. The total spin angular momentum (\(j_{T}\)) is the sum of the individual spins of the electrons.
02

Applying Hund's Rule to Carbon

The 2p orbitals of a carbon atom in its ground state contain two electrons in each of the three orbitals, for a total of six electrons. Given Hund's Rule, each of these electrons will fill the orbitals singly before pairing up, meaning their spins will be aligned. When two electrons occupy the same orbital, their spin angular momenta cancel out, resulting in a total spin angular momentum of 0.
03

Determining Total Orbital Angular Momentum

Given that the total spin angular momentum (\(j_{T}\)) is 0, this implies that the total orbital angular momentum must also be 0, as these two quantities are related by the vector sum of the individual orbital and spin angular momenta of all electrons within the atom. Thus, the total orbital angular momentum of carbon's 2p electrons in their ground state must be 0.

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Key Concepts

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

Orbital Angular Momentum
Orbital angular momentum is a fundamental concept in quantum mechanics that helps us understand how electrons move within an atom. It arises due to the electron's motion around the nucleus and is quantized, meaning it can only take on certain discrete values. The orbital angular momentum of an electron in an atom is given by the formula:
  • \[ L = ext{n(n-1)} \ ext{where} \ L ext{ is the orbital angular momentum, and n is the principal quantum number.}\]
For carbon's 2p electrons, the principal quantum number \(n = 2\) and the azimuthal quantum number \(l = 1\), which means the electrons have angular momentum quantized as \(l(l+1)\). In this particular state, the total orbital angular momentum is represented by the sum of the angular momenta vectors of the individual electrons. In carbon's ground state, however, all the individual electron angular momenta cancel each other out due to the symmetrical filling of the orbitals, leading to a total orbital angular momentum of 0.
This cancellation aligns with Hund's rule and contributes to our understanding of atomic stability and electron interactions.
Spin Angular Momentum
Spin angular momentum refers to the intrinsic form of angular momentum carried by particles like electrons. Each electron has a spin, which is an intrinsic property represented by a quantum number, \(s = \frac{1}{2}\), and its spin angular momentum is given by:
  • \[ S = \sqrt{s(s+1)} \hbar \ ext{where} \ \hbar ext{ is the reduced Planck's constant.}\]
This property of electrons is quantized and can have one of two possible orientations: "up" or "down." These correspond to the spin quantum numbers \(m_s = +\frac{1}{2}\) or \(m_s = -\frac{1}{2}\).
When applying Hund's rule, carbon's 2p electrons initially fill each of the three 2p orbitals singly with parallel spins to maximize the total spin angular momentum. In the ground state, each electron pairs up with another having opposite spin within the same orbital, resulting in their net spin angular momentum canceling out, producing a total spin angular momentum of 0.
Carbon 2p Electrons
Carbon is an element with atomic number 6, and its electron configuration includes electrons in the 2p orbitals. These orbitals have a characteristic "dumbbell" shape and can hold up to six electrons distributed across three orbitals.
In its ground state, carbon's 2p orbital configuration is explained by Hund's rule, which dictates that electrons preferentially occupy separate orbitals with aligned spins before pairing. Specifically, carbon's 2p orbitals will have two singly occupied orbitals and one empty orbital when considering just the two electrons in the 2p orbitals. These electrons maintain parallel spins to maximize spin angular momentum before any pairing occurs.
Finally, when the 2p electrons fully pair, they occupy the orbitals in a way that any spin or orbital angular momentum present cancels out completely, resulting in the characteristic zero net angular momentum commonly observed in the ground state of many atoms, including carbon.

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Most popular questions from this chapter

Show that unless \(\ell=s . L\) and \(S\) cannot be exactly opposite: that is, show that at its minimum possible value, for which \(j=\ell-s\), the magnitude \(J\) of the total angular momenturn is strictly greater than the difference \(L-S \mid\) berween the magnitudes of the orbital and intrinsic angular momentum vectors.

What is the angle between the spins in a triplet state?

Exercise 44 gives an antisymmetric multiparticle state for two particles in a box with opposite spins. Another antisymmetric state with spins opposite and the same quantum numbers is $$ \psi_{n}\left(x_{1}\right) \downarrow \psi_{n}\left(x_{2}\right) \uparrow-\psi_{n}\left(x_{1}\right) \uparrow \psi_{n}\left(x_{2}\right) \downarrow $$ Refer to these states as \(\mid\) and 11 . We have tended to characterize exchange symmetry as to whether the state's sign changes when we swap particle labels. but we could achieve the same result by instead swapping the particles' stares, specifically the \(n\) and \(n^{\prime}\) in equation \((8-22)\). In this exercise. we look at swapping only parts of the state-spatial or spin. (a) What is the exchange symmeiry - symmetric (unchanged). antisymmetric (switching sign), or neither-of multiparticle states 1 and \(\mathrm{II}\) with respect to swapping spatial states alone? (b) Answer the same question. but with respect to swapping spin states/arrows alone. (c) Show that the algebraic sum of states 1 and \(\mathrm{II}\) may be written \(\left(\psi_{n}\left(x_{1}\right) \psi_{n}\left(x_{2}\right)-\psi_{n}\left(x_{1}\right) \psi_{n}\left(x_{2}\right)\right)(\downarrow T+\uparrow \downarrow)\) where the left arrow in any couple represents the spin of particle 1 and the right arrow that of particle 2 (d) Answer the same questions as in parts \((a)\) and (b). but for this algebraic sum. (e) Is the sum of states I and 11 still antis ymmetric if we swap the particles? total-spatial plus spin -states? (f) If the two particles repel each other, would any of the three multiparticle states - l. II, and the sum - be preferred? Explain.

The Zeeman effect occurs in sodium just as in hydrogen - sodium's lone 3 s valence electron behaves much as hydrogen's 1.5. Suppose sodium atoms are immersed in a \(0.1 \mathrm{~T}\) magnetic field. (a) Into how many levels is the \(3 p_{1 / 2}\) level split? (b) Determine the energy spacing between these states. (c) Into how many lines is the \(3 p_{1 / 2}\) to \(3 s_{1 / 2}\) spectral line split by the field? (d) Describe quantitatively the spacing of these lines. (e) The sodium doublet \((589.0 \mathrm{nm}\) and \(589.6 \mathrm{nm}\) ) is two spectral lines. \(3 p_{3 n} \rightarrow 3 s_{1 / 2}\) and \(3 p_{1 / 2} \rightarrow 3 s_{1 / 2}\) which are split according to the two different possible spin-orbit ener gies in the \(3 p\) state (see Exercise 60 ). Detemine the splitting of the sodium doublet (the energy diff erence between the two photons). How does it compare with the line splitting of part (d), and why?

The \(K_{a}\) line in copper is a very common one to use in \(X\) -ray ciystallography. To produce it, electrons are accelerated through a potential difference and smashed into a copper target. Section 7.8 gives the energies in a hydrogenlike atom as \(Z^{2}\left(-13.6 \mathrm{eV} / \mathrm{n}^{2}\right)\). Making the reasonable approximation that an \(n=1\) electron in copper orbils the nucleus and half of its fellow \(n=1\) electron. being unaff ected by the roughly spherical cloud of other electrons around it, estimate the minimum accelerating potential needed to make a hole in copper's \(K\) shell.
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