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Chapter 5: Q14P (page 218)

The ground state of dysprosium (element 66, in the 6th row of the Periodic Table)

is listed as Is5. What are the total spin, total orbital, and grand total angular

momentum quantum numbers? Suggest a likely electron configuration for

dysprosium.

Short Answer

Expert verified

The total spin, orbital spin and total angular momentum are:

S=2;L=6;J=8.

Electronic configuration:

(1s)2(2s)2(2p)6(3s)2(3p)6(3d)10(4s)2(4p)6铃�definite(36electrons)(4d)10(5s)2(5p)6(4f)10(6s)2铃�likely(30electrons)

Step by step solution

01

Definition of total spin, total orbital and total angular momentum quantum number

Spin quantum number, the fourth quantum number introduced to explain the rotation and direction of the electron spin in space. By combining a particle's orbital and intrinsic angular momentum, the total angular momentum quantum number in quantum mechanics parameterizes the total angular momentum of a particular particle.

02

Determining the total spin, total orbital and grand total angular momentum quantum numbers

Ground state of dysprosium is 5l8. That means 2s + 1 = 5 and J = 8. So total spin , orbital spin and total angular momentum are:

S = 2, L = 6, J = 8.

Electronic configuration :

(1s)2(2s)2(2p)6(3s)2(3p)6(3d)10(4s)2(4p)6铃�definite (36 electrons)(4d)10(5s)2(5p)6(4f)10(6s)2铃�likely (30 electrons)

30 electrons which fill all orbits from4d10 to6s2 are likely there.

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Most popular questions from this chapter

Imagine two noninteracting particles, each of mass m, in the infinite square well. If one is in the state蠄n(Equation 2.28 ), and the other in state 蠄1(l鈮�n), calculate localid="1658214464999" (x1-x2)2, assuming (a) they are distinguishable particles, (b) they are identical bosons, and (c) they are identical fermions.

(a) If 蠄aand蠄b are orthogonal, and both normalized, what is the constant A in Equation 5.10?

(b) Ifrole="math" localid="1658225858808" 蠄a=蠄b (and it is normalized), what is A ? (This case, of course, occurs only for bosons.)

Discuss (qualitatively) the energy level scheme for helium if (a) electrons were identical bosons, and (b) if electrons were distinguishable particles (but with the same mass and charge). Pretend these 鈥渆lectrons鈥� still have spin 1/2, so the spin configurations are the singlet and the triplet.

Suppose you had three particles, one in state蠄a(x), one in state蠄b(x), and one in state蠄c(x). Assuming 蠄a,蠄b, and蠄c are orthonormal, construct the three-particle states (analogous to Equations 5.15,5.16, and 5.17) representing

(a) distinguishable particles,

(b) identical bosons, and

(c) identical fermions.

Keep in mind that (b) must be completely symmetric, under interchange of any pair of particles, and (c) must be completely antisymmetric, in the same sense. Comment: There's a cute trick for constructing completely antisymmetric wave functions: Form the Slater determinant, whose first row is蠄a(x1),蠄b(x1),蠄c(x1) , etc., whese second row is蠄a(x2),蠄b(x2),蠄c(x2) , etc., and so on (this device works for any number of particles).

Find the energy at the bottom of the first allowed band, for the case尾=10 , correct to three significant digits. For the sake of argument, assume 伪a=1eV.
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