Physics 3304 Assignment 8 solutions

Grading: The problems that will be graded in detail for this assignment are Ch. 8, problems 3 and 12 (each worth 5 points total). You will receive 1 point for each of the other problems if you have made a reasonable attempt at a solution. (Total points for this assignment=18)

Ch. 8, Problem 2

a) [Ne]

b) [Ar]

c) [Kr]

d) [Xe]

Ch. 8, Problem 3

For sodium the ionization energy is 5.14 eV and the outermost electron is in an n=3 state. To estimate the effective nuclear charge we can picture this as a single-electron atom with central charge . Then we have:

For potassium the ionization energy is 4.34 eV and the electron is in an n=4 state.

In the Bohr model the orbit of the outermost electron is completely outside the inner electrons, so we would expect the inner electrons (Z-1 of them) to provide complete screening, which means .

Ch. 8, Problem 4

Applying Equation 6.42 to an atom with Z-1 nuclear charge, we obtain

using Equation 6.40 for . The frequency of the emitted radiation is

Ch. 8, Problem 5

Solving Equation 8.1 with gives:

Thus Z=26 and the element is iron.

Ch. 8, Problem 10

Cr: [Ar]

There is nothing that prevents all 6 electrons from having , and thus . The Pauli principle forces the five 3d electrons to have different values, which are +2,+1,0,-1,-2; the 4s electron has only. The total L is therefore L=+2+1+0-1-2+0=0. The ground state therefore has L=0 and S=3.

Ch. 8, Problem 11

a) Ce: [Xe]

The 6s subshell is full. The 4f and 5d electron can each have for a total S of 1. The Pauli principle does not prevent the values of these two electrons from having their maximum possible values of +3 (for 4f) and +2 (for 5d). Thus L=2+3=5. The ground state is thus L=5,S=1.

b) Gd: [Xe]

The 6s subshell is full. To maximize S, we give all seven 4f electrons (which is possible, because the 4f shell has a capacity of 14) and we also give to the 5d electron. Thus S=8(1/2)=4. The Pauli principle requires that all seven of the 4f electrons have different values, which are for a total of 0. The only contribution to L comes from the 5d electron, which contributes . The ground state is therefore L=2,S=4.

c) Pt: [Xe]

The 4f subshell is full. The spin of the nine 5d electrons is maximized by giving five of them and the remaining four . The 6s electron also contibutes , for a total of S=1. The 6s electron doesn't contibute to L, and the five 5d electrons with have and give no net contribution to L. We therefore maximize the contribution to L of the remaining four 5d electrons by assigning them of +2,+1,0,-2 for a net total of L=2. The ground state is thus L=2,S=1.

Ch. 8, Problem 12

a) F: [He]

The 2s subshell is full. For the five 2p electrons, the total spin is maximized by having three with and the remaining two with which gives . The Pauli principle requires that the three electrons have for a total of zero. The two electrons maximize L if assigned . Thus for the ground state.

b) Mg: [Ne]

The 2s subshell is full so L=0,S=0 for the ground state.

c) Ti: [Ar]

The 4s subshell is full. Maximizing S for the 3d electrons gives . The 3d electrons must have different , and so L is maximized with . Thus L=3,S=1 for the ground state.

d) Fe: [Ar]

The 4s subshell is full. Maximizing S for the 3d electrons gives . The five electrons with are assigned , and the electron with has . Thus L=2,S=2 for the ground state.

Ch. 8, Problem 13

Each electron has , so the total possible S is 0 or 1. The two d electrons, each with l=2, can couple to give L=0,1,2,3,4.

Ch. 8, Problem 19

a) For a 2p electron n=2,l=1; thus and . So the six possible sets of quantum numbers are:

b) Since there are 6 possible sets for each electron, the total number of possibilities for 2 electrons is .

c) The Pauli principle prevents the two sets from being identical. There will be 6 combinations in which the sets are identical, which must be eliminated, leaving 30 allowed combinations.

d) Since the n values are different, the Pauli principle does not restrict the number of possible cominations, so there will be 36 possible sets.

Ch. 8, Problem 22

a) For ,