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In a multi-electron atom, inner electrons partially screen the nuclear charge so that outerelectrons feel a weakened Coulomb force. One can define an effective nuclear charge seenby outer electrons by using the hydrogenic energy equation E = (−13.6 eV)(Zeff /n)2 forthe outer electron’s energy. When the outer electron’s energy is measured by chemists, it iscalled the ionization energy which is the amount of energy required to remove the electronfrom the atom.

a) Lithium (Z = 3) has a single 2s electron outside the full K shell core. The ionization energyis 5.37 eV. Determine Zeff . Notice how effective the two core electrons are at screening thenuclear charge.

b) Singly ionized lithium (Li+) has only the two 1s electrons having lost the 2s electron.

The ionization energy is 76.3 eV. Determine Zeff . Notice that the other 1s electron is notas effective at screening the nuclear charge.

c) Sodium (Z = 11) like lithium has a single electron outside a full core, [Ne]3s1. Theionization energy is 5.12 eV. Determine Zeff . Notice that the core electrons are not aseffective at screening the nuclear charge in sodium as in lithium.

d) Copper (Z = 29) has the electronic configuration [Ar]3d104s1. Removing the 4s electronrequires an ionization energy of 7.69 eV. Determine Zeff . Notice that the 3d electrons arenot nearly as effective at screening the nuclear charge.

e) Gallium (Z = 31) has the electronic configuration [Ar]3d104s24p1. Removing the 4pelectron requires an ionization energy of 5.97 eV. Determine Zeff . Notice that the screeningis better than for copper.

a) Zeff = n Eion/(13.6 eV) = 2 (5.37 eV)(13.6 eV) = 1.26. Notice how the two inner shellelectrons have quite effectively screened the +3 nuclear charge.

b) Zeff = 1 (76.3 eV)(13.6 eV) = 2.37. The other 1s electron provides some screening ofthe +3 nuclear charge but it is not as effective being in the same orbital.

c) Zeff = 3 (5.12 eV)(13.6 eV) = 1.84. The K and L shell electrons screen most of the +11nuclear charge.

d) Zeff = 4 (7.69 eV)(13.6 eV) =3.01. The 3d electrons are not as effective at screening thenuclear charge from the 4s electron because the mean radius for the 3d orbital is not thatmuch less than that of the 4s orbital. And so the 4s electron has a fair probability of beingcloser to the nucleus where the effective nuclear charge is larger.

e) Zeff = 4 (5.97 eV)(13.6 eV) = 2.65. The 4s electrons provide somewhat better screeningfor the 4p electron.

A fluorine atom has the electronic configuration [He]2s22p5. The F atom can capture an
additional electron into the 2p state to become a F− ion; the energy released by capturingthe electron is the electron affinity which for fluorine is 3.4 eV.

a) Calculate the effective nuclear charge seen by this extra electron. The two 1s electrons inthe closed K shell (very close to the nucleus) and the two 2s electrons will effectively shieldfour positive charges. Of the remaining nuclear charge of +5e, how much is screened by the5 2p electrons for the sixth electron?
b) The F atom does not enjoy losing an electron. The ionization energy of the F atom is17.4 eV. What is the Zeff that is experienced by a 2p electron?
a) For the captured electron, Zeff = n Eaff/(13.6 eV) = 2 (3.4 eV)(13.6 eV) = 1.00. Theother five 2p electrons effectively screen +4 of the nuclear charge.

b) For ionization, the lost 2p electron experienced an effective nuclear charge of Zeff =n Eion/(13.6 eV) = 2 (17.4 eV)(13.6 eV) = 2.26.

For each of the following atoms, sketch the electronic structure, using a box for an orbitalwavefunction and an arrow (indicating up or down spin) for an electron as is done in thetext.

The carbon atom has the electronic structure 2s22p2 in its ground state. The ground state
and some excited states of C are shown below. The excited states have the energies 7.3 eV,4.1 eV, 7.9 eV, and 1.2 eV above the ground state. Using reasonable arguments, determinewhich configuration has each energy. Hint: use Hunds rule to determine the ground state.

Flipping a spin costs some energy. Moving an electron from an s to a higher p sub-orbitalor from a p to a higher s sub-orbital costs a lot of energy. Two electrons in the same box(obviously with opposite spin) has substantial Coulombic repulsion energy.

Diagram a obeys Hund’s rule and has the electrons in the lowest possible states consistent
with the Pauli exclusion principle. Clearly this diagram represents the ground state.

Diagram b differs from the ground state by moving the two 2p electrons into the same
box, thus violating Hund’s rule. The change costs some energy but not as much as movingan electron to a higher sub-orbital. This state is 1.2 eV above the ground state.

In Diagram d, a 2s electron has been promoted to the 2p state. This change requires
considerable energy but not as much as promoting an electron to the next shell. This stateis 4.1 eV above the ground state.

Diagram c is like Diagram d except that Hund’s rule is violated as well which requires
additional energy. This state is 7.3 eV above the ground state.

In Diagram e, a 2p electron has been promoted to the M shell which requires a lot of
energy. This state is 7.9 eV above the ground state.

A HeNe laser operating at 632.8 nm has a tube that is 40 cm long. The tube temperatureis 130◦C.

a) Calculate the mode number n that satisfies the resonant cavity condition nearest thecenter wavelength.

b) Calculate the frequency separation and the wavelength separation of the laser modes.

How do these change as the tube warms up during operation taking the linear expansioncoefficient of the tube to be 10−6K−1?
a) The resonance condition is 2L = nλ. So the mode number is n = 2L/λ = 1, 264, 223 ± 1.

b) The frequency separation between modes is ∆ν = c/2L = 3.75 × 108 Hz. The wavelengthseparation is ∆λ = λ2/2L = 5.0 × 10−13 m = 5.0 × 10−4 nm. As the tube warms from roomtemperature (20◦C) to 130◦C, it will expand by ∆L = αL(T − T0) = 4.4 × 10−3 cm. ∆ν willchange by (c/2L)(∆L/L) ≈ 41 KHz.

Source: http://kasap13.usask.ca/ee372/solutions/ee372_solutions5.pdf

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