AP - Advanced Placement Atomic Structure and Periodicity 1

Question 1

Light of frequency 8.00 × 10¹⁴ Hz strikes a metal surface that has a work function of 2.50 × 10⁻¹⁹ J. What is the maximum kinetic energy of the ejected photoelectrons? (h = 6.626 × 10⁻³⁴ J·s)

Accepted Answer

2.80 × 10⁻¹⁹ J

Answer

Using the photoelectric effect equation, KE_max = hf − φ = (6.626 × 10⁻³⁴)(8.00 × 10¹⁴) − 2.50 × 10⁻¹⁹ = 5.301 × 10⁻¹⁹ − 2.50 × 10⁻¹⁹ = 2.80 × 10⁻¹⁹ J. The work function represents the minimum energy needed to eject an electron; any excess photon energy becomes kinetic energy.

Question 2

Which of the following represents the correct ground-state electron configuration of chromium (Cr, Z = 24)?

Accepted Answer

[Ar] 3d⁵ 4s¹

Answer

Chromium adopts [Ar] 3d⁵ 4s¹ rather than the expected [Ar] 3d⁴ 4s² because a half-filled 3d subshell (five electrons, one per orbital) provides extra stability through exchange energy. One electron is therefore promoted from 4s to 3d.

Question 3

An electron's position is known with an uncertainty of 1.00 × 10⁻¹⁰ m. What is the minimum uncertainty in its momentum? (h = 6.626 × 10⁻³⁴ J·s)

Accepted Answer

5.27 × 10⁻²⁵ kg·m/s

Answer

The Heisenberg uncertainty principle gives Δp ≥ h/(4πΔx) = 6.626 × 10⁻³⁴ / (4π × 1.00 × 10⁻¹⁰) = 6.626 × 10⁻³⁴ / 1.2566 × 10⁻⁹ ≈ 5.27 × 10⁻²⁵ kg·m/s. The smaller the position uncertainty, the larger the minimum momentum uncertainty must be.

Question 4

Using Slater's rules, what is the effective nuclear charge (Z_eff) experienced by a 3p electron in chlorine (Cl, Z = 17)?

Accepted Answer

6.10

Answer

For a 3p electron in Cl (config 1s² 2s² 2p⁶ 3s² 3p⁵): the 6 other electrons in the 3s/3p group each contribute 0.35 (6 × 0.35 = 2.10); the 8 electrons in the n = 2 shell each contribute 0.85 (8 × 0.85 = 6.80); the 2 electrons in the n = 1 shell each contribute 1.00 (2 × 1.00 = 2.00). Total shielding σ = 10.90, so Z_eff = 17 − 10.90 = 6.10.

Question 5

Which of the following elements has the most negative (largest magnitude) electron affinity?

Accepted Answer

Chlorine (Cl)

Answer

Although F is the most electronegative element, Cl has a more negative electron affinity (−349 kJ/mol vs. F's −328 kJ/mol). Fluorine's compact 2p orbitals cause significant electron–electron repulsion when a second electron is added, reducing the energy released. Cl's larger, less congested 3p orbitals accommodate the incoming electron more readily.

AP Practice Test

AP Practice Test

AP - Advanced Placement Atomic Structure and Periodicity 1