AP - Advanced Placement Atomic Structure and Periodicity 1 — Questions and Answers
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)
- 2.80 × 10⁻¹⁹ J (Correct answer)
- 5.30 × 10⁻¹⁹ J
- 7.80 × 10⁻¹⁹ J
- 1.26 × 10⁻¹⁸ J
Correct answer: 2.80 × 10⁻¹⁹ J
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)?
- [Ar] 3d⁴ 4s²
- [Ar] 3d⁵ 4s¹ (Correct answer)
- [Ar] 3d⁶ 4s⁰
- [Ar] 3d³ 4s² 4p¹
Correct answer: [Ar] 3d⁵ 4s¹
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)
- 5.27 × 10⁻²⁵ kg·m/s (Correct answer)
- 1.05 × 10⁻²⁴ kg·m/s
- 2.11 × 10⁻²⁴ kg·m/s
- 6.63 × 10⁻²⁴ kg·m/s
Correct answer: 5.27 × 10⁻²⁵ kg·m/s
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)?
- 6.10 (Correct answer)
- 7.00
- 5.75
- 12.85
Correct answer: 6.10
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?
- Fluorine (F)
- Chlorine (Cl) (Correct answer)
- Bromine (Br)
- Oxygen (O)
Correct answer: Chlorine (Cl)
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.
Question 6: What is the de Broglie wavelength of an electron traveling at 2.00 × 10⁶ m/s? (mₑ = 9.109 × 10⁻³¹ kg, h = 6.626 × 10⁻³⁴ J·s)
- 3.64 × 10⁻¹⁰ m (Correct answer)
- 7.27 × 10⁻¹⁰ m
- 1.82 × 10⁻¹⁰ m
- 3.64 × 10⁻⁷ m
Correct answer: 3.64 × 10⁻¹⁰ m
Using λ = h/(mv): λ = 6.626 × 10⁻³⁴ / (9.109 × 10⁻³¹ × 2.00 × 10⁶) = 6.626 × 10⁻³⁴ / 1.822 × 10⁻²⁴ ≈ 3.64 × 10⁻¹⁰ m (0.364 nm). This wavelength is on the order of atomic dimensions, confirming that wave behavior is significant for electrons.
AP - Advanced Placement Atomic Structure and Periodicity 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)