Exercises
1. (i) Calculate the number of electrons which will together weigh one gram. (ii) Calculate the mass and charge of one mole of electrons.
Answer
(i) Mass of one electron = 9.11 × 10⁻²⁸ g; Number = 1/(9.11 × 10⁻²⁸) = 1.098 × 10²⁷ electrons
(ii) Mass = 6.022 × 10²³ × 9.11 × 10⁻²⁸ g = 5.49 × 10⁻⁴ g; Charge = 6.022 × 10²³ × 1.6 × 10⁻¹⁹ C = 9.65 × 10⁴ C (1 Faraday)
2. (i) Total electrons in 1 mol methane (CH₄). (ii)(a) total number and (b) total mass of neutrons in 7 mg of ¹⁴C. (iii)(a) total number and (b) total mass of protons in 34 mg of NH₃.
Answer
(i) CH₄ has 6+4 = 10 electrons; 1 mol CH₄ has 10 × 6.022 × 10²³ = 6.022 × 10²⁴ electrons
(ii) ¹⁴C has 8 neutrons. Moles ¹⁴C = 7×10⁻³/14 = 5×10⁻⁴ mol; (a) neutrons = 8 × 5×10⁻⁴ × 6.022×10²³ = 2.409 × 10²¹; (b) mass = 2.409×10²¹ × 1.675×10⁻²⁷ kg = 4.035 × 10⁻⁶ kg
(iii) NH₃ has 10 protons; Moles = 2×10⁻³ mol; (a) protons = 1.2044 × 10²²; (b) mass = 2.015 × 10⁻⁵ kg. Answer does not change with T and P.
3. Neutrons and protons in: ¹³C₆, ¹⁶O₈, ²⁴Mg₁₂, ⁵⁶Fe₂₆, ⁸⁸Sr₃₈
Answer
¹³C: p=6, n=7; ¹⁶O: p=8, n=8; ²⁴Mg: p=12, n=12; ⁵⁶Fe: p=26, n=30; ⁸⁸Sr: p=38, n=50
4. Write complete symbols: (i) Z=17, A=35 (ii) Z=92, A=233 (iii) Z=4, A=9
Answer
(i) ³⁵₁₇Cl; (ii) ²³³₉₂U; (iii) ⁹₄Be
5. Yellow light λ = 580 nm. Calculate frequency and wavenumber.
Answer
ν = c/λ = 3×10⁸/580×10⁻⁹ = 5.17 × 10¹⁴ Hz; wavenumber = 1/λ = 1.724 × 10⁶ m⁻¹
6. Energy of photons: (i) ν = 3×10¹⁵ Hz (ii) λ = 0.50 Å
Answer
(i) E = hν = 6.626×10⁻³⁴ × 3×10¹⁵ = 1.988 × 10⁻¹⁸ J
(ii) E = hc/λ = (6.626×10⁻³⁴ × 3×10⁸)/(0.50×10⁻¹⁰) = 3.976 × 10⁻¹⁵ J
7. Period = 2.0×10⁻¹⁰ s. Wavelength, frequency, wavenumber.
Answer
ν = 1/T = 5.0 × 10⁹ Hz; λ = c/ν = 0.06 m; wavenumber = 16.67 m⁻¹
8. Photons of λ = 4000 pm providing 1 J. How many?
Answer
E/photon = hc/λ = 4.97×10⁻¹⁷ J; N = 1/4.97×10⁻¹⁷ = 2.01 × 10¹⁶ photons
9. Photon λ = 4×10⁻⁷ m, work function = 2.13 eV: (i) energy in eV (ii) KE (iii) velocity.
Answer
(i) E = 4.97×10⁻¹⁹/1.602×10⁻¹⁹ = 3.10 eV; (ii) KE = 3.10−2.13 = 0.97 eV = 1.555×10⁻¹⁹ J; (iii) v = √(2KE/m) = 5.84 × 10⁵ m s⁻¹
10. λ = 242 nm just ionises Na. IE in kJ mol⁻¹.
Answer
E/photon = 8.21×10⁻¹⁹ J; IE = 8.21×10⁻¹⁹ × 6.022×10²³ = 494.6 kJ mol⁻¹
11. 25 W bulb, λ = 0.57 µm. Rate of quanta emission per second.
Answer
E/photon = 3.488×10⁻¹⁹ J; Rate = 25/3.488×10⁻¹⁹ = 7.17 × 10¹⁹ quanta/s
12. Electrons emitted with zero KE at λ = 6800 Å. Threshold frequency and work function.
Answer
ν₀ = 3×10⁸/6800×10⁻¹⁰ = 4.41 × 10¹⁴ Hz; W₀ = hν₀ = 2.92 × 10⁻¹⁹ J
13. Wavelength emitted for H atom n=4 → n=2 transition.
Answer
1/λ = 1.097×10⁷(1/4−1/16) = 2.057×10⁶ m⁻¹; λ = 486 nm (Balmer, blue-green)
14. Energy to ionise H from n=5. Compare with n=1.
Answer
IE from n=5 = 2.18×10⁻¹⁸/25 = 8.72 × 10⁻²⁰ J; IE from n=1 = 2.18×10⁻¹⁸ J; ratio = 25.
15. Maximum emission lines for H excited at n=6 → ground state.
Answer
Lines = n(n−1)/2 = 6×5/2 = 15
16. (i) Energy of 5th orbit (E₁ = −2.18×10⁻¹⁸ J). (ii) Radius of 5th orbit.
Answer
(i) E₅ = −2.18×10⁻¹⁸/25 = −8.72 × 10⁻²⁰ J; (ii) r₅ = 25×52.9 pm = 1322.5 pm
17. Wavenumber for longest wavelength Balmer transition.
Answer
n=2
⇒ 3: ν̄ = 1.097×10⁷×(1/4−1/9) = 1.523 × 10⁶ m⁻¹
18. Energy to shift H electron from n=1 to n=5. Wavelength on return.
Answer
ΔE = 2.09×10⁻¹⁸ J; λ = hc/ΔE = 95 nm (UV)
19. Energy to remove electron from n=2. Longest wavelength for transition.
Answer
E = 5.45×10⁻¹⁹ J; λ = hc/E = 3.65×10⁻⁷ m = 3.65 × 10⁻⁵ cm
20. Wavelength of electron, v = 2.05×10⁷ m s⁻¹.
Answer
λ = h/(mv) = 6.626×10⁻³⁴/(9.11×10⁻³¹×2.05×10⁷) = 3.55 × 10⁻¹¹ m = 355 pm
21. Electron KE = 3.0×10⁻²⁵ J. Wavelength.
Answer
v = √(2KE/m) = 812 m/s; λ = h/(mv) = 8.96 × 10⁻⁷ m = 896 nm
22. Isoelectronic species from Na⁺, K⁺, Mg²⁺, Ca²⁺, S²⁻, Ar.
Answer
Na⁺ and Mg²⁺ have 10 electrons each; K⁺, Ca²⁺, S²⁻ and Ar have 18 electrons each.
23. (i) Electronic configs of H⁻, Na⁺, O²⁻, F⁻. (ii) Z for 3s¹, 2p³, 3p⁵. (iii) Atoms for [He]2s¹, [Ne]3s²3p³, [Ar]4s²3d¹.
Answer
(i) H⁻: 1s²; Na⁺: 1s²2s²2p⁶; O²⁻: 1s²2s²2p⁶; F⁻: 1s²2s²2p⁶
(ii) Na (Z=11); N (Z=7); Cl (Z=17)
(iii) Li; P; Sc
24. Lowest n for g orbitals.
Answer
g has l=4; n ≥ l+1 = 5
25. Electron in 3d: values of n, l, ml.
Answer
n=3, l=2, ml = −2, −1, 0, +1, +2
26. Atom with 29e and 35n: (i) protons (ii) electronic configuration.
Answer
(i) Z = 29 (Cu); (ii) [Ar]3d¹⁰4s¹ (anomalous due to stability of filled 3d)
27. Electrons in H₂⁺ and H₂.
Answer
H₂⁺: 1; H₂: 2
28. (i) l and ml for n=3. (ii) QN for 3d. (iii) Possible: 1p, 2s, 2p, 3f?
Answer
(i) l=0,1,2; ml ranges accordingly. (ii) n=3, l=2, ml=−2 to +2. (iii) 2s and 2p are possible; 1p and 3f are not possible.
29. Describe orbitals: (a) n=1,l=0 (b) n=3,l=1 (c) n=4,l=2 (d) n=4,l=3
Answer
(a) 1s; (b) 3p; (c) 4d; (d) 4f
30. Which quantum number sets are not possible?
Answer
(a) Not possible: n=0 is not allowed. (b) Possible. (c) Not possible: l=1 for n=1 is not allowed. (d) Possible. (e) Not possible: l=3 for n=3 is not allowed (l must be < n). (f) Possible.
31. Electrons with (a) n=4, ms=−½ (b) n=3, l=0.
Answer
(a) n=4 has 32 electrons; half = 16. (b) 3s orbital holds 2 electrons.
32. Show Bohr orbit circumference = integral multiple of de Broglie wavelength.
Answer
From mvr = nh/2π and λ = h/(mv): 2πr = nλ. The circumference equals n de Broglie wavelengths.
33. H transition with same λ as Balmer n=4→n=2 of He⁺.
Answer
He⁺: 1/λ = 4RH(1/4−1/16) = 3RH/4. For H: same gives n₁=1, n₂=2
⇒ Lyman series
34. Energy for He⁺(g) → He²⁺(g) + e⁻.
Answer
E = Z²×IEH = 4×2.18×10⁻¹⁸ = 8.72 × 10⁻¹⁸ J atom⁻¹
35. Diameter C atom = 0.15 nm; atoms in 20 cm line.
Answer
N = 0.20/(0.15×10⁻⁹) = 1.33 × 10⁹ atoms
36. 2×10⁸ C atoms in 2.4 cm. Radius.
Answer
Diameter = 1.2×10⁻⁸ cm = 1.2×10⁻¹⁰ m; radius = 60 pm
37. Zn atom diameter = 2.6 Å: (a) radius in pm (b) atoms in 1.6 cm.
Answer
(a) 130 pm; (b) 1.6×10⁻²/(2.6×10⁻¹⁰) = 6.15 × 10⁷ atoms
38. Charge 2.5×10⁻¹⁶ C. Electrons.
Answer
N = 2.5×10⁻¹⁶/1.6×10⁻¹⁹ = 1563 electrons
39. Oil drop charge −1.282×10⁻¹⁸ C. Electrons.
Answer
N = 1.282×10⁻¹⁸/1.6×10⁻¹⁹ = 8 electrons
40. Using light atoms (Al) in Rutherford's experiment instead of gold?
Answer
Fewer large-angle deflections and backscattering — the lighter nucleus has smaller positive charge, weaker electrostatic repulsion for α-particles, so most particles pass through with less deflection.
41. Why ³⁵₇₉Br and ⁷⁹Br acceptable, ⁷⁹₃₅Br and ³⁵Br not?
Answer
Correct notation places mass number A as superscript and atomic number Z as subscript on the left of the symbol. ⁷⁹₃₅Br reverses A and Z; ³⁵Br uses Z as superscript — both violate standard notation.
42. A=81, 31.7% more neutrons than protons. Atomic symbol.
Answer
p + 1.317p = 81
⇒ p=35; Z=35 is Br
⇒ ⁸¹₃₅Br
43. Ion A=37, charge −1, 11.1% more neutrons than electrons. Symbol.
Answer
Electrons = Z+1; Z=17 (Cl); A=37
⇒ ³⁷₁₇Cl⁻
44. Ion A=56, charge 3+, 30.4% more neutrons than electrons. Symbol.
Answer
Electrons = Z−3; Z=26 (Fe) → ⁵⁶₂₆Fe³⁺
45. Increasing frequency: (a) microwave (b) amber light (c) FM radio (d) cosmic rays (e) X-rays.
Answer
FM radio < microwave < amber light < X-rays < cosmic rays
46. N₂ laser λ=337.1 nm, 5.6×10²⁴ photons. Power.
Answer
E/photon = 5.9×10⁻¹⁹ J; Total E = 5.9×10⁻¹⁹ × 5.6×10²⁴ = 3.3 × 10⁶ J
47. Neon, λ=616 nm: (a) frequency (b) distance in 30 s (c) energy per quantum (d) quanta for 2 J.
Answer
(a) 4.87 × 10¹⁴ Hz; (b) 9×10⁹ m; (c) 3.23 × 10⁻¹⁹ J; (d) 6.19 × 10¹⁸
48. Detector receives 3.15×10⁻¹⁸ J from λ=600 nm. Number of photons.
Answer
E/photon = 3.31×10⁻¹⁹ J; N = 3.15×10⁻¹⁸/3.31×10⁻¹⁹ ≈ 10 photons
49. Radiation source: 2 ns duration, 2.5×10¹⁵ photons. Energy of source.
Answer
ν = 1/(2×10⁻⁹) = 5×10⁸ Hz; E/photon = hν = 3.31×10⁻²⁵ J; Total = 3.31×10⁻²⁵ × 2.5×10¹⁵ = 8.28 × 10⁻¹⁰ J
50. Doublet absorption at 589 and 589.6 nm. Frequencies and energy difference.
Answer
ν₁ = 5.093 × 10¹⁴ Hz; ν₂ = 5.088 × 10¹⁴ Hz; ΔE = h×Δν = 6.626×10⁻³⁴ × 5×10¹¹ = 3.31 × 10⁻²² J
51. Cs work function = 1.9 eV: (a) threshold λ (b) threshold ν. At 500 nm: KE and velocity.
Answer
(a) λ₀ = hc/W₀ = 654 nm; (b) ν₀ = 4.59 × 10¹⁴ Hz; KE = (2.483−1.9)eV = 0.583 eV = 9.33 × 10⁻²⁰ J; v = 4.52 × 10⁵ m s⁻¹
52. Sodium: λ=500,450,400 nm; v=2.55,4.35,5.35 (×10⁵ cm/s). Find (a) threshold λ and (b) Planck's constant.
Answer
(b) Using two data points: h = ΔKE/(cΔ(1/λ)) ≈ 6.57 × 10⁻³⁴ J·s; (a) λ₀ ≈ 540 nm
53. Ag photoelectron stopped at 0.35 V, λ=256.7 nm. Work function.
Answer
KE = eV = 5.6×10⁻²⁰ J; E = 7.75×10⁻¹⁹ J; W₀ = E−KE = 7.19 × 10⁻¹⁹ J = 4.49 eV
54. Photon λ=150 pm ejects electron at v=1.5×10⁷ m s⁻¹. Binding energy.
Answer
Ephoton = 1.325×10⁻¹⁵ J; KE = 1.025×10⁻¹⁶ J; BE = 1.22 × 10⁻¹⁵ J
55. Paschen series at λ=1285 nm. Find n and spectral region.
Answer
1/λ = RH(1/9−1/n²); solving: 1/n² = 0.04; n=5; region: infrared
56. Emission from r=1.3225 nm to r=211.6 pm. Series and region.
Answer
n₁=5 (r=1322.5 pm), n₂=2 (r=211.6 pm); transition 5
⇒ 2: Balmer series, visible region
57. Electron v=1.6×10⁶ m s⁻¹. de Broglie wavelength.
Answer
λ = h/(mv) = 6.626×10⁻³⁴/(9.11×10⁻³¹×1.6×10⁶) = 4.55 × 10⁻¹⁰ m = 455 pm
58. Neutron diffraction λ=800 pm. Velocity.
Answer
v = h/(mλ) = 6.626×10⁻³⁴/(1.675×10⁻²⁷×800×10⁻¹²) = 494.5 m s⁻¹
59. Electron in Bohr's first orbit, v=2.19×10⁶ m s⁻¹. de Broglie wavelength.
Answer
λ = 6.626×10⁻³⁴/(9.11×10⁻³¹×2.19×10⁶) = 3.32 × 10⁻¹⁰ m = 332 pm
60. Hockey ball m=0.1 kg at v=4.37×10⁵ m s⁻¹. de Broglie wavelength.
Answer
λ = h/(mv) = 6.626×10⁻³⁴/(0.1×4.37×10⁵) = 1.52 × 10⁻³⁸ m (unmeasurably small)
61. Δx = 0.002 nm. Δp and can p = h/(4π×m×0.05 nm) be defined?
Answer
Δp ≥ h/(4πΔx) = 6.626×10⁻³⁴/(4π×2×10⁻¹²) = 2.637 × 10⁻²³ kg m s⁻¹. The stated p is smaller than Δp, so it cannot be defined simultaneously with the given position — there is a fundamental problem.
62. Arrange six electrons in increasing energy order.
Answer
Use n+l rule: (5) n=3,l=1
⇒ n+l=4 (3p); (2,4) n=3,l=2
⇒ n+l=5 (3d); (3,6) n=4,l=1
⇒ n+l=5 (4p); (1) n=4,l=2
⇒ n+l=6 (4d). For same n+l, lower n has lower energy: 3d < 4p. Order: 5 < 2 = 4 < 3 = 6 < 1. Note: 3 and 6 share identical quantum numbers, violating Pauli exclusion.
63. Br: 2p, 3p, 4p electrons. Which has lowest effective nuclear charge?
Answer
4p electrons — they are outermost and most shielded.
64. Larger effective nuclear charge: (i) 2s or 3s (ii) 4d or 4f (iii) 3d or 3p?
Answer
(i) 2s; (ii) 4d; (iii) 3p
65. Al and Si 3p electrons: which sees higher effective nuclear charge?
Answer
Si (Z=14 vs Al Z=13, similar shielding)
66. Unpaired electrons in: (a) P (b) Si (c) Cr (d) Fe (e) Kr
Answer
(a) 3; (b) 2; (c) 6; (d) 4; (e) 0
67. (a) Subshells for n=4. (b) Electrons with ms=−½ for n=4.
Answer
(a) 4 subshells (4s, 4p, 4d, 4f); (b) Total 32 electrons; half = 16
Important Extra Questions
What are the limitations of Bohr's model?
Answer
Bohr's model fails to explain: spectra of multi-electron atoms; fine structure of spectral lines; the Zeeman effect; de Broglie's wave nature of electrons; and the Heisenberg uncertainty principle. It incorrectly assumes electrons move in fixed circular orbits.
State Heisenberg's uncertainty principle and its significance.
Answer
Δx·Δp ≥ h/4π — the position and momentum of a particle cannot both be precisely known simultaneously. For electrons, this means we cannot speak of fixed orbits; instead we describe probability regions (orbitals).
What is the aufbau principle?
Answer
Electrons fill orbitals in order of increasing energy. The (n+l) rule gives energy order; for same (n+l), lower n is lower energy. Sequence: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p…
State Hund's rule with an example.
Answer
Pairing occurs only after each orbital in a subshell is singly occupied. Nitrogen (Z=7): 2px¹2py¹2pz¹ with parallel spins, not 2px²2py¹.
Electronic configuration of Cu (Z=29) and reason for anomaly.
Answer
Cu: [Ar]3d¹⁰4s¹ (not 3d⁹4s²). A completely filled 3d subshell is extra stable due to symmetrical electron distribution and maximum exchange energy. One electron shifts from 4s to complete the 3d subshell.