NCERT Solutions for Class 11 Chemistry Chapter 2: Structure of Atom

Concept. The mass of a single electron is m_e = 9.109 × 10⁻³¹ kg (= 9.109 × 10⁻²⁸ g) and its charge is e = 1.602 × 10⁻¹⁹ C. The number of electrons in a given mass is (total mass)/(mass of one electron); one mole contains Avogadro’s number N_A = 6.022 × 10²³ particles, so the mass and charge of a mole of electrons are N_A times those of one electron.

Number = (1 g)/(9.109 × 10⁻²⁸ g)     Mass of 1 mol = N_A m_e     Charge of 1 mol = N_A e

1. (i) Number of electrons in 1 g: 1 g = 10⁻³ kg, so N = 10⁻³ kg / 9.109 × 10⁻³¹ kg = 1.098 × 10²⁷ electrons.
2. (ii) Mass of one mole of electrons: 6.022 × 10²³ × 9.109 × 10⁻³¹ kg = 5.485 × 10⁻⁷ kg (= 5.485 × 10⁻⁴ g).
3. Charge of one mole of electrons: 6.022 × 10²³ × 1.602 × 10⁻¹⁹ C = 9.647 × 10⁴ C (negative; with e = 1.602176 × 10⁻¹⁹ C it is 96,485 C).

Extra depth. The charge of one mole of electrons, about 96,485 C, is called one faraday (F) and is the key quantity of electrolysis you will meet in Class XII. The mass of a mole of electrons, 5.485 × 10⁻⁴ g, is only about 1/1836 of the mass of a mole of hydrogen atoms (1.008 g), which is why the mass of an atom is practically the mass of its nucleus.

Conclusion: One gram contains 1.098 × 10²⁷ electrons; one mole of electrons has a mass of 5.485 × 10⁻⁷ kg and a charge of 9.647 × 10⁴ C.

Concept. The number of sub-atomic particles in a sample = (number of moles) × N_A × (particles per atom or molecule). Per atom the number of protons is Z, electrons Z (neutral) and neutrons A − Z. Mass of particles = number × mass of one particle.

(i) One mole of methane, CH₄. Electrons per molecule = 6 (C) + 4 × 1 (H) = 10. Electrons in one mole = 10 × 6.022 × 10²³ = 6.022 × 10²⁴.

(ii) 7 mg of ¹⁴C (Z = 6, A = 14, so 8 neutrons per atom; molar mass 14 g mol⁻¹).

1. Moles = 7 × 10⁻³ g / 14 g mol⁻¹ = 5.0 × 10⁻⁴ mol.
2. Atoms = 5.0 × 10⁻⁴ × 6.022 × 10²³ = 3.011 × 10²⁰.
3. (a) Neutrons = 8 × 3.011 × 10²⁰ = 2.409 × 10²¹.
4. (b) Mass of neutrons = 2.409 × 10²¹ × 1.675 × 10⁻²⁷ kg = 4.03 × 10⁻⁶ kg.

(iii) 34 mg of NH₃ (molar mass 17 g mol⁻¹; protons per molecule = 7 + 3 = 10).

1. Moles = 34 × 10⁻³ / 17 = 2.0 × 10⁻³ mol; molecules = 2.0 × 10⁻³ × 6.022 × 10²³ = 1.2044 × 10²¹.
2. (a) Protons = 10 × 1.2044 × 10²¹ = 1.2044 × 10²².
3. (b) Mass of protons = 1.2044 × 10²² × 1.6726 × 10⁻²⁷ kg = 2.01 × 10⁻⁵ kg (proton mass from Table 2.1).

Will the answer change with temperature and pressure? No. The number of particles and their mass depend only on the amount of substance (number of moles), not on temperature or pressure. T and P change only the volume of the gas (22.4 L per mole at STP), not the number of protons, neutrons or electrons in a given mass.

Extra depth. Notice that the neutron/proton mass is used only in the final step; the counting uses the mole concept. The same logic gives, e.g., the number of electrons in 1.8 g of water: 1.8/18 = 0.1 mol × 10 e⁻ × N_A = 6.022 × 10²³.

Conclusion: (i) 6.022 × 10²⁴ electrons; (ii) 2.409 × 10²¹ neutrons of mass 4.03 × 10⁻⁶ kg; (iii) 1.2044 × 10²² protons of mass 2.01 × 10⁻⁵ kg; no change with temperature or pressure.

Concept. In the symbol ^A_ZX, the atomic number Z (lower left) equals the number of protons and the mass number A (upper left) equals protons + neutrons. Hence neutrons = A − Z (equations 2.3 and 2.4 of the chapter).

Protons = Z     Neutrons = A − Z

Extra depth. Light stable nuclei such as ¹⁶O and ²⁴Mg have equal numbers of neutrons and protons; heavier nuclei need a progressively larger neutron excess (Fe: 30 n for 26 p; Sr: 50 n for 38 p) to offset the proton–proton repulsion. ¹³C is an isotope of carbon (7 neutrons) used in NMR; ⁸⁸Sr has the magic number of 50 neutrons.

Conclusion: ¹³C: 6 p, 7 n; ¹⁶O: 8 p, 8 n; ²⁴Mg: 12 p, 12 n; ⁵⁶Fe: 26 p, 30 n; ⁸⁸Sr: 38 p, 50 n.

Concept. The complete symbol of an atom is ^A_ZX: the symbol X of the element (identified from Z using the periodic table) with the mass number A as left superscript and the atomic number Z as left subscript. The element is fixed by Z alone (number of protons).

1. Z = 17 is chlorine (Cl); A = 35 → ³⁵₁₇Cl (neutrons = 35 − 17 = 18).
2. Z = 92 is uranium (U); A = 233 → ²³³₉₂U (neutrons = 141).
3. Z = 4 is beryllium (Be); A = 9 → ⁹₄Be (neutrons = 5).

Extra depth. ³⁵Cl is the more abundant isotope of chlorine (the other being ³⁷Cl, Chapter 1 and Section 2.2.4); ²³³U is a fissile isotope made by neutron capture in thorium reactors; ⁹Be is the only stable isotope of beryllium and is the target in Chadwick’s neutron discovery (Section 2.1.4).

Conclusion: (i) ³⁵₁₇Cl; (ii) ²³³₉₂U; (iii) ⁹₄Be.

Concept. For an electromagnetic wave in vacuum, the speed of light c, frequency ν and wavelength λ are related by c = νλ (equation 2.5). The wavenumber is the number of wavelengths per unit length, ν̄ = 1/λ. Always convert the wavelength to metres first.

ν = c/λ     ν̄ = 1/λ

1. λ = 580 nm = 580 × 10⁻⁹ m = 5.80 × 10⁻⁷ m.
2. Frequency: ν = (3.0 × 10⁸ m s⁻¹)/(5.80 × 10⁻⁷ m) = 5.17 × 10¹⁴ s⁻¹ (Hz).
3. Wavenumber: ν̄ = 1/(5.80 × 10⁻⁷ m) = 1.724 × 10⁶ m⁻¹ = 1.724 × 10⁴ cm⁻¹.

Extra depth. The value lies inside the visible range (4.0–7.5 × 10¹⁴ Hz) — it is the sodium D line (589 nm in Ex 2.50). It is also the same as Problem 2.5 of the chapter (5800 Å), which is the same wavelength.

Conclusion: ν = 5.17 × 10¹⁴ Hz and ν̄ = 1.724 × 10⁶ m⁻¹ (= 1.724 × 10⁴ cm⁻¹).

Concept. By Planck’s quantum theory the energy of one photon is E = hν where h = 6.626 × 10⁻³⁴ J s. Since ν = c/λ, a photon of wavelength λ has E = hc/λ.

E = hν = hc/λ

1. (i) ν = 3 × 10¹⁵ Hz: E = (6.626 × 10⁻³⁴ J s)(3 × 10¹⁵ s⁻¹) = 1.988 × 10⁻¹⁸ J.
2. (ii) λ = 0.50 Å = 0.50 × 10⁻¹⁰ m = 5.0 × 10⁻¹¹ m: E = (6.626 × 10⁻³⁴)(3.0 × 10⁸)/(5.0 × 10⁻¹¹) = 3.976 × 10⁻¹⁵ J.

Extra depth. The wavelength in (ii) corresponds to ν = c/λ = 6 × 10¹⁸ Hz, so the photon is an X-ray photon with about 2000 times the energy of the ultraviolet photon in (i) — energy increases with frequency and decreases with wavelength. (1 eV = 1.602 × 10⁻¹⁹ J was used only for the comparison.)

Conclusion: (i) E = 1.988 × 10⁻¹⁸ J; (ii) E = 3.976 × 10⁻¹⁵ J.

Concept. The period T of a wave is the time for one complete oscillation; it is the reciprocal of the frequency, ν = 1/T. The wavelength then follows from c = νλ and the wavenumber from ν̄ = 1/λ = ν/c.

ν = 1/T     λ = c/ν = cT     ν̄ = 1/λ

1. Frequency: ν = 1/(2.0 × 10⁻¹⁰ s) = 5.0 × 10⁹ s⁻¹ (Hz).
2. Wavelength: λ = c/ν = (3.0 × 10⁸ m s⁻¹)/(5.0 × 10⁹ s⁻¹) = 6.0 × 10⁻² m (6 cm).
3. Wavenumber: ν̄ = 1/λ = 1/(6.0 × 10⁻² m) = 16.67 m⁻¹ (= 0.1667 cm⁻¹).

Extra depth. A frequency of 5 × 10⁹ Hz and a wavelength of 6 cm place the wave in the microwave region (radar, about 10¹⁰ Hz in Section 2.3.1). Period, frequency and wavelength are three descriptions of the same wave: T = 1/ν = λ/c.

Conclusion: ν = 5.0 × 10⁹ Hz, λ = 6.0 × 10⁻² m, ν̄ = 16.67 m⁻¹.

Concept. Total energy = (number of photons) × (energy of one photon). The energy of one photon of wavelength λ is E = hc/λ. So the number of photons providing 1 J is N = 1 J / (hc/λ).

N = (total energy)/(hc/λ) = λ × (total energy)/(hc)

1. λ = 4000 pm = 4000 × 10⁻¹² m = 4.0 × 10⁻⁹ m.
2. Energy of one photon: E = (6.626 × 10⁻³⁴ J s)(3.0 × 10⁸ m s⁻¹)/(4.0 × 10⁻⁹ m) = 4.969 × 10⁻¹⁷ J.
3. Number of photons: N = 1 J / (4.969 × 10⁻¹⁷ J) = 2.01 × 10¹⁶ photons.

Extra depth. A 4000 pm (= 4 nm) photon is an X-ray/extreme-UV photon; the shorter the wavelength, the more energy each photon carries, hence the fewer photons are needed for 1 J. Compare Problem 2.7 (visible light of 400 nm needs 2 × 10¹⁸ photons for 1 J).

Conclusion: About 2.01 × 10¹⁶ photons of wavelength 4000 pm are needed to provide 1 J of energy.

Concept. In the photoelectric effect Einstein’s equation states: photon energy hν = work function W₀ + kinetic energy of the ejected electron. So K.E. = hν − W₀ = ½ m_ev², valid only if hν > W₀.

E_photon = hc/λ     K.E. = E_photon − W₀     v = √(2 K.E./m_e)

1. (i) Photon energy: E = hc/λ = (6.626 × 10⁻³⁴)(3.0 × 10⁸)/(4 × 10⁻⁷) = 4.9695 × 10⁻¹⁹ J. In eV: 4.9695 × 10⁻¹⁹/1.6020 × 10⁻¹⁹ = 3.10 eV.
2. (ii) Kinetic energy: K.E. = 3.102 − 2.13 = 0.972 eV = 0.972 × 1.6020 × 10⁻¹⁹ J = 1.557 × 10⁻¹⁹ J.
3. (iii) Velocity: v = √(2 × 1.557 × 10⁻¹⁹ J / 9.109 × 10⁻³¹ kg) = √(3.419 × 10¹¹) = 5.85 × 10⁵ m s⁻¹.

Extra depth. The electron speed of 5.85 × 10⁵ m s⁻¹ is only 0.2% of the speed of light, so the non-relativistic formula K.E. = ½ m_ev² is justified. If the photon energy were smaller than 2.13 eV (i.e. λ > 583 nm), no electron would be emitted at all.

Conclusion: (i) 3.10 eV; (ii) K.E. = 0.972 eV = 1.557 × 10⁻¹⁹ J; (iii) v = 5.85 × 10⁵ m s⁻¹.

Concept. “Just sufficient” means the photon energy equals the minimum energy needed to remove one electron — the ionisation energy per atom. Multiplying by Avogadro’s number gives the molar ionisation energy.

IE per atom = hc/λ     IE per mole = N_A × hc/λ

1. λ = 242 nm = 2.42 × 10⁻⁷ m.
2. Energy per atom: E = (6.626 × 10⁻³⁴ J s)(3.0 × 10⁸ m s⁻¹)/(2.42 × 10⁻⁷ m) = 8.214 × 10⁻¹⁹ J.
3. Per mole: 8.214 × 10⁻¹⁹ J × 6.022 × 10²³ mol⁻¹ = 4.946 × 10⁵ J mol⁻¹ = 494.6 kJ mol⁻¹.

Extra depth. The accepted first ionisation enthalpy of sodium is 496 kJ mol⁻¹, so the calculation reproduces the experimental value — an example of how spectroscopic data give ionisation energies (Chapter 3 on periodic trends). The radiation (242 nm) is in the ultraviolet.

Conclusion: The ionisation energy of sodium is about 494.6 kJ mol⁻¹ (≈ 495 kJ mol⁻¹).

Concept. Power is energy emitted per second (1 W = 1 J s⁻¹). The number of quanta (photons) emitted per second is therefore the power divided by the energy of one photon, E = hc/λ (as in Problem 2.7 of the chapter).

Rate of emission = Power/(hc/λ)

1. λ = 0.57 µm = 0.57 × 10⁻⁶ m = 5.7 × 10⁻⁷ m.
2. Energy of one photon: E = (6.626 × 10⁻³⁴)(3.0 × 10⁸)/(5.7 × 10⁻⁷) = 3.487 × 10⁻¹⁹ J.
3. Rate = 25 J s⁻¹ / 3.487 × 10⁻¹⁹ J = 7.17 × 10¹⁹ quanta s⁻¹.

Extra depth. About 7 × 10¹⁹ photons leave the lamp each second — a huge number because each visible photon carries only 3.5 × 10⁻¹⁹ J. The calculation assumes that all the electrical power is converted to yellow light; a real bulb is far less efficient.

Conclusion: The bulb emits about 7.17 × 10¹⁹ quanta per second.

Concept. Electrons are ejected with zero kinetic energy exactly when the photon energy equals the work function. Thus the given radiation is at the threshold: its frequency is ν₀ and its energy is W₀ = hν₀.

ν₀ = c/λ     W₀ = hν₀

1. λ = 6800 Å = 6800 × 10⁻¹⁰ m = 6.8 × 10⁻⁷ m.
2. Threshold frequency: ν₀ = (3.0 × 10⁸ m s⁻¹)/(6.8 × 10⁻⁷ m) = 4.41 × 10¹⁴ s⁻¹.
3. Work function: W₀ = (6.626 × 10⁻³⁴ J s)(4.41 × 10¹⁴ s⁻¹) = 2.92 × 10⁻¹⁹ J (≈ 1.82 eV).

Extra depth. A 6800 Å threshold lies in the red region, so the metal is photoelectrically sensitive even to red light — typical of the alkali metals with low work functions (compare K = 2.25 eV and Cs = 1.9 eV in Table 2.2 / Ex 2.51). Radiation of longer wavelength than 6800 Å cannot eject electrons at all.

Conclusion: Threshold frequency ν₀ = 4.41 × 10¹⁴ Hz and work function W₀ = 2.92 × 10⁻¹⁹ J.

Concept. In Bohr’s model the energy of the n^th level of hydrogen is E_n = −2.18 × 10⁻¹⁸/n² J. When the electron falls from n_i to n_f a photon is emitted whose energy equals the energy difference, ΔE = hν = hc/λ.

ΔE = 2.18 × 10⁻¹⁸ J (1/n_f² − 1/n_i²)     λ = hc/ΔE

1. n_i = 4, n_f = 2: ΔE = 2.18 × 10⁻¹⁸(1/4 − 1/16) = 2.18 × 10⁻¹⁸ × 3/16 = 4.0875 × 10⁻¹⁹ J (energy emitted).
2. λ = hc/ΔE = (6.626 × 10⁻³⁴ J s)(3.0 × 10⁸ m s⁻¹)/(4.0875 × 10⁻¹⁹ J) = 4.863 × 10⁻⁷ m = 486 nm.
3. Check with the Rydberg formula: ν̄ = 1.09677 × 10⁷ m⁻¹(1/4 − 1/16) = 2.056 × 10⁶ m⁻¹, λ = 1/ν̄ = 486 nm.

Extra depth. This is the second line (H_β) of the Balmer series, the only series in the visible region. The first line (3 → 2) is red at 656 nm (Ex 2.17) and the third line (5 → 2) is at 434 nm (Problem 2.10).

Conclusion: The emitted light has wavelength λ = 4.86 × 10⁻⁷ m (486 nm), in the visible (Balmer) region.

Concept. Ionisation means taking the electron from level n to n = ∞ (where E = 0). The energy required is therefore |E_n| = 2.18 × 10⁻¹⁸/n² J. For the ground state (n = 1) this is the ionisation enthalpy of hydrogen.

Energy required = E_∞ − E_n = 0 − (−2.18 × 10⁻¹⁸/n²) = 2.18 × 10⁻¹⁸/n² J

1. From n = 5: ΔE = 2.18 × 10⁻¹⁸/25 = 8.72 × 10⁻²⁰ J per atom (≈ 52.5 kJ mol⁻¹).
2. From n = 1: ΔE = 2.18 × 10⁻¹⁸ J per atom (≈ 1312 kJ mol⁻¹).
3. Comparison: (2.18 × 10⁻¹⁸)/(8.72 × 10⁻²⁰) = 25. Ionising an atom whose electron is in n = 5 needs only 1/25 (one twenty-fifth, i.e. n² times less) of the ground-state ionisation energy.

Extra depth. An electron in a higher orbit is farther from the nucleus and less tightly bound, so it is easier to remove — the energy needed falls as 1/n². This is why excited atoms are ionised more easily than ground-state atoms.

Conclusion: Ionising H from n = 5 needs 8.72 × 10⁻²⁰ J, which is 1/25 of the ionisation enthalpy from n = 1 (2.18 × 10⁻¹⁸ J).

Concept. An electron in n = 6 can return to n = 1 either directly or in steps through any intermediate level. Every different pair (n_i > n_f) is a different transition and gives a different spectral line. Counting all possible pairs from n levels gives n(n − 1)/2 lines.

Number of lines = n(n − 1)/2 = 6 × 5/2 = 15

Counting shows 5 + 4 + 3 + 2 + 1 = 15 lines.

Extra depth. A single atom can show only one transition at a time, but a sample contains a huge number of atoms in all excited states, so all 15 lines appear together (the chapter notes that a collection of atoms contains all possible excited states). The lines belong to the Lyman (UV), Balmer (visible) and Paschen, Brackett, Pfund (infrared) series.

Conclusion: A maximum of 15 emission lines can be observed when the electron falls from n = 6 to the ground state.

Concept. In Bohr’s theory, for the hydrogen atom, E_n = E₁/n² (equation 2.13) and the radius of the n^th orbit is r_n = n²a₀ where a₀ = 52.9 pm is the radius of the first (Bohr) orbit.

1. (i) Energy of the fifth orbit: E₅ = −2.18 × 10⁻¹⁸ J/5² = −2.18 × 10⁻¹⁸/25 = −8.72 × 10⁻²⁰ J atom⁻¹.
2. (ii) Radius of the fifth orbit: r₅ = 5² × 52.9 pm = 25 × 52.9 pm = 1322.5 pm = 1.3225 nm (= 1.3225 × 10⁻⁹ m).

Extra depth. As n increases the orbit becomes larger (r ∝ n²) and the energy less negative (E ∝ −1/n²), i.e. higher in energy and less tightly bound. The radius 1322.5 pm reappears in Ex 2.56.

Conclusion: E₅ = −8.72 × 10⁻²⁰ J atom⁻¹ and r₅ = 1322.5 pm (1.3225 nm).

Concept. The Balmer series ends at n_f = 2 and begins at n = 3, 4, 5 … (Table 2.3). The longest wavelength corresponds to the smallest energy difference, i.e. the transition 3 → 2. The wavenumber is given by Rydberg’s equation.

ν̄ = 109,677 cm⁻¹ (1/n₁² − 1/n₂²)  with n₁ = 2, n₂ = 3

1. ν̄ = 109,677 (1/2² − 1/3²) = 109,677 (1/4 − 1/9) = 109,677 × 5/36.
2. ν̄ = 1.523 × 10⁴ cm⁻¹ (= 1.523 × 10⁶ m⁻¹).
3. The corresponding wavelength is λ = 1/ν̄ = 6.565 × 10⁻⁵ cm = 656.5 nm (red).

Extra depth. The longest wavelength means the smallest wavenumber, so the answer is the first line of the series (H_α, red). The series limit (n = ∞ → 2) has ν̄ = 109,677/4 = 27,419 cm⁻¹ (λ = 364.6 nm) and is the shortest wavelength of the series.

Conclusion: The wavenumber of the longest-wavelength Balmer line (3 → 2) is 1.523 × 10⁴ cm⁻¹ (1.523 × 10⁶ m⁻¹).

Concept. The energy needed to raise the electron from n₁ to n₂ is ΔE = E₂ − E₁ with E_n = E₁/n². When the electron returns to the ground state the same energy is released as a photon of wavelength λ = hc/ΔE. Convert ergs to joules first: 1 erg = 10⁻⁷ J.

1. E₁ = −2.18 × 10⁻¹¹ erg = −2.18 × 10⁻¹¹ × 10⁻⁷ J = −2.18 × 10⁻¹⁸ J.
2. E₅ = −2.18 × 10⁻¹⁸/25 = −8.72 × 10⁻²⁰ J.
3. Energy required: ΔE = E₅ − E₁ = −8.72 × 10⁻²⁰ + 2.18 × 10⁻¹⁸ = 2.093 × 10⁻¹⁸ J.
4. Wavelength on returning (5 → 1): λ = hc/ΔE = (6.626 × 10⁻³⁴)(3.0 × 10⁸)/(2.093 × 10⁻¹⁸) = 9.50 × 10⁻⁸ m = 95 nm (ultraviolet, Lyman series).

Extra depth. The direct 5 → 1 return gives the 95 nm Lyman line. If the electron returns in steps (e.g. 5 → 2 and then 2 → 1) several lines of longer wavelength are emitted instead (see Ex 2.15); the textbook answer refers to the direct return to the ground state.

Conclusion: The electron needs 2.093 × 10⁻¹⁸ J to go from n = 1 to n = 5, and on falling back to n = 1 it emits light of wavelength about 9.5 × 10⁻⁸ m (95 nm).

Concept. Removing the electron completely from n = 2 means taking it to n = ∞ where E = 0. The minimum energy required is ΔE = 0 − E₂. The photon that supplies exactly this energy has the longest wavelength that can still cause the transition (longer wavelengths carry less energy): λ_max = hc/ΔE.

1. E₂ = −2.18 × 10⁻¹⁸/2² = −5.45 × 10⁻¹⁹ J.
2. Energy required: ΔE = E_∞ − E₂ = 5.45 × 10⁻¹⁹ J per atom.
3. Longest wavelength: λ = hc/ΔE = (6.626 × 10⁻³⁴ J s)(3.0 × 10⁸ m s⁻¹)/(5.45 × 10⁻¹⁹ J) = 3.647 × 10⁻⁷ m = 3.647 × 10⁻⁵ cm (≈ 365 nm).

Extra depth. 365 nm is the Balmer series limit (n = ∞ → 2) found in Ex 2.17. Any light with λ ≤ 365 nm can ionise an H atom from n = 2; light of longer wavelength cannot, however intense it is — the same threshold idea as in the photoelectric effect.

Conclusion: Removing the electron from n = 2 needs 5.45 × 10⁻¹⁹ J; the longest wavelength that can do this is 3.647 × 10⁻⁵ cm.

Concept. According to de Broglie, a moving particle of mass m and velocity v has an associated wave of wavelength λ = h/(mv) (equation 2.22). The mass of the electron is 9.109 × 10⁻³¹ kg.

λ = h/(m_e v)

1. λ = 6.626 × 10⁻³⁴ J s / (9.109 × 10⁻³¹ kg × 2.05 × 10⁷ m s⁻¹).
2. Denominator = 9.109 × 10⁻³¹ × 2.05 × 10⁷ = 1.867 × 10⁻²³ kg m s⁻¹.
3. λ = 6.626 × 10⁻³⁴/1.867 × 10⁻²³ = 3.55 × 10⁻¹¹ m (= 35.5 pm = 3.55 × 10⁻⁹ cm).

Extra depth. The wavelength (35.5 pm) is comparable to atomic spacings (about 100–300 pm) and is why electron beams are diffracted by crystals and why the electron microscope can resolve very small features. A 0.1 kg ball moving at 10 m s⁻¹ has λ = 6.6 × 10⁻³⁴ m (Problem 2.12), far too small to detect.

Conclusion: The de Broglie wavelength of the electron is 3.55 × 10⁻¹¹ m.

Concept. The wavelength needs the velocity, which follows from the kinetic energy K.E. = ½ mv². Then λ = h/mv. (Equivalent shortcut: λ = h/√(2m K.E.).)

v = √(2 K.E./m)     λ = h/(mv)

1. v = √(2 × 3.0 × 10⁻²⁵ J / 9.1 × 10⁻³¹ kg) = √(6.59 × 10⁵) = 812 m s⁻¹.
2. λ = 6.626 × 10⁻³⁴/(9.1 × 10⁻³¹ × 812) = 6.626 × 10⁻³⁴/7.39 × 10⁻²⁸ = 8.97 × 10⁻⁷ m (= 896.7 nm).

Extra depth. This slow electron has a wavelength (897 nm) longer than visible light — the slower the electron, the longer the wave. Fast electrons (Ex 2.20) have picometre wavelengths. This is exactly Problem 2.13 of the chapter.

Conclusion: The electron has a speed of 812 m s⁻¹ and de Broglie wavelength 8.97 × 10⁻⁷ m (897 nm).

Concept. Isoelectronic species contain the same number of electrons. For a neutral atom electrons = Z; for a cation electrons = Z − charge; for an anion electrons = Z + charge.

• Group 1 (10 electrons, like Ne): Na⁺ and Mg²⁺.
• Group 2 (18 electrons, like Ar): K⁺, Ca²⁺, S²⁻ and Ar.

Extra depth. Isoelectronic species have the same electron configuration but different nuclear charges, so their sizes differ: among the 18-electron species the radius decreases as Z increases, S²⁻ > Ar > K⁺ > Ca²⁺ (a theme of Chapter 3). The 10-electron pair is Na⁺ and Mg²⁺, isoelectronic with Ne, F⁻ and O²⁻.

Conclusion: Na⁺ and Mg²⁺ are isoelectronic (10 electrons); K⁺, Ca²⁺, S²⁻ and Ar are isoelectronic (18 electrons).

Concept. Write the configuration of the neutral atom with the aufbau order (1s, 2s, 2p, 3s, 3p, 4s, 3d …), Pauli exclusion principle and Hund’s rule, then add electrons for an anion or remove them (from the outermost shell) for a cation. For (ii) and (iii) count all the electrons in the configuration; for a neutral atom this number is the atomic number Z.

(i) Configurations of ions

(ii) Atomic numbers from the outermost electron. The outermost electron(s) together with the filled inner shells give the full configuration:

1. 3s¹: 1s²2s²2p⁶3s¹ → 11 electrons → Z = 11 (Na).
2. 2p³: 1s²2s²2p³ → 7 electrons → Z = 7 (N).
3. 3p⁵: 1s²2s²2p⁶3s²3p⁵ → 17 electrons → Z = 17 (Cl).

(iii) Atoms from the configuration (core [He] = 2, [Ne] = 10, [Ar] = 18 electrons):

1. [He] 2s¹: 2 + 1 = 3 electrons → Li.
2. [Ne] 3s² 3p³: 10 + 2 + 3 = 15 electrons → P.
3. [Ar] 4s² 3d¹: 18 + 2 + 1 = 21 electrons → Sc (scandium).

Extra depth. Na⁺, O²⁻ and F⁻ are all isoelectronic with neon (1s²2s²2p⁶, a stable octet); H⁻ has the helium configuration. In (ii) the answers assume a neutral atom with the outermost electrons shown. Note that in scandium the 4s orbital is written before 3d in [Ar]4s²3d¹ because 4s fills first, though 3d is written first in the conventional order [Ar]3d¹4s².

Conclusion: (i) H⁻: 1s²; Na⁺, O²⁻, F⁻: 1s²2s²2p⁶. (ii) 3s¹ → Z = 11; 2p³ → Z = 7; 3p⁵ → Z = 17. (iii) Li, P and Sc respectively.

Concept. Azimuthal quantum number l takes values 0 to (n − 1); the subshell notation is l = 0 (s), 1 (p), 2 (d), 3 (f), 4 (g), 5 (h). A g subshell therefore needs l = 4, and since l ≤ n − 1, we need n − 1 ≥ 4.

1. g orbitals: l = 4.
2. Condition: l ≤ n − 1 ⇒ 4 ≤ n − 1 ⇒ n ≥ 5.
3. Lowest n = 5 (the 5g subshell, with 2l + 1 = 9 orbitals).

Extra depth. The 5g subshell holds up to 18 electrons (9 orbitals), but no known ground-state atom has electrons in g orbitals; they would appear only in very heavy elements (Z ≈ 121). For the same reason d orbitals first appear in n = 3 (Section 2.6.2) and f orbitals in n = 4.

Conclusion: g orbitals first occur in the fifth shell; the lowest value of n is 5.

Concept. The label “3d” carries two of the quantum numbers: the number 3 is the principal quantum number n, and the letter d means azimuthal quantum number l = 2 (s = 0, p = 1, d = 2, f = 3). The magnetic quantum number m_l can take (2l + 1) values from −l to +l, one for each of the five d orbitals.

n = 3     l = 2     m_l = −2, −1, 0, +1, +2

So the electron may be in any one of the five orbitals (n, l, m_l) = (3, 2, −2), (3, 2, −1), (3, 2, 0), (3, 2, +1) or (3, 2, +2); in addition its spin quantum number m_s is +½ or −½.

Extra depth. The five 3d orbitals (d_xy, d_yz, d_xz, d_x²−y², d_z²) are degenerate, and the particular m_l value does not correspond one-to-one to an axis (Section 2.6.2). The shorthand “3d” is therefore the compact way of writing n = 3 and l = 2.

Conclusion: For a 3d electron n = 3, l = 2 and m_l = −2, −1, 0, +1 or +2.

Concept. The question refers to an atom, which is electrically neutral, so the number of protons equals the number of electrons (= atomic number Z). The mass number A = protons + neutrons identifies the isotope. The configuration follows from the aufbau principle.

1. (i) Protons = electrons = 29 (neutral atom), so Z = 29 — the element is copper; mass number A = 29 + 35 = 64, i.e. ⁶⁴₂₉Cu.
2. (ii) Electronic configuration: by the aufbau order 1s²2s²2p⁶3s²3p⁶4s²3d⁹ would be expected, but a completely filled 3d subshell is extra stable, so one electron moves from 4s to 3d: 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹⁰ 4s¹ = [Ar]3d¹⁰4s¹.

Extra depth. Copper is one of the exceptional configurations marked * in Table 2.6, along with chromium (3d⁵4s¹). Section 2.6.6 explains that the extra stability comes from the symmetrical distribution and the larger exchange energy of a completely filled 3d¹⁰. Copper therefore has one unpaired electron (in 4s).

Conclusion: The element has 29 protons (copper, ⁶⁴Cu) and the configuration 1s²2s²2p⁶3s²3p⁶3d¹⁰4s¹.

Concept. The number of electrons in a molecule or ion = (sum of the atomic numbers of all atoms) − (positive charge) or + (negative charge). Atomic numbers: H = 1, O = 8.

1. H₂⁺ — 2 protons, 1 electron (the hydrogen molecule-ion).
2. H₂ — 2 protons, 2 electrons.
3. O₂⁺ — 16 protons, 16 − 1 = 15 electrons (an odd number of electrons: the species has an unpaired electron).

Extra depth. The positive charge means fewer electrons than protons, so subtract the charge. H₂⁺ is the simplest molecular ion: one electron holds two protons together. With 15 electrons, O₂⁺ is paramagnetic (Chapter 4 on chemical bonding).

Conclusion: H₂⁺ has 1 electron, H₂ has 2 electrons and O₂⁺ has 15 electrons.

Concept. The quantum numbers are related: n = 1, 2, 3 …; l = 0, 1, … (n − 1); m_l = −l, …, 0, …, +l. An orbital “nl” exists only if l ≤ n − 1.

(i) n = 3: l = 0, 1, 2 and for each l the m_l values are:

Total orbitals = 1 + 3 + 5 = 9 = n².

(ii) 3d orbitals: l = 2 and m_l = −2, −1, 0, +1, +2 (five orbitals).

(iii) Which orbitals are possible?

Extra depth. An f subshell first appears at n = 4 and a p subshell at n = 2. The same rule makes sets such as n = 1, l = 1 forbidden in Ex 2.30.

Conclusion: (i) l = 0, 1, 2 with m_l = 0; −1, 0, +1; −2 to +2; (ii) for 3d: l = 2, m_l = −2, −1, 0, +1, +2; (iii) 2s and 2p are possible; 1p and 3f are not.

Concept. An orbital is described by the principal quantum number n followed by the letter for l: l = 0 → s, 1 → p, 2 → d, 3 → f.

All four sets are allowed because l ≤ n − 1 in each case (0 ≤ 0, 1 ≤ 2, 2 ≤ 3, 3 ≤ 3).

Extra depth. The letters come from the descriptions of spectral lines — sharp, principal, diffuse and fundamental; after f the letters continue alphabetically (g, h …). The number of orbitals in these subshells is 1, 3, 5 and 7 and their shapes were described in Section 2.6.2.

Conclusion: (a) 1s; (b) 3p; (c) 4d; (d) 4f.

Concept. A valid set must satisfy four rules: (1) n is a positive integer (n ≥ 1); (2) l = 0, 1, …, (n − 1); (3) m_l runs from −l to +l; (4) m_s = +½ or −½. Test each set against these rules in order.

• (a) n cannot be zero — the principal quantum number starts at 1.
• (c) For n = 1 only l = 0 is allowed (l = 0 to n − 1); l = 1 is not.
• (e) For n = 3 the maximum l is 2; l = 3 (f) cannot occur in the third shell.

Extra depth. Sets (b), (d) and (f) correspond to the 1s, 2p and 3p orbitals. Each m_s value is acceptable, because the spin can be +½ or −½.

Conclusion: Sets (a), (c) and (e) are not possible; (b), (d) and (f) are allowed.

Concept. Each orbital (fixed n, l, m_l) can hold at most two electrons, one with m_s = +½ and one with m_s = −½ (Pauli exclusion principle). The number of orbitals in the n^th shell is n² and in a subshell 2l + 1.

1. (a) n = 4 and m_s = −½: the fourth shell has n² = 16 orbitals (one 4s, three 4p, five 4d, seven 4f). Each orbital can contain exactly one electron with m_s = −½, so at most 16 electrons (when the shell is full: 2n² = 32 electrons in total, half of them with m_s = −½).
2. (b) n = 3 and l = 0: this is the 3s subshell, which has only one orbital (m_l = 0); it can hold 2 electrons (m_s = +½ and −½).

Extra depth. These are maximum numbers; they apply only if the orbitals are filled. In a real ground-state atom the number is smaller: potassium (Z = 19), for example, has only its single 4s electron in the fourth shell. (Part (b) of Ex 2.67 asks the same question for the 4th shell.)

Conclusion: (a) 16 electrons (at most); (b) 2 electrons.

Concept. Bohr’s fourth postulate says that the angular momentum of the electron in a stationary orbit is quantised, m_evr = nh/2π (n = 1, 2, 3 …). De Broglie’s relation gives the wavelength of the electron as λ = h/(m_ev). Combining the two shows that an orbit fits a whole number of electron waves.

1. Bohr’s quantum condition: m_evr = n h/2π.
2. Multiply both sides by 2π and divide by m_ev: 2πr = n h/(m_ev).
3. De Broglie: h/(m_ev) = λ. Substituting, 2πr = nλ.
4. The left side, 2πr, is the circumference of the orbit. Hence the circumference of the Bohr orbit = n × (de Broglie wavelength), an integral multiple (n = 1, 2, 3 …). ∎

2πr = nλ     (n = 1, 2, 3 …)

Extra depth. This shows that Bohr’s allowed orbits are exactly those on which the electron wave is a standing wave — the wave closes on itself in phase. If the circumference were not a whole number of wavelengths the wave would interfere destructively after each revolution and die out. This idea (standing electron waves) was a stepping stone to the Schrödinger equation; the circumference of the first orbit (2π × 52.9 pm = 332 pm) equals the electron’s wavelength there (Ex 2.59).

Conclusion: Using Bohr’s quantisation condition and de Broglie’s relation, 2πr = nλ: the orbit circumference is an integral multiple of the electron’s wavelength.

Concept. For a hydrogen-like species the transition energy is ΔE = 2.18 × 10⁻¹⁸ Z²(1/n₁² − 1/n₂²) J (equation 2.14), and λ = hc/ΔE. The same wavelength therefore means the same energy difference.

1. He⁺ (Z = 2), n = 4 → 2: ΔE = 2.18 × 10⁻¹⁸ × 4 × (1/4 − 1/16) = 2.18 × 10⁻¹⁸ × 4 × 3/16 = 1.635 × 10⁻¹⁸ J.
2. Hydrogen (Z = 1): 2.18 × 10⁻¹⁸(1/n₁² − 1/n₂²) = 1.635 × 10⁻¹⁸ ⇒ (1/n₁² − 1/n₂²) = 0.75 = 3/4.
3. The only integers giving 3/4 are n₁ = 1 and n₂ = 2 (1 − 1/4 = 3/4).
4. Wavelength (both): λ = hc/ΔE = 1.988 × 10⁻²⁵/1.635 × 10⁻¹⁸ = 1.216 × 10⁻⁷ m = 121.6 nm.

Extra depth. A general result: because E ∝ Z²/n², a transition (n₁ → n₂) in He⁺ has the same energy as the transition (n₁/2 → n₂/2) in H when n₁ and n₂ are even. Here 4 → 2 in He⁺ equals 2 → 1 in H — the first line of the Lyman (UV) series.

Conclusion: The n = 2 → n = 1 transition of hydrogen (Lyman series) has the same wavelength (121.6 nm) as the n = 4 → 2 transition of He⁺.

Concept. He⁺ is a hydrogen-like (one-electron) ion, so Bohr’s theory applies with nuclear charge Z = 2: E_n = −2.18 × 10⁻¹⁸ Z²/n² J. The energy needed to remove the electron from the ground state (n = 1) to infinity is the negative of E₁.

1. For He⁺: E₁ = −2.18 × 10⁻¹⁸ × (2)²/(1)² = −8.72 × 10⁻¹⁸ J.
2. Ionisation energy = E_∞ − E₁ = 0 − (−8.72 × 10⁻¹⁸) = 8.72 × 10⁻¹⁸ J per ion.

Extra depth. Doubling the nuclear charge makes the 1s electron four times more tightly bound (Z²). Per mole this is 8.72 × 10⁻¹⁸ × 6.022 × 10²³ = 5251 kJ mol⁻¹, which matches the experimental second ionisation enthalpy of helium (5250 kJ mol⁻¹).

Conclusion: The energy required for He⁺ → He²⁺ + e⁻ is 8.72 × 10⁻¹⁸ J per ion.

Concept. When atoms are placed side by side in a line, each occupies a length equal to its diameter. The number of atoms = (length of the line)/(diameter of one atom). Convert both lengths to the same unit (metres).

1. Diameter of carbon atom = 0.15 nm = 0.15 × 10⁻⁹ m = 1.5 × 10⁻¹⁰ m.
2. Length of scale = 20 cm = 20 × 10⁻² m = 0.20 m.
3. Number of atoms = 0.20 m / 1.5 × 10⁻¹⁰ m = 1.33 × 10⁹.

Extra depth. More than a billion carbon atoms would fit along a 20 cm scale — an illustration of how small atoms are (their diameter is about 10⁻¹⁰ m, Section 2.2.2). Cross-check: 1.33 × 10⁹ × 0.15 nm = 2.0 × 10⁸ nm = 20 cm.

Conclusion: About 1.33 × 10⁹ carbon atoms can be placed side by side along the 20 cm scale.

Concept. For atoms in a row, the total length = (number of atoms) × (diameter of one atom). The radius is half the diameter.

1. Length = 2.4 cm = 2.4 × 10⁻² m; number of atoms = 2 × 10⁸.
2. Diameter = (2.4 × 10⁻² m)/(2 × 10⁸) = 1.2 × 10⁻¹⁰ m.
3. Radius = diameter/2 = 6.0 × 10⁻¹¹ m = 60 pm (0.060 nm = 0.6 Å).

Extra depth. Note the difference from Ex 2.35: there the diameter was given and the number of atoms was asked; here the number is given and the radius is asked. The value 60 pm is smaller than the covalent radius of carbon (77 pm), because the numbers in the exercise are idealised.

Conclusion: The radius of the carbon atom is 6.0 × 10⁻¹¹ m (60 pm).

Concept. Radius = diameter/2; 1 Å = 10⁻¹⁰ m = 100 pm. The number of atoms in a line = length/diameter.

1. (a) Radius: diameter = 2.6 Å = 260 pm, so radius = 260/2 = 130 pm (1.3 Å).
2. (b) Number of atoms: length = 1.6 cm = 1.6 × 10⁻² m; diameter = 2.6 × 10⁻¹⁰ m. Number = 1.6 × 10⁻²/2.6 × 10⁻¹⁰ = 6.15 × 10⁷.

Extra depth. The radius of 130 pm agrees with the order of magnitude of metallic radii (Zn: about 134 pm). The same unit conversions (Å, pm, nm, cm) appear throughout this chapter: 1 Å = 0.1 nm = 100 pm.

Conclusion: (a) Radius of zinc atom = 130 pm; (b) about 6.15 × 10⁷ zinc atoms fit in 1.6 cm.

Concept. Charge is quantised: the total charge q = n e where e = 1.602 × 10⁻¹⁹ C is the charge of one electron (Millikan). So the number of electrons n = q/e.

1. n = (2.5 × 10⁻¹⁶ C)/(1.602 × 10⁻¹⁹ C per electron).
2. n = 1.56 × 10³ ≈ 1561 electrons (about 1.56 × 10³).

Extra depth. With e = 1.6 × 10⁻¹⁹ C the answer is 1562.5, so the result is correctly quoted as about 1.56 × 10³. A real particle carries a whole number of electrons; the slight non-integer value arises because the given charge is rounded to two significant figures.

Conclusion: The particle contains about 1.56 × 10³ (≈ 1561) electrons.

Concept. Millikan found that the charge on an oil drop is always an integral multiple of the electronic charge: q = n e, n = 1, 2, 3 …, with e = 1.602 × 10⁻¹⁹ C. The X-rays ionise the air; the drop acquires electrons by collision with gaseous ions.

n = q/e = 1.282 × 10⁻¹⁸ C / 1.602 × 10⁻¹⁹ C

1. n = 12.82 × 10⁻¹⁹/1.602 × 10⁻¹⁹ = 8.002.
2. Since n must be an integer, n = 8 electrons.

Extra depth. The fact that the ratio comes out so close to a whole number (8.00) is the experimental evidence for the quantisation of charge; had charge been continuous, every drop would give an arbitrary ratio. Millikan used the charges on many drops and found each was a multiple of the same smallest charge, which he took as e (Section 2.1.3).

Conclusion: The oil drop carries 8 excess electrons (q = 8e).

Concept. The deflection of an α-particle (charge +2e, mass 4 u) arises from the electrostatic repulsion between it and the positive nucleus, and that force is proportional to the product of the charges, 2e × Ze. A heavier nucleus also stays in place when struck, whereas a light nucleus is pushed aside (recoils).

• Much less deflection: the nuclear charge of Al (+13) is much smaller than that of Au (+79), so the repulsive force on the α-particle is about six times weaker; most α-particles pass almost straight through.
• Few or no large-angle deflections: the very rare rebounds at ~180° would essentially disappear, because the α-particle cannot be turned back by such a weakly charged, light nucleus (the nucleus is also pushed aside).
• Consequence for the model: with an aluminium foil the experiment would not so clearly reveal that the positive charge and mass are concentrated in a tiny, massive nucleus — which is why Rutherford, Geiger and Marsden chose a heavy metal such as gold, which can also be beaten into a very thin foil.

Extra depth. In Rutherford scattering the number of particles scattered through a given angle is proportional to Z², so the scattering from Al is smaller than from Au by a factor of (13/79)² ≈ 1/37 per nucleus. The qualitative result — “mostly undeflected” — is what the question expects.

Conclusion: With an aluminium (light-atom) foil the α-particles would be deflected much less — most would pass straight through and large-angle (back) scattering would be practically absent.

Concept. In the notation ^A_ZX the superscript is the mass number A (protons + neutrons) and the subscript is the atomic number Z (number of protons). The atomic number of an element is fixed — it is what defines the element — and always A ≥ Z. For bromine Z = 35.

1. The atomic number of bromine is 35 (35 protons). Mass number = 35 + neutrons; the known isotopes have 44 and 46 neutrons, i.e. A = 79 and 81.
2. The atomic number (subscript) is optional because the symbol of the element fixes it, but the mass number must be correct for an isotope of that element.
3. ³⁵₇₉Br interchanges A and Z; ³⁵Br gives a mass number that does not exist for bromine.

Extra depth. The position matters: A is written at the top left, Z at the bottom left. A symbol such as ³⁵Cl (chlorine-35) is perfectly acceptable because 35 is the mass number of chlorine (Z = 17); the same number 35 appears as the atomic number of bromine.

Conclusion: ⁷⁹₃₅Br and ⁷⁹Br are correct because Z = 35 for Br and A = 79; ³⁵₇₉Br and ³⁵Br are wrong because the atomic number and mass number are interchanged / A = 35 is not a mass number of bromine.

Concept. Mass number A = protons (Z) + neutrons (n). If the neutrons are 31.7% more than the protons, then n = Z + 0.317 Z = 1.317 Z.

A = Z + n = Z + 1.317 Z = 2.317 Z

1. 2.317 Z = 81 ⇒ Z = 81/2.317 = 34.96 ≈ 35.
2. Neutrons n = 81 − 35 = 46.
3. Check: 46/35 = 1.314, i.e. about 31.4–31.7% more neutrons than protons (rounding of the data).
4. Atomic number 35 is bromine. Symbol: ⁸¹₃₅Br.

Extra depth. ⁸¹Br is one of the two stable isotopes of bromine (the other is ⁷⁹Br); the pair occurs in nearly equal abundance (about 50.7% and 49.3%), giving bromine its average atomic mass of 79.9 u.

Conclusion: The element is bromine and the atomic symbol is ⁸¹₃₅Br.

Concept. For an ion carrying a charge, the number of electrons differs from the number of protons: for an anion with one negative charge, electrons = protons + 1. Let the number of electrons be e; then neutrons n = 1.111 e and protons Z = e − 1.

1. Mass number A = Z + n = (e − 1) + 1.111 e = 2.111 e − 1 = 37.
2. 2.111 e = 38 ⇒ e = 38/2.111 = 18.0.
3. Electrons = 18; protons Z = 18 − 1 = 17; neutrons = 37 − 17 = 20 (check: 20/18 = 1.111, i.e. 11.1% more).
4. Z = 17 is chlorine. The species has one extra electron, so it is the chloride ion: ³⁷₁₇Cl⁻.

Extra depth. The chloride ion has 18 electrons, the argon configuration 1s²2s²2p⁶3s²3p⁶ (it is isoelectronic with Ar, K⁺ and Ca²⁺ in Ex 2.22). ³⁷Cl is the heavier stable isotope of chlorine (about 24% natural abundance).

Conclusion: The ion is the chloride ion, ³⁷₁₇Cl⁻.

Concept. For a cation with charge +3, the electrons are three fewer than the protons: protons Z = electrons + 3. Let electrons = e; then neutrons n = 1.304 e and Z = e + 3.

1. A = Z + n = (e + 3) + 1.304 e = 2.304 e + 3 = 56.
2. 2.304 e = 53 ⇒ e = 53/2.304 = 23.0.
3. Electrons = 23; protons Z = 23 + 3 = 26; neutrons = 56 − 26 = 30 (check: 30/23 = 1.304).
4. Z = 26 is iron, so the ion is ⁵⁶₂₆Fe³⁺.

Extra depth. Fe³⁺ has the configuration [Ar]3d⁵ — a half-filled d subshell, which is why iron(III) is a particularly stable oxidation state (Section 2.6.6). ⁵⁶Fe is the most abundant isotope of iron (91.7%).

Conclusion: The ion is ⁵⁶₂₆Fe³⁺ (iron(III) ion).

Concept. In the electromagnetic spectrum (Fig. 2.7) the order of increasing frequency (decreasing wavelength) is: radio waves < microwaves < infrared < visible light < ultraviolet < X-rays < γ-rays/cosmic rays. Place each radiation in its region.

(c) FM radio < (a) microwave oven < (b) amber light < (e) X-rays < (d) cosmic rays

Extra depth. Frequency and energy go together (E = hν), so cosmic rays are the most energetic and FM radio waves the least; wavelength decreases in the same order. Microwave ovens operate near 2.45 GHz, FM radio at about 100 MHz, amber light at about 5 × 10¹⁴ Hz, X-rays at about 3 × 10¹⁸ Hz (λ ≈ 0.1 nm, Section 2.2).

Conclusion: Increasing order of frequency: (c) FM radio < (a) microwave oven < (b) amber light < (e) X-rays < (d) cosmic rays.

Concept. The energy of one photon is E = hc/λ; the total energy of N photons is N hc/λ. Power is energy per unit time (1 W = 1 J s⁻¹). The question does not state the time; as in the standard solution we take the photons to be emitted in one second, so that the numerical value of the energy emitted equals the power (W).

1. λ = 337.1 nm = 3.371 × 10⁻⁷ m.
2. Energy of one photon: E = (6.626 × 10⁻³⁴)(3.0 × 10⁸)/(3.371 × 10⁻⁷) = 5.897 × 10⁻¹⁹ J.
3. Total energy of 5.6 × 10²⁴ photons: 5.6 × 10²⁴ × 5.897 × 10⁻¹⁹ = 3.30 × 10⁶ J.
4. If this is emitted per second: power = 3.30 × 10⁶ J s⁻¹ = 3.30 × 10⁶ W (≈ 3.3 MW).

Extra depth. Data note: the exercise gives no emission time, so the power of 3.3 × 10⁶ W is valid when the photons are emitted in 1 s; for a time t seconds the power would be 3.30 × 10⁶/t W. Such megawatt powers are reached by pulsed lasers during very short pulses. 337.1 nm is an ultraviolet line.

Conclusion: Total energy = 3.30 × 10⁶ J, i.e. a power of about 3.3 × 10⁶ W if the photons are emitted in one second.

Concept. Use c = νλ for the frequency, distance = speed × time for (b), E = hν for the quantum energy, and N = (total energy)/(energy of one quantum) for (d).

1. (a) Frequency: ν = c/λ = 3.0 × 10⁸/(616 × 10⁻⁹) = 4.87 × 10¹⁴ s⁻¹.
2. (b) Distance in 30 s: d = c t = 3.0 × 10⁸ m s⁻¹ × 30 s = 9.0 × 10⁹ m.
3. (c) Energy of a quantum: E = hν = 6.626 × 10⁻³⁴ × 4.87 × 10¹⁴ = 3.23 × 10⁻¹⁹ J.
4. (d) Number of quanta in 2 J: N = 2 J/3.227 × 10⁻¹⁹ J = 6.2 × 10¹⁸.

Extra depth. 616 nm is orange-red, the colour of neon signs. All electromagnetic radiation travels at c in vacuum, so the distance in (b) depends only on time and not on the wavelength. The result N = 6.2 × 10¹⁸ shows how many photons carry just 2 J of light.

Conclusion: (a) 4.87 × 10¹⁴ Hz; (b) 9.0 × 10⁹ m; (c) 3.23 × 10⁻¹⁹ J; (d) 6.2 × 10¹⁸ quanta.

Concept. Number of photons = (total energy received)/(energy of one photon), with E = hc/λ.

1. λ = 600 nm = 6.0 × 10⁻⁷ m.
2. Energy of one photon: E = (6.626 × 10⁻³⁴)(3.0 × 10⁸)/(6.0 × 10⁻⁷) = 3.313 × 10⁻¹⁹ J.
3. N = 3.15 × 10⁻¹⁸ J/3.313 × 10⁻¹⁹ J = 9.5 ≈ 10 photons.

Extra depth. Only about ten photons give a measurable signal — modern detectors can respond to individual photons. The calculation gives 9.5 rather than a whole number because the data are rounded (3.15 × 10⁻¹⁸ J corresponds to between 9 and 10 photons); since photons cannot be split, the answer is quoted as about 10 photons.

Conclusion: The detector receives about 10 photons (calculated value 9.5).

Concept. The energy of the source = (number of photons) × (energy of one photon) = N hν. The question gives no wavelength or frequency, so the standard NCERT solution takes the frequency of the radiation to be the reciprocal of the pulse duration: ν = 1/t.

1. t = 2 ns = 2 × 10⁻⁹ s; ν = 1/t = 1/(2 × 10⁻⁹ s) = 5 × 10⁸ s⁻¹.
2. Energy = N hν = (2.5 × 10¹⁵)(6.626 × 10⁻³⁴ J s)(5 × 10⁸ s⁻¹).
3. Energy = 8.28 × 10⁻¹⁰ J.

Extra depth. Data note: a frequency of 5 × 10⁸ Hz is a radio-frequency value, so this exercise is best read as a dimensional-analysis exercise rather than a realistic optical pulse; the exercise as printed does not provide the wavelength, and the answer 8.28 × 10⁻¹⁰ J follows the textbook’s convention.

Conclusion: Taking ν = 1/(2 ns) = 5 × 10⁸ Hz, the energy of the source is N hν = 8.28 × 10⁻¹⁰ J.

Concept. Each wavelength gives a frequency ν = c/λ. The two closely spaced lines (a doublet) correspond to two slightly different excited-state energies; the energy difference is ΔE = h(ν₁ − ν₂). These are the sodium D₁ and D₂ lines, caused by the spin of the electron (Section 2.6.1).

1. ν₁ = c/λ₁ = 3.0 × 10⁸/(589 × 10⁻⁹) = 5.093 × 10¹⁴ Hz.
2. ν₂ = 3.0 × 10⁸/(589.6 × 10⁻⁹) = 5.088 × 10¹⁴ Hz.
3. Δν = ν₁ − ν₂ = 5.18 × 10¹¹ Hz.
4. ΔE = h Δν = 6.626 × 10⁻³⁴ × 5.18 × 10¹¹ = 3.43 × 10⁻²² J.

Extra depth. The small energy difference of 3.43 × 10⁻²² J (about 2 meV) is the fine-structure splitting of the 3p level of sodium — two energy levels so close that the lines look like one yellow line to the eye. Bohr’s model could not explain such doublets (Section 2.4.2); the electron-spin quantum number m_s does.

Conclusion: ν₁ = 5.093 × 10¹⁴ Hz, ν₂ = 5.088 × 10¹⁴ Hz; the energy difference is ≈ 3.43 × 10⁻²² J.

Concept. The work function W₀ = hν₀ = hc/λ₀. For radiation of wavelength 500 nm the kinetic energy of the photoelectron is K.E. = hc/λ − W₀ and v = √(2 K.E./m_e). Use 1 eV = 1.602 × 10⁻¹⁹ J.

1. W₀ = 1.9 eV = 1.9 × 1.602 × 10⁻¹⁹ = 3.044 × 10⁻¹⁹ J.
2. (b) Threshold frequency: ν₀ = W₀/h = 3.044 × 10⁻¹⁹/6.626 × 10⁻³⁴ = 4.59 × 10¹⁴ Hz.
3. (a) Threshold wavelength: λ₀ = c/ν₀ = 3.0 × 10⁸/4.594 × 10¹⁴ = 6.53 × 10⁻⁷ m = 653 nm.
4. Photon of 500 nm: E = hc/λ = 1.988 × 10⁻²⁵/5.0 × 10⁻⁷ = 3.976 × 10⁻¹⁹ J (2.48 eV > 1.9 eV, so emission occurs).
5. Kinetic energy: K.E. = 3.976 × 10⁻¹⁹ − 3.044 × 10⁻¹⁹ = 9.3 × 10⁻²⁰ J (0.58 eV).
6. Velocity: v = √(2 × 9.32 × 10⁻²⁰/9.109 × 10⁻³¹) = 4.52 × 10⁵ m s⁻¹.

Extra depth. Caesium has the lowest work function among the common metals (Table 2.2 lists K = 2.25 eV and Na = 2.3 eV), so it responds even to red light — hence its use in photocells. Radiation with λ > 653 nm cannot eject electrons however intense it is.

Conclusion: λ₀ = 653 nm, ν₀ = 4.59 × 10¹⁴ Hz; for 500 nm light K.E. = 9.3 × 10⁻²⁰ J and v = 4.52 × 10⁵ m s⁻¹.

Concept. Einstein’s equation: hc/λ = W₀ + ½ m_ev², i.e. ½ m_ev² = hc(1/λ) − W₀. A graph (or any two points) of K.E. against 1/λ is a straight line with slope hc and intercept −W₀. The threshold wavelength is where K.E. = 0, λ₀ = hc/W₀.

Data note on units. The velocities are printed as “v × 10⁻⁵ (cm s⁻¹)”. Read literally (2.55 × 10⁵ cm s⁻¹ = 2.55 × 10³ m s⁻¹) the kinetic energies are 10⁴ times smaller and give an impossible Planck constant (≈ 10⁻³⁸ J s). The data give the correct h and a sensible work function only if v is taken as 2.55 × 10⁵, 4.35 × 10⁵, 5.35 × 10⁵ m s⁻¹, which is how it is solved below.

1. K.E. = ½ m_ev² with m_e = 9.109 × 10⁻³¹ kg: 500 nm → 2.96 × 10⁻²⁰ J; 450 nm → 8.62 × 10⁻²⁰ J; 400 nm → 1.304 × 10⁻¹⁹ J.
2. 1/λ: 2.000 × 10⁶, 2.222 × 10⁶, 2.500 × 10⁶ m⁻¹.
3. Slope of K.E. against 1/λ (least-squares line through the three points) = hc = 1.998 × 10⁻²⁵ J m ⇒ h = slope/c = 6.66 × 10⁻³⁴ J s (the accepted value is 6.626 × 10⁻³⁴ J s).
4. Intercept = −W₀ = −3.66 × 10⁻¹⁹ J, i.e. W₀ = 3.66 × 10⁻¹⁹ J (2.28 eV).
5. Threshold wavelength: λ₀ = hc/W₀ = 1.998 × 10⁻²⁵/3.66 × 10⁻¹⁹ = 5.46 × 10⁻⁷ m = 546 nm.

Extra depth. Two-point check with the 500 nm and 400 nm data: slope = (13.04 − 2.96) × 10⁻²⁰/(0.500 × 10⁶) = 2.02 × 10⁻²⁵ J m ⇒ h = 6.7 × 10⁻³⁴ J s. The result agrees with the accepted h and gives a work function of 2.3 eV, matching Table 2.2 (Na: 2.3 eV) — this confirms the intended unit of v.