Constants used: h = 6.63 × 10⁻³⁴ J s, e = 1.6 × 10⁻¹⁹ C, c = 3 × 10⁸ m/s, ε₀ = 8.85 × 10⁻¹² C² N⁻¹ m⁻², R = 1.097 × 10⁷ m⁻¹.
Exercises
(12.1) Choose the correct alternative from the clues given at the end of each statement:
(a) The size of the atom in Thomson's model is .......... the atomic size in Rutherford's model. (much greater than/no different from/much less than.)
(b) In the ground state of .......... electrons are in stable equilibrium, while in .......... electrons always experience a net force. (Thomson's model/Rutherford's model.)
(c) A classical atom based on .......... is doomed to collapse. (Thomson's model/Rutherford's model.)
(d) An atom has a nearly continuous mass distribution in a .......... but has a highly non-uniform mass distribution in .......... (Thomson's model/Rutherford's model.)
Answer
(e) The positively charged part of the atom possesses most of the mass in .......... (Rutherford's model/both the models.)
Ans. (a) no different from — both models take the atom to be about 10⁻¹⁰ m in size.
(b) Thomson's model; Rutherford's model — in Thomson's model electrons sit at rest at equilibrium points inside the positive sphere, while in Rutherford's model they must revolve, so a centripetal (net) force always acts on them.
(c) Rutherford's model — an orbiting electron is accelerated, radiates energy classically and spirals into the nucleus.
(d) Thomson's model; Rutherford's model — Thomson spreads mass uniformly over the atom, Rutherford concentrates it in a tiny nucleus.
(e) both the models.
(12.2) Suppose you are given a chance to repeat the alpha-particle scattering experiment using a thin sheet of solid hydrogen in place of the gold foil. (Hydrogen is a solid at temperatures below 14 K.) What results do you expect?
Answer
The hydrogen nucleus is a single proton, about four times lighter than an alpha particle, and carries only charge +e (gold carries +79e). A heavy projectile striking a lighter target cannot bounce back; it pushes the proton forward and itself continues nearly undeflected.
So we would observe almost no large-angle scattering and no back-scattering — only small deflections. The experiment would therefore not reveal the small, massive nucleus, and Rutherford's nuclear model could not be confirmed this way.
(12.3) A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom makes a transition from the upper level to the lower level?
Answer
Given ΔE = 2.3 eV = 2.3 × 1.6 × 10⁻¹⁹ J = 3.68 × 10⁻¹⁹ J.
hν = ΔE
⇒ ν = ΔE/h = 3.68 × 10⁻¹⁹/6.63 × 10⁻³⁴
⇒ ν ≈ 5.6 × 10¹⁴ Hz
(12.4) The ground state energy of hydrogen atom is –13.6 eV. What are the kinetic and potential energies of the electron in this state?
Answer
In a Bohr orbit, K = −E and U = 2E (since U = −2K and E = K + U = −K).
K = −(−13.6 eV) → K = 13.6 eV
U = 2 × (−13.6 eV) → U = −27.2 eV
(12.5) A hydrogen atom initially in the ground level absorbs a photon, which excites it to the n = 4 level. Determine the wavelength and frequency of photon.
Answer
Eₙ = −13.6/n² eV
⇒ E₁ = −13.6 eV, E₄ = −13.6/16 = −0.85 eV
ΔE = E₄ − E₁ = 12.75 eV = 12.75 × 1.6 × 10⁻¹⁹ = 2.04 × 10⁻¹⁸ J
ν = ΔE/h = 2.04 × 10⁻¹⁸/6.63 × 10⁻³⁴
⇒ ν ≈ 3.1 × 10¹⁵ Hz
λ = c/ν = 3 × 10⁸/3.08 × 10¹⁵
⇒ λ ≈ 9.7 × 10⁻⁸ m ≈ 97 nm (ultraviolet)
(12.6) (a) Using the Bohr's model calculate the speed of the electron in a hydrogen atom in the n = 1, 2, and 3 levels. (b) Calculate the orbital period in each of these levels.
Answer
(a) In Bohr's model vₙ = e²/(2ε₀hn).
v₁ = (1.6 × 10⁻¹⁹)²/(2 × 8.85 × 10⁻¹² × 6.63 × 10⁻³⁴ × 1)
⇒ v₁ ≈ 2.18 × 10⁶ m/s
v₂ = v₁/2 → v₂ ≈ 1.09 × 10⁶ m/s
v₃ = v₁/3 → v₃ ≈ 7.27 × 10⁵ m/s
(b) Orbital period Tₙ = 2πrₙ/vₙ with rₙ = n²r₁, r₁ = 0.53 × 10⁻¹⁰ m. Since r ∝ n² and v ∝ 1/n, T ∝ n³.
T₁ = 2π × 0.53 × 10⁻¹⁰/2.18 × 10⁶
⇒ T₁ ≈ 1.53 × 10⁻¹⁶ s
T₂ = 8T₁ → T₂ ≈ 1.22 × 10⁻¹⁵ s
T₃ = 27T₁ → T₃ ≈ 4.12 × 10⁻¹⁵ s
(12.7) The radius of the innermost electron orbit of a hydrogen atom is 5.3 × 10⁻¹¹ m. What are the radii of the n = 2 and n = 3 orbits?
Answer
rₙ = n²r₁, with r₁ = 5.3 × 10⁻¹¹ m.
r₂ = 4 × 5.3 × 10⁻¹¹
⇒ r₂ = 2.12 × 10⁻¹⁰ m
r₃ = 9 × 5.3 × 10⁻¹¹
⇒ r₃ = 4.77 × 10⁻¹⁰ m
(12.8) A 12.5 eV electron beam is used to bombard gaseous hydrogen at room temperature. What series of wavelengths will be emitted?
Answer
At room temperature the atoms are in the ground state (−13.6 eV). Energy levels: E₂ = −3.4 eV, E₃ = −1.51 eV, E₄ = −0.85 eV.
Energy needed to reach n = 3: 13.6 − 1.51 = 12.09 eV (≤ 12.5 eV, possible). To reach n = 4: 12.75 eV (> 12.5 eV, not possible). So atoms are excited at most to n = 3.
Possible transitions and wavelengths, using 1/λ = R(1/n₁² − 1/n₂²):
n = 3
⇒ 1: 1/λ = R(1 − 1/9)
⇒ λ ≈ 102.6 nm (Lyman series)
n = 2
⇒ 1: 1/λ = R(1 − 1/4)
⇒ λ ≈ 121.6 nm (Lyman series)
n = 3
⇒ 2: 1/λ = R(1/4 − 1/9)
⇒ λ ≈ 656.3 nm (Balmer series)
Hence two lines of the Lyman series and one line of the Balmer series are emitted.
(12.9) In accordance with the Bohr's model, find the quantum number that characterises the earth's revolution around the sun in an orbit of radius 1.5 × 10¹¹ m with orbital speed 3 × 10⁴ m/s. (Mass of earth = 6.0 × 10²⁴ kg.)
Answer
Bohr's quantisation: mvr = nh/2π
⇒ n = 2πmvr/h
n = 2π × 6.0 × 10²⁴ × 3 × 10⁴ × 1.5 × 10¹¹/6.63 × 10⁻³⁴
⇒ n ≈ 2.6 × 10⁷⁴
Such an enormous quantum number means the allowed orbits are practically continuous, so quantisation of the earth's motion is unobservable and classical mechanics applies.
Important Extra Questions
What did Rutherford conclude from the fact that a very small fraction of alpha particles were scattered through large angles?
Answer
That the positive charge and almost all the mass of the atom are concentrated in a tiny central region, the nucleus, while most of the atom is empty space.
State Bohr's three postulates.
Answer
(i) Electrons revolve in certain stationary orbits without radiating energy. (ii) Only those orbits are allowed in which angular momentum is an integral multiple of h/2π (mvr = nh/2π). (iii) A photon is emitted or absorbed only when the electron jumps between orbits, with hν = Eᵢ − E_f.
What is the distance of closest approach?
Answer
It is the minimum distance d from the nucleus reached by an alpha particle in a head-on collision, where its kinetic energy becomes entirely electrostatic potential energy: K = (1/4πε₀)(2e × Ze)/d.
What is the ionisation energy of hydrogen in its ground state and in the first excited state?
Answer
Ground state: 13.6 eV. First excited state (n = 2): 13.6/4 = 3.4 eV.
Explain how de Broglie's hypothesis explains Bohr's second postulate.
Answer
If the electron is a standing wave around the orbit, the circumference must hold a whole number of wavelengths: 2πr = nλ = nh/mv, which gives mvr = nh/2π, Bohr's quantisation condition.