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The Vault
RowQ
The Vault
CBSE Class 12 Physics · 11 questions · 26 marks
This chapter traces how physicists figured out what an atom actually looks like inside, from Rutherford's startling scattering experiment to Bohr's audacious rule that angular momentum comes in fixed steps. The reward for accepting Bohr's postulates is a formula that predicts, almost exactly, every spectral line of hydrogen ever measured.
In Rutherford's alpha-particle scattering experiment, the small fraction of alpha particles that bounce back at angles close to 180° indicate that:
Answer
Most of the atom's mass and all its positive charge are concentrated in a tiny nucleus — only a very close, essentially head-on encounter with a small, massive, positively charged core could exert enough repulsive Coulomb force to reverse the path of a fast, relatively heavy alpha particle; a spread-out positive charge (as in Thomson's earlier model) could never produce such large-angle scattering.
According to Bohr's model, the radius of the nth orbit of hydrogen is proportional to:
Answer
n² — from r_n = n² × 0.529 Å, the orbit radius grows with the square of the principal quantum number. So the n = 2 orbit has four times the radius of the ground state, and higher orbits are spaced increasingly far apart from one another.
The energy required to excite a hydrogen atom from its ground state (n = 1) to the first excited state (n = 2) is closest to:
Answer
10.2 eV — E_n = -13.6/n² eV, so E₁ = -13.6 eV and E₂ = -13.6/4 = -3.4 eV. The excitation energy is E₂ - E₁ = -3.4 - (-13.6) = 10.2 eV, the energy that must be absorbed for the electron to jump from the ground state to the first excited state.
A hydrogen atom in an excited state emits a photon belonging to the Balmer series. This means the electron has made a transition ending at:
Answer
n = 2 — the Balmer series consists of all transitions from higher levels (n = 3, 4, 5, …) down to the n = 2 level, and these transitions produce spectral lines mostly in the visible range, which is why the Balmer series was the first to be observed and catalogued historically.
Assertion (A): Bohr's model could not be extended to explain accurately the spectra of atoms with more than one electron, such as helium. Reason (R): In multi-electron atoms, mutual repulsion between electrons and more complex electron-electron interactions are not accounted for by the simple Coulomb attraction and angular momentum quantisation rule used for a single electron orbiting a nucleus.
Answer
Both A and R are true and R is the correct explanation of A — Bohr's model was built around a single electron moving under the Coulomb attraction of a nucleus, with an ad hoc quantisation rule. Multi-electron systems involve extra electron-electron repulsions that this simple picture ignores, and the model gives incorrect results for such atoms, a limitation resolved only by full quantum mechanics.
State Bohr's quantisation condition for angular momentum and use it, along with the Coulomb force providing the centripetal force, to write the two starting equations needed to derive the radius of the nth orbit (derivation of the final formula is not required).
Answer
Bohr's quantisation condition: the angular momentum of the electron in a stationary orbit is an integral multiple of h/2π, that is L = mvr_n = nh/2π, where n = 1, 2, 3, … is the principal quantum number. The two starting equations: (i) the Coulomb attraction supplies the centripetal force for the electron's circular orbit, (1/4πε₀)(e²/r_n²) = mv²/r_n; (ii) the quantisation condition itself, mvr_n = nh/2π. Solving these two equations simultaneously for r_n and v eliminates v and yields the formula for the allowed orbit radius r_n.
Calculate the wavelength of the spectral line emitted when a hydrogen electron falls from n = 3 to n = 2 (this is the first line of the Balmer series, Rydberg constant R = 1.097×10⁷ m⁻¹).
Answer
Using 1/λ = R(1/n_f² - 1/n_i²) with n_f = 2, n_i = 3: 1/λ = 1.097×10⁷ × (1/4 - 1/9) 1/4 - 1/9 = 9/36 - 4/36 = 5/36. 1/λ = 1.097×10⁷ × 5/36 = 1.097×10⁷ × 0.1389 = 1.524×10⁶ m⁻¹. λ = 1/1.524×10⁶ = 6.56×10⁻⁷ m = 656 nm. This is the well-known red H-alpha line of the Balmer series.
Define ionisation energy. Calculate the ionisation energy, in electron volts, needed to remove the electron from a hydrogen atom that is already in its first excited state (n = 2).
Answer
Ionisation energy is the minimum energy needed to completely remove an electron from an atom, taking it from its present bound orbit to n = ∞ (E = 0), leaving the atom ionised. For n = 2, E₂ = -13.6/4 = -3.4 eV. Ionisation energy from this state = E_∞ - E₂ = 0 - (-3.4) = 3.4 eV, considerably less than the 13.6 eV needed to ionise from the ground state, since the electron is already less tightly bound in an excited state.
Using Bohr's postulates, derive the expression for the total energy of an electron in the nth orbit of a hydrogen atom, E_n = -13.6/n² eV. Hence find the energy released, in eV, when an electron transitions from n = 4 to n = 1.
Answer
From the Coulomb force providing centripetal force: (1/4πε₀)(e²/r_n²) = mv_n²/r_n, giving mv_n² = e²/(4πε₀r_n). From Bohr's quantisation: mv_n r_n = nh/2π, so v_n = nh/(2πmr_n). Substituting v_n into the force equation and solving for r_n gives r_n = n²h²ε₀/(πme²) (radius formula, obtained by combining the two above equations). Total energy is kinetic plus potential: E_n = ½mv_n² + [-e²/(4πε₀r_n)] (Coulomb potential energy is negative, attractive). From the force equation, ½mv_n² = e²/(8πε₀r_n), so E_n = e²/(8πε₀r_n) - e²/(4πε₀r_n) = -e²/(8πε₀r_n). Substituting r_n = n²h²ε₀/(πme²): E_n = -e²/(8πε₀) × πme²/(n²h²ε₀) = -me⁴/(8ε₀²n²h²). Evaluating the constants for hydrogen gives the standard numerical result: E_n = -13.6/n² eV. Numerical: E₄ = -13.6/16 = -0.85 eV. E₁ = -13.6 eV. Energy released, ΔE = E₄ - E₁ in magnitude for the photon = E₁ (final, lower) taken correctly as |E_i - E_f|: photon energy = E_i - E_f = -0.85 - (-13.6) = 12.75 eV. So a photon of energy 12.75 eV is emitted as the electron falls from n = 4 to the ground state.
(a) Explain the major shortcomings of Rutherford's nuclear model of the atom that Bohr's postulates were designed to resolve. (b) A hydrogen atom initially in its ground state absorbs a photon of energy 12.09 eV. Determine the principal quantum number of the excited state to which the electron jumps, and list all the possible wavelengths of light that could then be emitted as the atom returns to the ground state (give the transitions, not full numerical wavelengths).
Answer
(a) Rutherford's model pictured electrons orbiting the nucleus like planets around the sun, held by Coulomb attraction. Classically, however, an accelerating charge (and an electron in circular orbit is always accelerating, changing direction continuously) must continuously radiate electromagnetic energy. Losing energy this way, the electron would spiral inward and crash into the nucleus within a fraction of a second, making stable atoms impossible according to classical physics. Further, since the electron would radiate at a continuously changing frequency as it spiralled in, classical theory predicted a continuous emission spectrum, whereas real atoms are observed to emit and absorb light only at sharp, discrete wavelengths (line spectra). Bohr resolved both issues with his postulates: certain orbits are simply declared stable and non-radiating (stationary states), and radiation is emitted or absorbed only in discrete jumps between these orbits, naturally producing a line spectrum rather than a continuous one and preventing the predicted collapse. (b) E₁ = -13.6 eV. After absorbing 12.09 eV, the electron's new energy is E = -13.6 + 12.09 = -1.51 eV. Since E_n = -13.6/n², n² = 13.6/1.51 = 9.0, so n = 3. The electron jumps to n = 3. Returning to the ground state, it can do so directly or via n = 2, giving three possible transitions and hence three possible emitted wavelengths: n = 3 → n = 1, n = 3 → n = 2, and n = 2 → n = 1.
Read the passage and answer the questions that follow: Astronomers analysing light from a distant hydrogen gas cloud detect an emission line they identify as belonging to the Lyman series (all transitions ending at n = 1). The observed line corresponds to the transition from n = 2 to n = 1, the same transition that would occur in a hydrogen atom at rest in the laboratory. They wish to use this to demonstrate the basic Bohr formula before analysing any Doppler shift. (a) Calculate the energy, in eV, of the photon emitted in the n = 2 to n = 1 transition. (b) Calculate the wavelength of this photon (hc = 1240 eV nm). (c) State which part of the electromagnetic spectrum this wavelength falls into. (d) Explain why the Lyman series lines all fall in the ultraviolet, while the Balmer series lines fall in the visible range, referring to the energy differences involved.
Answer
(a) E₁ = -13.6 eV, E₂ = -13.6/4 = -3.4 eV. Photon energy = E₂ - E₁ (magnitude of energy released) = -3.4 - (-13.6) = 10.2 eV. (b) λ = hc/E = 1240/10.2 = 121.6 nm. (c) This wavelength (121.6 nm) lies in the ultraviolet region of the electromagnetic spectrum, well below the visible range (roughly 400-700 nm). (d) Lyman series transitions end at n = 1, the most tightly bound level, so the energy differences E_i - E_1 are large (of order 10 eV or more), corresponding by E = hc/λ to short wavelengths in the ultraviolet. Balmer series transitions end at n = 2, a less tightly bound level, so the energy differences are smaller (a few eV), corresponding to longer wavelengths that happen to fall within the visible range — which is precisely why the Balmer lines are the ones historically observed first in laboratory spectroscopes without any special ultraviolet-sensitive equipment.
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