How Boards Test This Chapter

Atoms is a dependable 4-6 mark contributor to CBSE papers. The recurring demands:

  • 1 mark: define impact parameter / distance of closest approach; state any Bohr postulate; ratio of radii or energies of given levels; which series lies in the visible region.
  • 2 marks: why is Rutherford's atom unstable classically; two conclusions from the Geiger-Marsden experiment; calculate a wavelength or excitation energy; de Broglie's standing-wave condition.
  • 3 marks: state the three Bohr postulates and derive rnr_n or EnE_n; obtain the expression for the distance of closest approach; explain hydrogen's spectral series using the energy-level diagram; derive Bohr's quantisation from de Broglie's hypothesis.
  • 5 marks: full account of the alpha-scattering experiment (set-up, observations, conclusions) plus the nuclear model; or the complete Bohr treatment — postulates → radii → energies → spectra — with the energy-level diagram.

The questions below are Board-style previous-year questions with full step-by-step solutions embedded in the explanations. Attempt each before reading its solution. Years are attached only where attribution is certain; otherwise questions are tagged simply [CBSE Board].

The Definitions and Statements Boards Reward (Model Answers)

Impact parameter: the perpendicular distance of the initial velocity vector of the alpha particle from the centre of the nucleus.

Distance of closest approach: the centre-to-centre distance at which a head-on alpha particle momentarily comes to rest, its entire kinetic energy converted to electrostatic potential energy: d=2Ze24πε0Kd = \dfrac{2Ze^2}{4\pi\varepsilon_0 K}.

Bohr's postulates: (1) electrons revolve in certain stable (stationary) orbits without emitting radiant energy; (2) the stationary orbits are those with angular momentum L=nh2πL = \dfrac{nh}{2\pi}; (3) in a transition between orbits, a photon of energy hν=Ei−Efh\nu = E_i - E_f is emitted (or absorbed for the upward jump).

Key derived results: rn=n2a0r_n = n^2a_0, a0=5.29×10−11a_0 = 5.29 \times 10^{-11} m; En=−13.6n2E_n = -\dfrac{13.6}{n^2} eV; ionisation energy 13.6 eV; 1λ=R(1nf2−1ni2)\dfrac{1}{\lambda} = R\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right).

de Broglie's explanation: allowed orbits accommodate an integral number of de Broglie wavelengths, 2πrn=nλ2\pi r_n = n\lambda; with λ=h/mv\lambda = h/mv this reproduces mvrn=nh/2πmvr_n = nh/2\pi.

[Board Tip] Marks leak in predictable places: forgetting 'without emitting radiant energy' in postulate 1; writing L = nh (missing the 2π); drawing energy-level diagrams with equal spacing (levels must crowd upward toward E = 0); and confusing the impact parameter with the closest approach.

Board PYQ Set A: Short Answer (1-2 marks)

PYQ 1. Define the distance of closest approach. An alpha particle of kinetic energy K approaches a nucleus of charge Ze head-on. Write the expression for this distance. [CBSE Board]

Solution:

  1. It is the centre-to-centre separation at which the head-on alpha momentarily stops, all its kinetic energy converted to electrostatic potential energy.
  2. Energy conservation: K=2Ze24πε0dK = \dfrac{2Ze^2}{4\pi\varepsilon_0 d}, so d=2Ze24πε0Kd = \dfrac{2Ze^2}{4\pi\varepsilon_0 K}.

PYQ 2. What is the impact parameter for a head-on collision, and what is the corresponding scattering angle? [CBSE Board]

Solution:

  1. For a head-on collision the impact parameter is zero (minimum).
  2. The alpha rebounds along its incoming line: scattering angle θ≈π\theta \approx \pi (180 degrees).

PYQ 3. Write two important conclusions drawn from the Geiger-Marsden experiment. [CBSE Board]

Solution:

  1. Most of the atom is empty space (most alphas pass undeflected).
  2. The entire positive charge and most of the mass are concentrated in a tiny central nucleus of size ~10−1510^{-15}-10−1410^{-14} m (rare large-angle scattering demands an intense central repulsion).

PYQ 4. The radius of the innermost orbit of hydrogen is 5.3×10−115.3 \times 10^{-11} m. What is the radius of the third orbit? [CBSE Board]

Solution:

  1. rn=n2a0=9×5.3×10−11r_n = n^2a_0 = 9 \times 5.3 \times 10^{-11}.
  2. r3=4.77×10−10r_3 = 4.77 \times 10^{-10} m.

PYQ 5. State Bohr's quantisation condition for angular momentum. How did de Broglie explain it? [CBSE Board]

Solution:

  1. Condition: L=mvrn=nh2πL = mvr_n = \dfrac{nh}{2\pi}, n = 1, 2, 3…
  2. de Broglie: the orbiting electron is a particle wave forming a standing wave on the orbit; this requires 2πrn=nλ2\pi r_n = n\lambda (integral wavelengths fitting the circumference).
  3. Substituting λ=hmvn\lambda = \dfrac{h}{mv_n}: 2πrn=nhmvn2\pi r_n = \dfrac{nh}{mv_n}, i.e. mvnrn=nh2πmv_nr_n = \dfrac{nh}{2\pi} — Bohr's condition derived.

PYQ 6. Why is the classical (Rutherford) model unable to account for the stability of the atom? [CBSE Board]

Solution:

  1. The orbiting electron is centripetally accelerated; classical electromagnetic theory requires an accelerating charge to radiate energy continuously.
  2. Losing energy, the electron's orbit shrinks continuously and it spirals into the nucleus — the atom cannot be stable.
  3. (The changing revolution frequency would also give a continuous spectrum, contradicting observed line spectra.)

PYQ 7. The ground-state energy of hydrogen is -13.6 eV. What are the kinetic and potential energies of the electron in this state? [CBSE Board]

Solution:

  1. K = -E = +13.6 eV.
  2. U = 2E = -27.2 eV (NCERT Exercise 12.4).

Board PYQ Set B: Standard Numericals (2-3 marks)

PYQ 8. A difference of 2.3 eV separates two energy levels in an atom. What is the frequency of radiation emitted when the atom transits from the upper to the lower level? [CBSE Board]

Solution:

  1. hν=ΔE=2.3×1.6×10−19h\nu = \Delta E = 2.3 \times 1.6 \times 10^{-19} J.
  2. ν=3.68×10−196.63×10−34≈5.6×1014\nu = \dfrac{3.68 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 5.6 \times 10^{14} Hz.

PYQ 9. A hydrogen atom in the ground level absorbs a photon and rises to n = 4. Determine the photon's wavelength and frequency. [CBSE Board]

Solution:

  1. ΔE=E4−E1=−0.85+13.6=12.75\Delta E = E_4 - E_1 = -0.85 + 13.6 = 12.75 eV.
  2. λ=124012.75≈97.3\lambda = \dfrac{1240}{12.75} \approx 97.3 nm; ν=ΔEh≈3.1×1015\nu = \dfrac{\Delta E}{h} \approx 3.1 \times 10^{15} Hz.

PYQ 10. Calculate the shortest wavelength of the Balmer series. In which region of the spectrum does it lie? [CBSE Board]

Solution:

  1. Series limit: E=13.64=3.4E = \dfrac{13.6}{4} = 3.4 eV.
  2. λ=12403.4≈365\lambda = \dfrac{1240}{3.4} \approx 365 nm — at the violet edge, bordering the near ultraviolet.

PYQ 11. The electron in a hydrogen atom jumps from n = 3 to n = 2. Find the wavelength of the emitted line and name its series. [CBSE Board]

Solution:

  1. E=13.6(14−19)=1.89E = 13.6\left(\dfrac{1}{4} - \dfrac{1}{9}\right) = 1.89 eV.
  2. λ=12401.89≈656\lambda = \dfrac{1240}{1.89} \approx 656 nm — the first line (Hα\alpha) of the Balmer series, visible red.

PYQ 12. Find the ratio of the energies of the hydrogen levels n = 2 and n = 4. [CBSE Board]

Solution:

  1. En∝−1n2E_n \propto -\dfrac{1}{n^2}: E2E4=164=4\dfrac{E_2}{E_4} = \dfrac{16}{4} = 4.
  2. E2:E4=4:1E_2 : E_4 = 4 : 1 (both negative: -3.4 eV and -0.85 eV).

Board PYQ Set C: Long-Answer Patterns (3-5 marks)

PYQ 13. State Bohr's three postulates and use them to obtain the expression for the total energy of the electron in the nth orbit of hydrogen. [CBSE Board]

Solution (marking-scheme outline):

  1. Postulates (as in the model answers above) — 1½ marks.
  2. Force balance: e24πε0r2=mv2r\dfrac{e^2}{4\pi\varepsilon_0 r^2} = \dfrac{mv^2}{r}; quantisation: mvr=nh2πmvr = \dfrac{nh}{2\pi}.
  3. Eliminate v: rn=n2h2ε0πme2r_n = \dfrac{n^2h^2\varepsilon_0}{\pi me^2}.
  4. Substitute in E=−e28πε0rE = -\dfrac{e^2}{8\pi\varepsilon_0 r}: En=−me48n2ε02h2=−13.6n2E_n = -\dfrac{me^4}{8n^2\varepsilon_0^2h^2} = -\dfrac{13.6}{n^2} eV. ∎

PYQ 14. Using the energy-level diagram of hydrogen, explain the origin of the Lyman, Balmer and Paschen series. Which lies in the visible region? [CBSE Board]

Solution:

  1. Draw levels -13.6, -3.4, -1.51, -0.85 eV, crowding toward E = 0.
  2. Lyman: all downward transitions terminating on n = 1 (ultraviolet). Balmer: terminating on n = 2 (visible). Paschen: terminating on n = 3 (infrared).
  3. Arrows drawn from higher levels to each floor level; the Balmer series is the visible one.

PYQ 15. Describe the Geiger-Marsden experiment with a labelled diagram. State the observations and the conclusions leading to the nuclear model. [CBSE Board]

Solution (outline the examiner expects):

  1. Diagram: Bi-214 alpha source, lead collimator, thin gold foil (2.1×10−72.1 \times 10^{-7} m), rotatable ZnS-screen + microscope detector, vacuum chamber.
  2. Observations: most alphas undeviated; ~0.14% beyond 1 degree; ~1 in 8000 beyond 90 degrees.
  3. Conclusions: atom mostly empty; entire positive charge + most mass in a nucleus of ~10−1510^{-15}-10−1410^{-14} m; electrons revolve around it; Coulomb repulsion from this concentrated charge explains the rare backscattering.

PYQ 16. (a) Show that the total energy of the electron in the nth orbit is En=−13.6n2E_n = -\dfrac{13.6}{n^2} eV. (b) Hence find the energy required to move the electron from n = 1 to n = 2. What wavelength of light achieves this? [CBSE Board]

Solution:

  1. (a) as in PYQ 13 steps 2-4.
  2. (b) ΔE=−3.4+13.6=10.2\Delta E = -3.4 + 13.6 = 10.2 eV.
  3. λ=124010.2≈121.6\lambda = \dfrac{1240}{10.2} \approx 121.6 nm (ultraviolet, Lyman-α\alpha absorption).