How NEET Tests This Chapter

Atoms contributes 1-2 questions to nearly every NEET paper, from a compact, predictable pool:

  1. Direct plug-ins: En=−13.6/n2E_n = -13.6/n^2 eV; rn=n2a0r_n = n^2a_0; vn∝1/nv_n \propto 1/n; L=nh/2πL = nh/2\pi.
  2. Excitation & ionisation: 10.2 eV, 12.09 eV, 12.75 eV; ionisation from excited states; Z2Z^2 scaling for He+^+ and Li2+^{2+}.
  3. Spectral series: which series/region; longest/shortest lines; wavelength ratios; line counting.
  4. Scattering: conclusions of Geiger-Marsden; impact parameter vs angle; closest-approach scaling.
  5. Statement recall: Bohr postulates; de Broglie's standing-wave explanation; limitations of the model.

Everything below is NEET-style previous-year material with fully worked explanations. Years are attached only where attribution is certain; otherwise the tag is the generic [NEET].

Constants: a0=0.529a_0 = 0.529 angstrom; 13.6 eV; hc ≈ 1240 eV nm; R = 1.097×1071.097 \times 10^7 m−1^{-1}.

NEET PYQ Worked Set A: Levels & Scaling

PYQ 1. The ratio of the ionisation energy of H and He+^+ is: [NEET]

Solution:

  1. Ionisation ∝ Z2Z^2 (ground states): H : He+^+ = 1 : 4.
  2. Numerically 13.6 eV vs 54.4 eV.

PYQ 2. The total energy of the electron in hydrogen's second excited state is: [NEET]

Solution:

  1. Second excited state = n = 3.
  2. E3=−13.69=−1.51E_3 = -\dfrac{13.6}{9} = -1.51 eV.

PYQ 3. The radius of hydrogen's n = 2 orbit, given Bohr radius a0a_0: [NEET]

Solution:

  1. rn=n2a0=4a0≈2.12×10−10r_n = n^2a_0 = 4a_0 \approx 2.12 \times 10^{-10} m.

PYQ 4. The angular momentum of the electron in the third orbit: [NEET]

Solution:

  1. L=3h2π=3×6.63×10−346.28≈3.17×10−34L = \dfrac{3h}{2\pi} = \dfrac{3 \times 6.63 \times 10^{-34}}{6.28} \approx 3.17 \times 10^{-34} J s.

PYQ 5. The ratio of kinetic energy to total energy of an electron in a Bohr orbit: [NEET]

Solution:

  1. K = -E always → K : E = -1 : 1 (i.e. |K| = |E|, opposite signs).
  2. Also useful: K : U = -1 : 2.

NEET PYQ Worked Set B: Transitions & Series

PYQ 6. Which transition of hydrogen emits the photon of maximum energy? (a) 2→1, (b) 3→2, (c) 4→3, (d) ∞→1. [NEET]

Solution:

  1. Energies: 10.2, 1.89, 0.66, 13.6 eV.
  2. ∞ → 1 (the Lyman limit, 13.6 eV) wins.

PYQ 7. The wavelength of the first line of the Balmer series is 656 nm. The wavelength of the first Lyman line is: [NEET]

Solution:

  1. Ratio λLαλBα=527\dfrac{\lambda_{L\alpha}}{\lambda_{B\alpha}} = \dfrac{5}{27}.
  2. λLα=656×527≈121.5\lambda_{L\alpha} = 656 \times \dfrac{5}{27} \approx 121.5 nm.

PYQ 8. Hydrogen atoms are excited to n = 4. The number of emitted spectral lines is: [NEET]

Solution:

  1. N=n(n−1)2=4×32=6N = \dfrac{n(n-1)}{2} = \dfrac{4 \times 3}{2} = 6.

PYQ 9. The Brackett series of hydrogen lies in which region? [NEET]

Solution:

  1. Brackett ends on n = 4 — small energy gaps.
  2. Infrared (first line 4051 nm). Only Balmer is visible.

PYQ 10. In which Bohr transition of He+^+ is the emitted wavelength the same as hydrogen's 2 → 1 line? [NEET]

Solution:

  1. Preserve Z/n: hydrogen (2, 1) → He+^+ (4, 2).
  2. Check: Z2(1nf2−1ni2)=4(14−116)=34Z^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) = 4\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = \dfrac{3}{4} = hydrogen's bracket. ✔

NEET PYQ Worked Set C: Scattering, Postulates & Concepts

PYQ 11. In the alpha-scattering experiment, on decreasing the impact parameter the scattering angle: [NEET]

Solution:

  1. Smaller b brings the alpha closer to the nucleus → stronger repulsion.
  2. The scattering angle increases (b = 0 gives θ ≈ 180 degrees).

PYQ 12. The significance of the negative total energy of the orbiting electron: [NEET]

Solution:

  1. The electron is bound to the nucleus; energy |E| must be supplied to free it.
  2. Positive E would mean an unbound (open, escaping) trajectory.

PYQ 13. When the atom's electron jumps to a lower orbit, the atom's angular momentum change in a 4 → 2 transition is: [NEET]

Solution:

  1. ΔL=(4−2)h2π=hπ\Delta L = (4 - 2)\dfrac{h}{2\pi} = \dfrac{h}{\pi}.
  2. Angular momentum decreases in steps of h/2πh/2\pi per unit change in n; the difference is carried by the emitted photon and recoil.

PYQ 14. The minimum energy an electron beam must have to produce ALL lines of hydrogen's spectrum from ground-state atoms: [NEET Pattern]

Solution:

  1. 'All lines' requires reaching arbitrarily high n — effectively ionisation-level energy.
  2. Minimum = 13.6 eV (approaching it excites all levels; at or above it, recombination through every level lights every series).

PYQ 15. Bohr's model cannot explain: (a) hydrogen's line frequencies, (b) helium's spectrum, (c) hydrogen's ionisation energy, (d) the Balmer series. [NEET]

Solution:

  1. Frequencies, ionisation energy and the Balmer series are its successes.
  2. Helium's spectrum (two electrons — electron-electron forces omitted) defeats it. Answer: (b).