How JEE Tests This Chapter

Atoms supplies 1-2 questions to nearly every JEE Main paper, drawn from tight templates:

  1. Scaling drills: r∝n2Zr \propto \dfrac{n^2}{Z}, v∝Znv \propto \dfrac{Z}{n}, E∝Z2n2E \propto \dfrac{Z^2}{n^2}, T∝n3Z2T \propto \dfrac{n^3}{Z^2}, L∝nL \propto n — ratios across levels and across hydrogen-like ions.
  2. Transition energetics: E=13.6Z2(1nf2−1ni2)E = 13.6Z^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) eV with the 1240 eV nm converter; matched-wavelength puzzles between H, He+^+, Li2+^{2+}.
  3. Series arithmetic: first lines, limits, wavelength ratios (R always cancels), line counting n(n−1)2\dfrac{n(n-1)}{2}.
  4. Closest approach: d∝qZKd \propto \dfrac{qZ}{K} scalings with projectile swaps.
  5. Twists: photon vs electron-beam excitation, atom recoil on emission, de Broglie wavelength in orbit (λn=2πna0\lambda_n = 2\pi na_0), correspondence-principle checks.

All questions below are JEE-style previous-year questions with worked solutions. Years are attached only where attribution is certain; otherwise tags read [JEE Mains] / [JEE Advanced].

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}; v1=2.18×106v_1 = 2.18 \times 10^6 m/s = c/137.

JEE PYQ Worked Set A: Scaling and Levels

PYQ 1. The ratio of the de Broglie wavelengths of the electron in the first and second Bohr orbits of hydrogen is: [JEE Mains]

Solution:

  1. λn=2πna0∝n\lambda_n = 2\pi na_0 \propto n.
  2. λ1:λ2=1:2\lambda_1 : \lambda_2 = 1 : 2.

PYQ 2. The time period of the electron in hydrogen's nth orbit is proportional to: [JEE Mains]

Solution:

  1. T=2πrnvnT = \dfrac{2\pi r_n}{v_n} with rn∝n2r_n \propto n^2, vn∝1nv_n \propto \dfrac{1}{n}.
  2. T∝n3T \propto n^3 — the Bohr-Kepler law.

PYQ 3. The magnetic moment due to the electron's orbital motion in the nth orbit is proportional to: [JEE Advanced flavour]

Solution:

  1. Current i=eT∝1n3i = \dfrac{e}{T} \propto \dfrac{1}{n^3}; area A=πrn2∝n4A = \pi r_n^2 \propto n^4.
  2. Moment μ=iA∝n\mu = iA \propto n. (Indeed μ=neh4πm\mu = \dfrac{neh}{4\pi m} — n Bohr magnetons.)
  3. Takeaway: the current-loop picture of the orbit links this chapter to magnetism — a favourite crossover.

PYQ 4. If the electron in hydrogen is replaced by a muon (mass 207 mem_e), the ground-state radius becomes: [JEE Mains]

Solution:

  1. a0∝1ma_0 \propto \dfrac{1}{m} (mass sits in the denominator of rnr_n).
  2. r=0.529207r = \dfrac{0.529}{207} angstrom ≈2.56×10−13\approx 2.56 \times 10^{-13} m.
  3. Takeaway: muonic hydrogen is ~200× smaller; energies scale UP by the mass ratio (E∝mE \propto m) — both scalings are asked.

JEE PYQ Worked Set B: Transitions and Series

PYQ 5. Hydrogen's electron jumps from n = 4 to n = 1. The ratio of the emitted photon's energy to the n = 4 ionisation energy is: [JEE Mains]

Solution:

  1. Photon: 13.6(1−116)=13.6×151613.6\left(1 - \dfrac{1}{16}\right) = 13.6 \times \dfrac{15}{16} eV.
  2. Ionisation from n = 4: 13.616\dfrac{13.6}{16} eV.
  3. Ratio = 15 : 1.

PYQ 6. The wavelength of Lyman-alpha for a hydrogen-like ion is 30.4 nm. Identify the ion. [JEE Mains]

Solution:

  1. Hydrogen's Lyman-alpha: 121.6 nm. Scaling: λ∝1Z2\lambda \propto \dfrac{1}{Z^2}.
  2. Z2=121.630.4=4⇒Z=2Z^2 = \dfrac{121.6}{30.4} = 4 \Rightarrow Z = 2.
  3. Answer: He+^+.

PYQ 7. In a hydrogen sample, atoms are excited to n = 5. The number of Balmer lines observable is: [JEE Mains]

Solution:

  1. Balmer lines end on n = 2, starting from 3, 4, 5.
  2. Three lines (of the total 5×42=10\dfrac{5 \times 4}{2} = 10).
  3. Takeaway: 'lines of a particular series' = (n - floor) choices of upper level, not the full pair count.

PYQ 8. The wavelength of the first Balmer line of hydrogen is 656 nm. The wavelength of the second line (4 → 2) is closest to: [JEE Mains]

Solution:

  1. Brackets: 3→2: 536\dfrac{5}{36}; 4→2: 316\dfrac{3}{16}.
  2. λ4→2=656×5/363/16=656×2027≈486\lambda_{4\to2} = 656 \times \dfrac{5/36}{3/16} = 656 \times \dfrac{20}{27} \approx 486 nm.
  3. Takeaway: scale by bracket ratios — never recompute from R.

JEE PYQ Worked Set C: Multi-step and Advanced Flavour

PYQ 9. A ground-state hydrogen atom (at rest) absorbs a 12.75 eV photon. Find the atom's final state and its recoil speed (M = 1.67×10−271.67 \times 10^{-27} kg). [JEE Advanced pattern]

Solution:

  1. 12.75 eV = E4−E1E_4 - E_1 exactly → final state n = 4.
  2. Recoil momentum = photon momentum: p=Ec=12.75×1.6×10−193×108=6.8×10−27p = \dfrac{E}{c} = \dfrac{12.75 \times 1.6 \times 10^{-19}}{3 \times 10^8} = 6.8 \times 10^{-27} kg m/s.
  3. v=pM≈4.1v = \dfrac{p}{M} \approx 4.1 m/s.
  4. Takeaway: photon absorption transfers momentum too — the atom recoils at walking pace.

PYQ 10. The ionisation energy of a hydrogen-like ion is 54.4 eV. The wavelength of radiation emitted in its n = 3 to n = 2 transition is: [JEE Mains]

Solution:

  1. 13.6Z2=54.4⇒Z=213.6Z^2 = 54.4 \Rightarrow Z = 2 (He+^+).
  2. E=13.6×4×536=7.56E = 13.6 \times 4 \times \dfrac{5}{36} = 7.56 eV.
  3. λ=12407.56≈164\lambda = \dfrac{1240}{7.56} \approx 164 nm.

PYQ 11. In hydrogen, the electron's kinetic energy in a certain orbit is 1.51 eV. Its angular momentum is: [JEE Mains]

Solution:

  1. K = |E| → E = -1.51 eV → n2=13.61.51=9n^2 = \dfrac{13.6}{1.51} = 9, n = 3.
  2. L=3h2π=3.17×10−34L = \dfrac{3h}{2\pi} = 3.17 \times 10^{-34} J s.
  3. Takeaway: energy identifies n; n hands you L. Two formulas chained.

PYQ 12. An alpha particle of energy 5 MeV is scattered through 180 degrees by a fixed uranium nucleus (Z = 92). The distance of closest approach is of the order of: [JEE Mains]

Solution:

  1. d=2Ze24πε0K=2×92×9×109×(1.6×10−19)25×106×1.6×10−19d = \dfrac{2Ze^2}{4\pi\varepsilon_0 K} = \dfrac{2 \times 92 \times 9 \times 10^9 \times (1.6 \times 10^{-19})^2}{5 \times 10^6 \times 1.6 \times 10^{-19}}.
  2. d=4.24×10−268×10−13≈5.3×10−14d = \dfrac{4.24 \times 10^{-26}}{8 \times 10^{-13}} \approx 5.3 \times 10^{-14} m.
  3. Answer: order 10−1410^{-14} m ≈ 53 fm — a classic JEE order-of-magnitude item.

PYQ 13. The ratio of the speeds of the electron in the ground states of hydrogen and He+^+ is: [JEE Mains]

Solution:

  1. v∝Znv \propto \dfrac{Z}{n}; ground states have n = 1.
  2. vHvHe+=12\dfrac{v_H}{v_{He^+}} = \dfrac{1}{2} → ratio 1 : 2.

PYQ 14. Hydrogen atoms in the ground state are excited by radiation of wavelength 97.3 nm. The number of spectral lines emitted subsequently is: [JEE Mains]

Solution:

  1. E=124097.3=12.75E = \dfrac{1240}{97.3} = 12.75 eV → excitation to n = 4 (exact match).
  2. Lines = 4×32=6\dfrac{4 \times 3}{2} = 6.

PYQ 15. As the quantum number n increases, the frequency separation between adjacent hydrogen levels' emitted lines (n → n-1) varies as: [JEE Advanced pattern]

Solution:

  1. ν∝(1(n−1)2−1n2)=2n−1n2(n−1)2\nu \propto \left(\dfrac{1}{(n-1)^2} - \dfrac{1}{n^2}\right) = \dfrac{2n-1}{n^2(n-1)^2}.
  2. For large n: ν∝2n3\nu \propto \dfrac{2}{n^3} — falling as n−3n^{-3}, converging to the orbital frequency (correspondence principle).

PYQ 16. If the radius of hydrogen's n = 2 orbit is 2.12×10−102.12 \times 10^{-10} m, the de Broglie wavelength of the electron in that orbit is: [JEE Mains]

Solution:

  1. Standing-wave condition: λ=2πrnn=2π×2.12×10−102\lambda = \dfrac{2\pi r_n}{n} = \dfrac{2\pi \times 2.12 \times 10^{-10}}{2}.
  2. λ=6.66×10−10\lambda = 6.66 \times 10^{-10} m ≈ 0.67 nm.
  3. Takeaway: the orbit hands you the wavelength directly — no need for momentum at all.