How to Use This Problem Set

Your full workout for Atoms, grouped by theme: scattering & closest approach, orbit energetics, Bohr radii/speeds/periods, energy levels & excitation, spectral series wavelengths, and de Broglie connections.

Keep these handy:

  • Closest approach: d=2Ze24πε0Kd = \dfrac{2Ze^2}{4\pi\varepsilon_0 K} (alpha); general projectile charge q: d=qZe4πε0Kd = \dfrac{qZe}{4\pi\varepsilon_0 K}
  • Orbit energetics: K=−EK = -E, U=2EU = 2E, E=−e28πε0rE = -\dfrac{e^2}{8\pi\varepsilon_0 r}
  • Bohr (hydrogen-like, charge Z): rn=n2Za0r_n = \dfrac{n^2}{Z}a_0, a0=0.529a_0 = 0.529 angstrom; vn=c137Znv_n = \dfrac{c}{137}\dfrac{Z}{n}; En=−13.6Z2n2E_n = -13.6\dfrac{Z^2}{n^2} eV; Tn∝n3Z2T_n \propto \dfrac{n^3}{Z^2}; L=nh2πL = \dfrac{nh}{2\pi}
  • Transitions: E=13.6Z2(1nf2−1ni2)E = 13.6Z^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right) eV; λ=1240E(eV)\lambda = \dfrac{1240}{E(\text{eV})} nm; 1λ=RZ2(1nf2−1ni2)\dfrac{1}{\lambda} = RZ^2\left(\dfrac{1}{n_f^2} - \dfrac{1}{n_i^2}\right), R = 1.097×1071.097 \times 10^7 m−1^{-1}
  • Line count from level n: n(n−1)2\dfrac{n(n-1)}{2}; de Broglie fit: 2πrn=nλ2\pi r_n = n\lambda

State the condition, then substitute.

Solved Examples - Scattering & Closest Approach

Example 1. Closest approach of a 7.7 MeV alpha to gold (Z = 79)?

Solution: d=2Ze24πε0K=3.84×10−16×79=3.0×10−14d = \dfrac{2Ze^2}{4\pi\varepsilon_0 K} = 3.84 \times 10^{-16} \times 79 = 3.0 \times 10^{-14} m = 30 fm.

Example 2. Same alpha on silver (Z = 47)?

Solution: d∝Zd \propto Z: 30×4779≈1830 \times \dfrac{47}{79} \approx 18 fm.

Example 3. A 3.85 MeV alpha on gold: closest approach?

Solution: d∝1/Kd \propto 1/K: half the energy of 7.7 MeV → d=60d = 60 fm.

Example 4. A proton with 7.7 MeV heading at gold?

Solution: charge halves (e vs 2e): d=15d = 15 fm.

Example 5. What kinetic energy (in MeV) brings an alpha within 10 fm of gold?

Solution: K=2Ze24πε0d=2×79×9×109×(1.6×10−19)210−14K = \dfrac{2Ze^2}{4\pi\varepsilon_0 d} = \dfrac{2 \times 79 \times 9 \times 10^9 \times (1.6 \times 10^{-19})^2}{10^{-14}} J =3.64×10−12= 3.64 \times 10^{-12} J ≈ 22.8 MeV.

Solved Examples - Orbit Energetics

Example 6. E = -13.6 eV for hydrogen's ground state. Find K and U.

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

Example 7. An orbit has U = -6.8 eV. Find E and K.

Solution: E = U/2 = -3.4 eV; K = -E = +3.4 eV — hydrogen's n = 2 numbers.

Example 8. Ionisation energy from an orbit with K = 1.51 eV?

Solution: |E| = K = 1.51 eV (n = 3 of hydrogen).

Example 9. If r doubles in a classical Coulomb orbit, what happens to |E|?

Solution: ∣E∣∝1/r|E| \propto 1/r → halves.

Example 10. Speed of an electron orbiting at r = 5.3×10−115.3 \times 10^{-11} m?

Solution: v=e4πε0mr=2.2×106v = \dfrac{e}{\sqrt{4\pi\varepsilon_0 mr}} = 2.2 \times 10^6 m/s (NCERT Example 12.3).

Solved Examples - Bohr Radii, Speeds, Periods

Example 11. Radii of hydrogen's n = 2 and n = 3 orbits (a0=5.3×10−11a_0 = 5.3 \times 10^{-11} m)?

Solution: rn=n2a0r_n = n^2a_0: r2=2.12×10−10r_2 = 2.12 \times 10^{-10} m, r3=4.77×10−10r_3 = 4.77 \times 10^{-10} m (NCERT Exercise 12.7).

Example 12. Speeds in n = 1, 2, 3 (NCERT Exercise 12.6a)?

Solution: vn=2.18×106nv_n = \dfrac{2.18 \times 10^6}{n} m/s: 2.18×1062.18 \times 10^6, 1.09×1061.09 \times 10^6, 7.27×1057.27 \times 10^5 m/s.

Example 13. Orbital periods in n = 1, 2, 3 (NCERT Exercise 12.6b)?

Solution: Tn=T1n3T_n = T_1n^3 with T1=2πa0v1=1.53×10−16T_1 = \dfrac{2\pi a_0}{v_1} = 1.53 \times 10^{-16} s: T2=1.22×10−15T_2 = 1.22 \times 10^{-15} s, T3=4.13×10−15T_3 = 4.13 \times 10^{-15} s.

Example 14. Ground-state radius of He+^+?

Solution: r=a0Z=5.3×10−112=2.65×10−11r = \dfrac{a_0}{Z} = \dfrac{5.3 \times 10^{-11}}{2} = 2.65 \times 10^{-11} m.

Example 15. Angular momentum in hydrogen's n = 2 orbit?

Solution: L=2h2π=hπ=2.11×10−34L = \dfrac{2h}{2\pi} = \dfrac{h}{\pi} = 2.11 \times 10^{-34} J s.

Example 16. In which hydrogen orbit is the speed c/411?

Solution: vn=c137nv_n = \dfrac{c}{137n} → n = 3.

Solved Examples - Energy Levels & Excitation

Example 17. Energies of hydrogen's n = 1 to 4 levels?

Solution: En=−13.6/n2E_n = -13.6/n^2: -13.6, -3.4, -1.51, -0.85 eV.

Example 18. First and second excitation energies of hydrogen?

Solution: E2−E1E_2 - E_1 = 10.2 eV; E3−E1E_3 - E_1 = 12.09 eV.

Example 19. Ionisation energy of Li2+^{2+} from its ground state?

Solution: 13.6Z2=13.6×9=122.413.6Z^2 = 13.6 \times 9 = 122.4 eV.

Example 20. Energy of He+^+'s n = 2 level?

Solution: E=−13.644=−13.6E = -13.6\dfrac{4}{4} = -13.6 eV — numerically equal to hydrogen's ground state.

Example 21. A hydrogen atom absorbs 12.09 eV. Which state does it reach, and what happens with 12.5 eV of PHOTON energy instead?

Solution: 12.09 eV lifts it exactly to n = 3. A 12.5 eV photon matches NO level difference from n = 1 (needs exactly 10.2, 12.09, 12.75…) → it is NOT absorbed. (An electron BEAM at 12.5 eV can excite to n = 3, keeping the 0.41 eV change — the crucial photon/electron difference.)

Example 22. Minimum energy to ionise hydrogen already in n = 3?

Solution: |E₃| = 1.51 eV.

Solved Examples - Spectral Wavelengths

Example 23. Wavelength of Lyman-alpha (2 → 1)?

Solution: E = 13.6×3413.6 \times \dfrac{3}{4} = 10.2 eV; λ=124010.2≈121.6\lambda = \dfrac{1240}{10.2} \approx 121.6 nm.

Example 24. Wavelength of H-alpha (3 → 2)?

Solution: E = 13.6×53613.6 \times \dfrac{5}{36} = 1.89 eV; λ≈656\lambda \approx 656 nm.

Example 25. Balmer series limit?

Solution: E = 13.64\dfrac{13.6}{4} = 3.4 eV; λ=12403.4≈365\lambda = \dfrac{1240}{3.4} \approx 365 nm.

Example 26. Frequency for a 2.3 eV level gap (NCERT Exercise 12.3)?

Solution: ν=2.3×1.6×10−196.63×10−34≈5.6×1014\nu = \dfrac{2.3 \times 1.6 \times 10^{-19}}{6.63 \times 10^{-34}} \approx 5.6 \times 10^{14} Hz.

Example 27. Photon absorbed for n = 1 → 4 (NCERT Exercise 12.5)?

Solution: E = 12.75 eV; λ≈97.3\lambda \approx 97.3 nm; ν≈3.1×1015\nu \approx 3.1 \times 10^{15} Hz.

Example 28. First Paschen line (4 → 3)?

Solution: E = 13.6(19−116)=13.6×7144=0.6613.6\left(\dfrac{1}{9} - \dfrac{1}{16}\right) = 13.6 \times \dfrac{7}{144} = 0.66 eV; λ=12400.66≈1875\lambda = \dfrac{1240}{0.66} \approx 1875 nm.

Example 29. Number of spectral lines from hydrogen excited to n = 4, listed by series?

Solution: 4×32=6\dfrac{4 \times 3}{2} = 6: Lyman 3 (4→1, 3→1, 2→1), Balmer 2 (4→2, 3→2), Paschen 1 (4→3).

Example 30. The 12.5 eV electron-beam problem (NCERT Exercise 12.8): which wavelengths emerge?

Solution: Excitation up to n = 3 → three lines: 102.6 nm (3→1), 121.6 nm (2→1), 656 nm (3→2) — two Lyman + one Balmer.

Example 31. Ratio λLyman−α:λBalmer−α\lambda_{Lyman-\alpha} : \lambda_{Balmer-\alpha}?

Solution: brackets 3/4 vs 5/36 → ratio =5/363/4⋅...= \dfrac{5/36}{3/4}\cdot... careful: λ∝1/bracket\lambda \propto 1/\text{bracket}, so ratio =4/336/5=527= \dfrac{4/3}{36/5} = \dfrac{5}{27}.

Example 32. Which He+^+ transition matches hydrogen's 2 → 1 wavelength?

Solution: preserve Z/n: hydrogen (2,1) → He+^+ (4,2). Check: 4(14−116)=344\left(\dfrac{1}{4} - \dfrac{1}{16}\right) = \dfrac{3}{4}. ✔

Example 33. Shortest wavelength of He+^+'s Balmer-like series (to n = 2)?

Solution: E = 13.6×4×1413.6 \times 4 \times \dfrac{1}{4} = 13.6 eV; λ=124013.6≈91.2\lambda = \dfrac{1240}{13.6} \approx 91.2 nm — He+^+'s Balmer limit coincides with hydrogen's Lyman limit.

Solved Examples - de Broglie & Mixed Mastery

Example 34. De Broglie wavelength of hydrogen's electron in n = 1?

Solution: λ1=2πa0=3.3×10−10\lambda_1 = 2\pi a_0 = 3.3 \times 10^{-10} m — exactly one circumference.

Example 35. In n = 3?

Solution: λn=2πna0\lambda_n = 2\pi na_0: λ3=3λ1≈10−9\lambda_3 = 3\lambda_1 \approx 10^{-9} m.

Example 36. How many wavelengths fit the n = 5 orbit? Nodes?

Solution: 5 wavelengths; 10 nodes.

Example 37. Earth's orbital quantum number (NCERT Exercise 12.9)?

Solution: n=2πMvrh=2π×6×1024×3×104×1.5×10116.63×10−34≈2.6×1074n = \dfrac{2\pi Mvr}{h} = \dfrac{2\pi \times 6 \times 10^{24} \times 3 \times 10^4 \times 1.5 \times 10^{11}}{6.63 \times 10^{-34}} \approx 2.6 \times 10^{74} — quantisation utterly imperceptible.

Example 38. A hydrogen atom emits a 656 nm photon. Between which levels did the electron jump?

Solution: E = 1240656\dfrac{1240}{656} = 1.89 eV = 13.6×53613.6 \times \dfrac{5}{36} → the 3 → 2 transition.

Example 39. The recoil speed of a hydrogen atom emitting a 10.2 eV photon (mass 1.67×10−271.67 \times 10^{-27} kg)?

Solution: photon momentum p=Ec=10.2×1.6×10−193×108=5.44×10−27p = \dfrac{E}{c} = \dfrac{10.2 \times 1.6 \times 10^{-19}}{3 \times 10^8} = 5.44 \times 10^{-27} kg m/s; recoil v=pM≈3.3v = \dfrac{p}{M} \approx 3.3 m/s.

Example 40. An electron with de Broglie wavelength 3.3×10−103.3 \times 10^{-10} m circles a proton. Which Bohr orbit is it in?

Solution: λn=2πna0\lambda_n = 2\pi na_0 → n=3.3×10−102π×5.3×10−11=1n = \dfrac{3.3 \times 10^{-10}}{2\pi \times 5.3 \times 10^{-11}} = 1 — the ground state.