Class 12 Physics | Unit VIII
Chapter 12: Atoms
Thomson's Model • Rutherford's Model • Bohr's Model • Hydrogen Spectrum • Energy Levels
1. Thomson's Model (Plum Pudding Model)
J.J. Thomson (1898): Atom is a uniform sphere of positive charge with electrons embedded in
it like plums in a pudding. Atom is electrically neutral overall.
- Could explain electrical neutrality of atom.
- Failed: Could NOT explain Rutherford's α-particle scattering results. Could not explain spectral lines of hydrogen.
2. Rutherford's α-Particle Scattering Experiment
Geiger-Marsden experiment (1909): α-particles (He2+) from radioactive
source were fired at a thin gold foil (~2.1×10−7 m). Scattered α-particles were
detected by a ZnS screen (scintillation detector).
2.1 Key Observations
- Most α-particles passed straight through → atom is mostly empty space.
- Few were deflected at small angles → positive charge is concentrated.
- Very few (~1 in 8000) were deflected by more than 90°.
- Some were scattered back (θ ≈ 180°) → massive, positively charged nucleus exists.
2.2 Rutherford's Nuclear Model
- Entire positive charge and almost all mass concentrated in a tiny nucleus (~10−15 m = 1 fm).
- Atom size ~10−10 m. So nucleus is ~105 times smaller than atom!
- Electrons revolve around nucleus in orbits (like planets around sun).
- Electrostatic attraction between nucleus (+Ze) and electrons provides centripetal force.
Distance of closest approach:
d = (1/4πε0) × 2Ze²/(½mv²) = kZe²×4/(mv²)
All KE converts to PE at closest approach: ½mv² = kZe·2e/d
Impact parameter (b): b = (kZe²/KE) cot(θ/2)
θ = scattering angle. Small b → large θ (head-on → 180°). Large b → small θ (passes far away).
d = (1/4πε0) × 2Ze²/(½mv²) = kZe²×4/(mv²)
All KE converts to PE at closest approach: ½mv² = kZe·2e/d
Impact parameter (b): b = (kZe²/KE) cot(θ/2)
θ = scattering angle. Small b → large θ (head-on → 180°). Large b → small θ (passes far away).
2.3 Limitations of Rutherford's Model
- Stability: Orbiting electron is accelerating → should radiate EM waves → lose energy → spiral into nucleus. But atoms are stable! ✘
- Spectrum: Predicts continuous spectrum (as electron spirals in). But hydrogen gives discrete line spectrum! ✘
⚠️ NEET: Most α-particles go straight (empty space). ~1/8000 scatter
>90°. Nucleus: +charge, almost all mass, ~10⁻¹⁵ m. Atom ~10⁻¹⁰ m.
Distance of closest approach ∝ 1/KE. Rutherford model fails: stability & spectrum.
3. Bohr's Model of Hydrogen Atom (1913)
Bohr combined Rutherford's nuclear model with Planck's quantum theory and proposed three postulates
to explain the hydrogen atom's stability and line spectrum.
3.1 Bohr's Three Postulates
- Quantised orbits: Electrons revolve in specific circular orbits (called stationary orbits or allowed orbits) without radiating energy. Only certain discrete orbits are allowed.
- Angular momentum quantisation: L = mvr = nħ = nh/(2π), where n = 1, 2, 3, ... (principal quantum number). ħ = h/(2π) = reduced Planck constant.
- Frequency condition: When electron jumps from higher orbit (n2) to lower orbit (n1), a photon is emitted with energy: hν = En2 − En1.
3.2 Results for Hydrogen Atom (Z=1)
Radius of nth orbit:
rn = n²a0/Z where a0 = 0.529 Å (Bohr radius)
rn ∝ n²/Z
Velocity of electron:
vn = (Z/n) × v0 where v0 = 2.18×106 m/s = c/137
vn ∝ Z/n
Energy of nth orbit:
En = −13.6 Z²/n² eV (for hydrogen Z=1)
En ∝ −Z²/n²
For hydrogen (Z=1):
E1 = −13.6 eV (ground state) | E2 = −3.4 eV | E3 = −1.51 eV | E∞ = 0 (ionised)
rn = n²a0/Z where a0 = 0.529 Å (Bohr radius)
rn ∝ n²/Z
Velocity of electron:
vn = (Z/n) × v0 where v0 = 2.18×106 m/s = c/137
vn ∝ Z/n
Energy of nth orbit:
En = −13.6 Z²/n² eV (for hydrogen Z=1)
En ∝ −Z²/n²
For hydrogen (Z=1):
E1 = −13.6 eV (ground state) | E2 = −3.4 eV | E3 = −1.51 eV | E∞ = 0 (ionised)
3.3 Energy Level Diagram
| n | En (eV) | rn (Å) | vn (×106 m/s) | Name |
|---|---|---|---|---|
| 1 | −13.6 | 0.529 | 2.18 | K shell (ground state) |
| 2 | −3.4 | 2.12 | 1.09 | L shell (1st excited) |
| 3 | −1.51 | 4.76 | 0.727 | M shell (2nd excited) |
| 4 | −0.85 | 8.46 | 0.545 | N shell (3rd excited) |
| ∞ | 0 | ∞ | 0 | Ionised (free electron) |
- nth excited state = (n+1)th energy level. E.g., 1st excited = n=2.
- Ionisation energy (from ground state) = 13.6 eV for hydrogen.
- Energy is negative → electron is bound. At n=∞, E=0 → free.
4. Hydrogen Spectrum — Spectral Series
Rydberg formula:
1/λ = RH [1/n1² − 1/n2²]
RH = 1.097×107 m−1 (Rydberg constant)
n1 < n2; electron falls from n2 to n1.
1/λ = RH [1/n1² − 1/n2²]
RH = 1.097×107 m−1 (Rydberg constant)
n1 < n2; electron falls from n2 to n1.
| Series | n1 | n2 | Region | Key Line |
|---|---|---|---|---|
| Lyman | 1 | 2, 3, 4, ... | Ultraviolet | Ly-α: 1→2 (121.6 nm) |
| Balmer | 2 | 3, 4, 5, ... | Visible | H-α: 3→2 (656 nm, Red) |
| Paschen | 3 | 4, 5, 6, ... | Infrared | First IR series |
| Brackett | 4 | 5, 6, 7, ... | Infrared | Far IR |
| Pfund | 5 | 6, 7, 8, ... | Infrared | Far IR |
- Lyman: Only UV series. Contains the shortest wavelength line of the entire spectrum (n=∞→1).
- Balmer: Only visible series. H-α (red), H-β (blue-green), H-γ (blue), H-δ (violet).
- Total spectral lines from nth level: N = n(n−1)/2.
- Maximum wavelength in a series: smallest energy transition (n2 = n1+1).
- Minimum wavelength (series limit): largest energy transition (n2 = ∞).
⚠️ NEET (2014–2023): Lyman=UV, Balmer=Visible,
Paschen/Brackett/Pfund=IR. 1/λ=R[1/n1²−1/n2²]. Total lines from
n: n(n−1)/2. En=−13.6/n² eV. rn=0.529n² Å. Ionisation
energy of H=13.6 eV.
5. Limitations of Bohr's Model
- Works accurately only for hydrogen and hydrogen-like ions (He+, Li2+, etc.) — single electron systems.
- Cannot explain spectra of multi-electron atoms.
- Cannot explain fine structure of spectral lines (splitting in magnetic/electric fields).
- Cannot explain relative intensity of spectral lines.
- Does not incorporate wave nature of electron (no de Broglie wave concept).
- Violates Heisenberg's uncertainty principle (assumes definite orbit with known r and v).
5.1 de Broglie Explanation of Bohr's Quantisation
Electron orbits = standing waves. Circumference must be integer multiple of λ:
2πr = nλ = nh/(mv) ⇒ mvr = nh/(2π) = Bohr's quantisation condition! ✓
This explains WHY only certain orbits are allowed — only those where electron wave forms a standing wave.
2πr = nλ = nh/(mv) ⇒ mvr = nh/(2π) = Bohr's quantisation condition! ✓
This explains WHY only certain orbits are allowed — only those where electron wave forms a standing wave.
🎓 NEET Previous Year Questions
Q1. [NEET 2022] The radius of 3rd Bohr orbit of hydrogen is:
Answer rn=n²×0.529 Å.
r3 = 9×0.529 = 4.761 Å.
Q2. [NEET 2021] An electron in hydrogen jumps from n=4 to n=2. Find wavelength of
emitted photon.
Answer 1/λ = R(1/4−1/16) = R×3/16 =
1.097×107×3/16 = 2.057×106. λ = 1/2.057×106 =
486 nm (H-β line, blue-green, Balmer series).
Q3. [NEET 2020] Total number of spectral lines from n=4 to ground state:
Answer N = n(n−1)/2 = 4×3/2 = 6
lines (4→3, 4→2, 4→1, 3→2, 3→1, 2→1).
Q4. [NEET 2019] Energy needed to ionise hydrogen from 2nd excited state:
Answer 2nd excited = n=3. E3 =
−13.6/9 = −1.51 eV. Ionisation energy = |E3| = 1.51 eV.
Q5. [NEET 2017] In Bohr model, the ratio of kinetic energy to total energy is:
Answer KE = −En = 13.6/n². Total
E = −13.6/n². Ratio = KE/E = −1. (KE is positive, total E is negative,
magnitude same.)
💡 Rapid Revision
- Thomson: plum pudding (failed) | Rutherford: nucleus discovery (stability failed)
- Bohr: rn=0.529n²/Z Å | vn=2.18×10⁶×Z/n m/s | En=−13.6Z²/n² eV
- L=mvr=nh/(2π) | hν=En2−En1 | Ionisation of H=13.6 eV
- 1/λ=R[1/n₁²−1/n₂²] | R=1.097×10⁷ m⁻¹
- Lyman=UV | Balmer=Visible | Paschen/Brackett/Pfund=IR
- Total lines from n: n(n−1)/2 | 1st excited=n=2
- KE=−En | PE=2En | E=−KE=PE/2
CLASS 12 PHYSICS | NCERT SOLUTIONS
Chapter 12 — Atoms
22 NCERT Exercise & Exemplar Questions — Step-by-Step Solutions
Key Formulas: En=−13.6Z²/n² eV | rn=0.529n²/Z
Å | 1/λ=R(1/n1²−1/n2²) | R=1.097×107
m−1 | Total lines=n(n−1)/2
📝 NCERT Exercise Questions (Q1 – Q12)
3 MarksQ1 (Ex 12.1). In Rutherford's experiment,
α-particles of energy 5.5 MeV are scattered by gold (Z=79). Find distance of closest approach.
✓ Solution
At closest approach: ½mv² = kZe×2e/d ⇒ d = 2kZe²/KE
d = 2×9×109×79×(1.6×10−19)²/(5.5×106×1.6×10−19)
= 2×9×109×79×2.56×10−38/(8.8×10−13)
At closest approach: ½mv² = kZe×2e/d ⇒ d = 2kZe²/KE
d = 2×9×109×79×(1.6×10−19)²/(5.5×106×1.6×10−19)
= 2×9×109×79×2.56×10−38/(8.8×10−13)
d = 3.637×10−26/8.8×10−13 =
4.13×10−14 m ≈ 41.3 fm
3 MarksQ2 (Ex 12.2). The ground state energy of
hydrogen is −13.6 eV. Find (a) KE, (b) PE in ground state.
✓ Solution
In Bohr model: KE = −En; PE = 2En
(a) KE = −(−13.6) = +13.6 eV
In Bohr model: KE = −En; PE = 2En
(a) KE = −(−13.6) = +13.6 eV
(b) PE = 2×(−13.6) = −27.2
eV
Total = KE + PE = 13.6 − 27.2 = −13.6 eV ✓3 MarksQ3 (Ex 12.3). What is the energy of electron
in H-atom in (a) 1st excited state, (b) 2nd excited state?
✓ Solution
1st excited = n=2; 2nd excited = n=3
(a) E2 = −13.6/2² = −13.6/4 = −3.4 eV
1st excited = n=2; 2nd excited = n=3
(a) E2 = −13.6/2² = −13.6/4 = −3.4 eV
(b) E3 = −13.6/3² = −13.6/9 =
−1.51 eV
3 MarksQ4 (Ex 12.4). Find the wavelength of the first
line (H-α) of Balmer series.
✓ Solution
Balmer: n1=2. First line: n2=3
1/λ = R(1/2² − 1/3²) = 1.097×107(1/4 − 1/9) = 1.097×107×5/36
= 1.097×107×0.1389 = 1.524×106
Balmer: n1=2. First line: n2=3
1/λ = R(1/2² − 1/3²) = 1.097×107(1/4 − 1/9) = 1.097×107×5/36
= 1.097×107×0.1389 = 1.524×106
λ = 1/1.524×106 = 6.56×10−7
m = 656 nm (Red line)
3 MarksQ5 (Ex 12.5). How many spectral lines are
possible when electron de-excites from n=4?
✓ Solution
N = n(n−1)/2 = 4×3/2 = 6 lines
Lines: 4→3, 4→2, 4→1 (from n=4) + 3→2, 3→1 (from n=3) + 2→1 (from
n=2) = 6 total.3 MarksQ6 (Ex 12.6). Find the shortest wavelength in
the Balmer series.
✓ Solution
Shortest λ → maximum energy → n2=∞, n1=2 (series limit)
1/λ = R(1/4 − 0) = R/4 = 1.097×107/4 = 2.743×106
Shortest λ → maximum energy → n2=∞, n1=2 (series limit)
1/λ = R(1/4 − 0) = R/4 = 1.097×107/4 = 2.743×106
λ = 1/2.743×106 = 364.6 nm (near UV
— series limit of Balmer)
3 MarksQ7 (Ex 12.7). Find the longest wavelength in
the Lyman series.
✓ Solution
Longest λ → minimum energy → n2=2, n1=1
1/λ = R(1/1 − 1/4) = R×3/4 = 1.097×107×0.75 = 8.228×106
Longest λ → minimum energy → n2=2, n1=1
1/λ = R(1/1 − 1/4) = R×3/4 = 1.097×107×0.75 = 8.228×106
λ = 1/8.228×106 =
1.215×10−7 m = 121.5 nm (UV)
3 MarksQ8 (Ex 12.8). Find the radius and energy of
He+ ion in 2nd orbit.
✓ Solution
He+: Z=2, n=2
r = 0.529×n²/Z = 0.529×4/2 = 1.058 Å
He+: Z=2, n=2
r = 0.529×n²/Z = 0.529×4/2 = 1.058 Å
E = −13.6×Z²/n² = −13.6×4/4 =
−13.6 eV
Same energy as hydrogen ground state! (Z²/n² = 4/4 = 1)3 MarksQ9 (Ex 12.9). Photon of 12.09 eV strikes
H-atom in ground state. To which orbit does the electron jump?
✓ Solution
En = −13.6/n²; Ground: E1 = −13.6 eV
After absorption: Efinal = −13.6 + 12.09 = −1.51 eV
−13.6/n² = −1.51 ⇒ n² = 13.6/1.51 = 9.01
En = −13.6/n²; Ground: E1 = −13.6 eV
After absorption: Efinal = −13.6 + 12.09 = −1.51 eV
−13.6/n² = −1.51 ⇒ n² = 13.6/1.51 = 9.01
n = 3 → Electron jumps to 3rd orbit (2nd excited state)
3 MarksQ10 (Ex 12.10). Find the angular momentum of
electron in 3rd Bohr orbit.
✓ Solution
L = nh/(2π) = nħ
L = nh/(2π) = nħ
L = 3×6.63×10−34/(2π) =
3×1.055×10−34 = 3.16×10−34
J·s
3 MarksQ11 (Ex 12.11). Find ratio of maximum and
minimum wavelengths in Balmer series.
✓ Solution
Max λ (min energy): 3→2 ⇒ 1/λmax = R(1/4−1/9) = 5R/36
Min λ (max energy): ∞→2 ⇒ 1/λmin = R(1/4) = R/4 = 9R/36
Max λ (min energy): 3→2 ⇒ 1/λmax = R(1/4−1/9) = 5R/36
Min λ (max energy): ∞→2 ⇒ 1/λmin = R(1/4) = R/4 = 9R/36
λmax/λmin = (9R/36)/(5R/36) = 9/5 =
1.8
3 MarksQ12 (Ex 12.12). What is the frequency of
revolution of electron in 2nd Bohr orbit?
✓ Solution
vn = 2.18×106/n m/s; rn = 0.529×n² Å
f = v/(2πr) = (2.18×106/2)/(2π×0.529×4×10−10)
= 1.09×106/(1.329×10−9)
vn = 2.18×106/n m/s; rn = 0.529×n² Å
f = v/(2πr) = (2.18×106/2)/(2π×0.529×4×10−10)
= 1.09×106/(1.329×10−9)
f = 8.2×1014 Hz
🌟 Additional / Exemplar Questions (Q13 – Q22)
3 MarksQ13. Find the wavelength of transition n=5 to
n=2 in hydrogen.
✓ Solution
1/λ = R(1/4 − 1/25) = 1.097×107×(25−4)/100 = 1.097×107×21/100
= 1.097×107×0.21 = 2.304×106
1/λ = R(1/4 − 1/25) = 1.097×107×(25−4)/100 = 1.097×107×21/100
= 1.097×107×0.21 = 2.304×106
λ = 1/2.304×106 = 434 nm (H-γ line,
blue, Balmer series)
3 MarksQ14. Find ionisation energy of Li2+
(Z=3) from ground state.
✓ Solution
E1 = −13.6×Z²/n² = −13.6×9/1 = −122.4 eV
E1 = −13.6×Z²/n² = −13.6×9/1 = −122.4 eV
Ionisation energy = |E1| = 122.4 eV
9 times that of hydrogen (Z² = 9).3 MarksQ15. An electron in hydrogen undergoes
transition from n=3 to n=1. Is the photon emitted in UV, visible, or IR?
✓ Solution
n=3→n=1: this falls in Lyman series (n1=1)
n=3→n=1: this falls in Lyman series (n1=1)
Lyman series is in the ultraviolet (UV) region.
1/λ = R(1−1/9) = R×8/9 ⇒ λ = 9/(8R) = 9/(8×1.097×107) =
102.5 nm (UV ✓).3 MarksQ16. The ratio of radii of orbits in hydrogen
for n=1 and n=3 is:
✓ Solution
r ∝ n² ⇒ r1:r3 = 1²:3² = 1:9
r ∝ n² ⇒ r1:r3 = 1²:3² = 1:9
r1/r3 = 1/9
3 MarksQ17. Find the energy needed to excite hydrogen
from ground state to 3rd excited state.
✓ Solution
3rd excited = n=4
E4 = −13.6/16 = −0.85 eV
Energy needed = E4 − E1 = −0.85 − (−13.6) = 12.75 eV
3rd excited = n=4
E4 = −13.6/16 = −0.85 eV
Energy needed = E4 − E1 = −0.85 − (−13.6) = 12.75 eV
Excitation energy = 12.75 eV
3 MarksQ18. Find the ratio of speeds of electron in
n=1 and n=3 orbits of hydrogen.
✓ Solution
v ∝ 1/n (for same Z)
v ∝ 1/n (for same Z)
v1/v3 = 3/1 = 3:1
3 MarksQ19. Which orbit in He+ has same
radius as the 2nd orbit of hydrogen?
✓ Solution
r = 0.529n²/Z. For H (Z=1, n=2): r = 0.529×4 = 2.116 Å
For He+ (Z=2): 0.529n²/2 = 2.116 ⇒ n² = 2.116×2/0.529 = 8 ⇒ n = 2√2
Since n must be integer, let's check: n²/2 = 4 ⇒ n² = 8... not an integer n.
r = 0.529n²/Z. For H (Z=1, n=2): r = 0.529×4 = 2.116 Å
For He+ (Z=2): 0.529n²/2 = 2.116 ⇒ n² = 2.116×2/0.529 = 8 ⇒ n = 2√2
Since n must be integer, let's check: n²/2 = 4 ⇒ n² = 8... not an integer n.
Actually: rH(n=2) = 4a0. rHe+(n) =
n²a0/2 = 4a0 ⇒ n²=8. No integer solution.
For exact match: n=4 gives r = 16a0/2 = 8a0 (too big). Exact same radius not possible for integer n in He+. But n=4, Z=4 (Be3+) would give r=16a0/4=4a0 ✓
For exact match: n=4 gives r = 16a0/2 = 8a0 (too big). Exact same radius not possible for integer n in He+. But n=4, Z=4 (Be3+) would give r=16a0/4=4a0 ✓
3 MarksQ20. Find the de Broglie wavelength of
electron in 3rd Bohr orbit.
✓ Solution
From Bohr: 2πr = nλ ⇒ λ = 2πr/n
r3 = 0.529×9 = 4.761 Å
From Bohr: 2πr = nλ ⇒ λ = 2πr/n
r3 = 0.529×9 = 4.761 Å
λ = 2π×4.761/3 = 2π×1.587 = 9.97 Å
≈ 10 Å = 1 nm
3 MarksQ21. In which series does the H-atom
transition from n=6 to n=2 fall? Find its wavelength.
✓ Solution
n1=2 → Balmer series (visible region)
1/λ = R(1/4 − 1/36) = R(9−1)/36 = 8R/36 = 2R/9
= 2×1.097×107/9 = 2.438×106
n1=2 → Balmer series (visible region)
1/λ = R(1/4 − 1/36) = R(9−1)/36 = 8R/36 = 2R/9
= 2×1.097×107/9 = 2.438×106
λ = 1/2.438×106 = 410.2 nm (H-δ
line, violet)
3 MarksQ22. If the electron in hydrogen is in the 4th
excited state, what is the maximum number of transitions it can make to reach ground state?
✓ Solution
4th excited state = n=5
Maximum transitions (one step at a time): 5→4→3→2→1 = 4 transitions.
Total possible spectral lines from n=5: N = 5×4/2 = 10 lines.
4th excited state = n=5
Maximum transitions (one step at a time): 5→4→3→2→1 = 4 transitions.
Total possible spectral lines from n=5: N = 5×4/2 = 10 lines.
Maximum transitions for single electron = n−1 = 5−1 =
4.
Total possible lines (for many atoms) = n(n−1)/2 = 10.
Total possible lines (for many atoms) = n(n−1)/2 = 10.
✍ Score Guide — 22 Questions
NCERT Ex 12.1–12.12: 3 marks × 12 = 36 | Additional Q13–Q22: 3 marks × 10 = 30 | Total: 66 marks
NCERT Ex 12.1–12.12: 3 marks × 12 = 36 | Additional Q13–Q22: 3 marks × 10 = 30 | Total: 66 marks
CLASS 12 PHYSICS | FORMULA CAPSULE
Atoms
Chapter 12 — Complete Formula Sheet for NEET/JEE
📅 Section A — Bohr Model (Hydrogen & H-like)
| Quantity | Formula | Proportionality |
|---|---|---|
| Radius | rn = 0.529 n²/Z Å | r ∝ n²/Z |
| Velocity | vn = 2.18×106 Z/n m/s | v ∝ Z/n |
| Energy | En = −13.6 Z²/n² eV | E ∝ −Z²/n² |
| Angular momentum | L = nh/(2π) = nħ | L ∝ n |
| Time period | T = 2πr/v | T ∝ n³/Z² |
| Frequency of revolution | f = v/(2πr) | f ∝ Z²/n³ |
| Current (orbit) | I = ef = ev/(2πr) | I ∝ Z²/n³ |
📅 Section B — Energy Relations
| Quantity | Formula | Key Note |
|---|---|---|
| KE | KE = −En = 13.6Z²/n² | Always positive |
| PE | PE = 2En = −27.2Z²/n² | Always negative; |PE|=2KE |
| Total E | E = KE + PE = −KE | E = −13.6Z²/n² |
| Ionisation energy | IE = |En| = 13.6Z²/n² | From ground: 13.6 eV (H) |
| Excitation energy | Eexc = Efinal − Einitial | 1st exc of H = 10.2 eV |
📅 Section C — Spectral Series
| Series | n1 | n2 | Region | Longest λ |
|---|---|---|---|---|
| Lyman | 1 | 2, 3, 4... | UV | 121.6 nm (2→1) |
| Balmer | 2 | 3, 4, 5... | Visible | 656 nm (3→2, red) |
| Paschen | 3 | 4, 5, 6... | IR | 1875 nm (4→3) |
| Brackett | 4 | 5, 6, 7... | IR | 4050 nm (5→4) |
| Pfund | 5 | 6, 7, 8... | IR | 7460 nm (6→5) |
| Quick Reference | Formula |
|---|---|
| Rydberg formula | 1/λ = R(1/n1² − 1/n2²) |
| Total spectral lines | N = n(n−1)/2 |
| Max transitions (1 electron) | n − 1 |
📅 Section D — Rutherford Scattering
| Quantity | Formula | Note |
|---|---|---|
| Distance of closest approach | d = 2kZe²/KE | d ∝ 1/KE |
| Impact parameter | b = (kZe²/KE)cot(θ/2) | b=0 → θ=180° |
| Nucleus size | ~10−15 m (1 fm) | Atom: ~10−10 m |
🧠 Memory Tricks
“13.6 / n² for everything”
En=−13.6/n². KE=+13.6/n². PE=−27.2/n². Ionisation=+13.6/n².
All energies in hydrogen come from 13.6 divided by n². Just remember the signs!
“nth excited = (n+1)th level”
1st excited state = n=2, 2nd excited = n=3, 3rd excited = n=4. Ground state = n=1 (NOT excited). NEET
loves testing this confusion!
“L B P B P → U V I I I”
Lyman=UV, Balmer=Visible, Paschen=IR, Brackett=IR, Pfund=IR. Only Lyman is UV, only Balmer is visible.
All others are IR. The n1 values are simply 1, 2, 3, 4, 5.
“KE = −Total | PE = 2×Total”
KE is always positive and equals negative of total energy. PE is always negative and equals twice the
total energy. |PE| = 2|KE|. These relations hold true for any orbit.
“Lines from n: n(n−1)/2”
From n=4: 4×3/2=6 lines. From n=5: 5×4/2=10 lines. This counts ALL possible transitions for a
collection of atoms. For single electron: max n−1 jumps.
“r ∝ n², v ∝ 1/n, E ∝ 1/n²”
Radius grows as n². Velocity decreases as 1/n. Energy becomes less negative as 1/n². Higher n
= larger orbit, slower electron, less tightly bound.
🔢 Critical Values (Hydrogen)
E1 = −13.6 eV
E2 = −3.4 eV
E3 = −1.51 eV
E4 = −0.85 eV
a0 = 0.529 Å
R = 1.097×107 m−1
v0 = 2.18×106 m/s
IE(H) = 13.6 eV
1st exc = 10.2 eV
❌ Common Mistakes to Avoid
- Excited state numbering: 1st excited = n=2 (NOT n=1). 3rd excited = n=4. Ground state is n=1.
- Rydberg formula: 1/λ=R(1/n1²−1/n2²) with n1<n2. Don't swap them or result becomes negative!
- Longest λ ≠ series limit: Longest λ = smallest energy gap (n2=n1+1). Shortest λ (series limit) = largest energy gap (n2=∞).
- Energy is negative: More negative = more tightly bound. E1 is the most negative (most stable). Don't say E1 is “maximum energy”!
- Bohr model: H & H-like only: Works for H, He+, Li2+ (single electron). Does NOT work for He, Li, etc.
- Total vs single electron lines: n(n−1)/2 is for many atoms. Single electron makes at most (n−1) transitions.
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App
📱 Practice MCQs for this topic inside our App