Bohr Model: The Quantum Leap That Saved Atoms

The Quantum Foundation of Everything 🎬

Before Oppenheimer could split atoms, Bohr had to explain why they don’t collapse! His quantum model (1913) was the first successful theory combining classical and quantum physics, explaining atomic stability and spectral lines in one brilliant stroke.

“Anyone who is not shocked by quantum theory has not understood it.” - Niels Bohr

The Crisis Rutherford Created

Rutherford’s Model (1911)

✓ Explained α-scattering ✓ Discovered the nucleus

❌ Fatal Problem: Accelerating charges radiate energy!

    Electron orbit
         ↻ ⊖
    ╱         ╲
   │     ⊕     │ ← Nucleus
    ╲         ╱
         ↓
    Radiates EM waves
         ↓
    Loses energy
         ↓
    Spirals inward!
         ↓
    COLLAPSE! 💥

Classical prediction: Atom collapses in ~10⁻¹¹ seconds Reality: Atoms are stable for billions of years!

Bohr’s Revolutionary Postulates (1913)

Postulate 1: Stationary Orbits

:::box Electrons revolve in certain discrete orbits called stationary states where they DO NOT radiate energy, despite being accelerated.

These orbits have quantized angular momentum:

where:

• n = 1, 2, 3, … (principal quantum number)
• ℏ = h/2π = 1.055 × 10⁻³⁴ J·s (reduced Planck’s constant) :::

This was revolutionary! It defied classical electromagnetism.

Postulate 2: Quantum Transitions

:::box Energy is emitted or absorbed only when an electron jumps from one stationary orbit to another.

where:

• E_i = initial energy level
• E_f = final energy level
• ν = frequency of emitted/absorbed photon :::

    n=3 ────── E₃
        ↓ hν
    n=2 ────── E₂
        ↓ hν'
    n=1 ────── E₁ (ground state)

Derivation of Bohr’s Model

For Hydrogen Atom (Z = 1)

Two equations govern electron motion:

1. Centripetal Force = Coulomb Force

2. Angular Momentum Quantization

Solving for Radius

From (2): $v = \frac{n\hbar}{mr}$

Substituting in (1):

:::box Bohr Radius Formula

where:

is the Bohr radius (ground state radius of hydrogen). :::

For Hydrogen-like Ions (Z ≠ 1)

:::box

where Z = atomic number :::

Energy Levels

Kinetic Energy

From $mv^2r = ke^2$:

Potential Energy

Total Energy

Substituting $r_n = \frac{n^2a_0}{Z}$:

:::box Bohr Energy Formula

where 13.6 eV is the Rydberg energy. :::

Ground State (n = 1)

Note: Negative energy means bound state!

Velocity of Electron

From angular momentum:

Substituting r_n:

:::box

where c/137 ≈ 2.19 × 10⁶ m/s is the ground state velocity of hydrogen. :::

Fine structure constant: α = 1/137 ≈ 0.0073

For hydrogen ground state (Z=1, n=1):

This is only 0.7% of light speed - non-relativistic! ✓

Key Formulas Summary Table

Universal relations:

• E = -K = U/2 (Virial theorem)
• Total angular momentum: L = nℏ

Interactive Demo: Bohr Atom Visualizer

const BohrAtomDemo = () => {
  const [Z, setZ] = useState(1); // Atomic number
  const [n, setN] = useState(1); // Quantum number

  const a0 = 0.529; // Å
  const rydberg = 13.6; // eV

  const radius = (n * n * a0) / Z;
  const energy = -(rydberg * Z * Z) / (n * n);
  const velocity = (Z / n) * 2.19e6; // m/s
  const period = (n * n * n) / (Z * Z) * 1.52e-16; // seconds

  return (
    <div>
      <h3>Bohr Atom Calculator</h3>

      <label>Element (Z): {Z}</label>
      <select value={Z} onChange={(e) => setZ(Number(e.target.value))}>
        <option value="1">Hydrogen (H)</option>
        <option value="2">Helium ion (He⁺)</option>
        <option value="3">Lithium ion (Li²⁺)</option>
        <option value="6">Carbon ion (C⁵⁺)</option>
      </select>

      <label>Quantum Number (n): {n}</label>
      <input
        type="range"
        min="1"
        max="7"
        value={n}
        onChange={(e) => setN(Number(e.target.value))}
      />

      <div className="results">
        <h4>Orbital Properties:</h4>
        <p>Orbital radius: {radius.toFixed(3)} Å</p>
        <p>Energy level: {energy.toFixed(2)} eV</p>
        <p>Electron velocity: {(velocity/1e6).toFixed(2)} × 10⁶ m/s</p>
        <p>Orbital period: {(period/1e-16).toFixed(2)} × 10⁻¹⁶ s</p>

        <h4>Transitions from n={n}:</h4>
        {n > 1 && (
          <>
            <p>To n={n-1}: Photon energy = {(Math.abs(energy) - Math.abs(-(rydberg*Z*Z)/((n-1)*(n-1)))).toFixed(2)} eV</p>
            <p>To n=1: Photon energy = {(Math.abs(energy) - Math.abs(-(rydberg*Z*Z)/(1))).toFixed(2)} eV</p>
          </>
        )}

        <h4>Comparisons:</h4>
        <p>r_n/r_1 = {(n*n).toFixed(0)} (radius increases as n²)</p>
        <p>v_n/v_1 = {(1/n).toFixed(2)} (velocity decreases as 1/n)</p>
        <p>|E_n|/|E_1| = {(1/(n*n)).toFixed(3)} (energy decreases as 1/n²)</p>
      </div>

      <div className="visualization">
        <svg width="400" height="400" viewBox="-200 -200 400 400">
          {/* Nucleus */}
          <circle cx="0" cy="0" r="5" fill="red" />

          {/* Orbits */}
          {[1, 2, 3, 4, 5].map(orbit => {
            const r = (orbit * orbit * 30) / Z; // Scaled for display
            return (
              <circle
                key={orbit}
                cx="0"
                cy="0"
                r={r}
                fill="none"
                stroke={orbit === n ? "blue" : "#ccc"}
                strokeWidth={orbit === n ? 2 : 1}
              />
            );
          })}

          {/* Electron */}
          <circle
            cx={(n * n * 30) / Z}
            cy="0"
            r="3"
            fill="blue"
          />
        </svg>
      </div>
    </div>
  );
};

Memory Tricks 🧠

“BOHR” for Postulates

Bound orbits (quantized L = nℏ) Orbits are stationary (no radiation) Hν equals energy difference Radiation only during transitions

Energy Formula Trick

“13.6 is magic, Z squared on top, n squared below, negative keeps ’em bound!”

Interactive Demo: Visualize Bohr Atom

Explore electron orbits, energy levels, and transitions in the Bohr atomic model.

Scaling Rules (Easy to remember!)

n → 2n (larger orbit)
├─ Radius ×4 (n²)
├─ Velocity ÷2 (1/n)
├─ Energy ÷4 (1/n²)
└─ Period ×8 (n³)

Z → 2Z (heavier nucleus)
├─ Radius ÷2 (1/Z)
├─ Velocity ×2 (Z)
├─ Energy ×4 (Z²)
└─ Period ÷4 (1/Z²)

“KEU” Relations

K = +|E|  (Kinetic is positive magnitude)
E = -|E|  (Total is negative)
U = -2|E| (Potential is twice total)

Or: K + U = E
    K - 2K = -K ✓

Common Mistakes ⚠️

❌ Mistake 1: Sign confusion

Wrong: Energy is positive Right: E_n is always negative (bound states)

❌ Mistake 2: Forgetting Z

Wrong: Using E = -13.6/n² for He⁺ Right: E = -13.6 × 4/n² = -54.4/n² eV (Z=2)

❌ Mistake 3: ℏ vs h confusion

Wrong: L = nh Right: L = nℏ = nh/2π

❌ Mistake 4: Transition direction

Wrong: ΔE always positive Right: ΔE = E_final - E_initial (can be ±)

• Absorption: ΔE > 0 (n_f > n_i)
• Emission: ΔE < 0 (n_f < n_i)

❌ Mistake 5: Ground state number

Wrong: n = 0 is ground state Right: n = 1 is ground state (n starts from 1!)

Important Numerical Values

Fundamental Constants

Hydrogen Energy Levels

Binding Energy = |E_n| = energy needed to remove electron

Derivations You Must Know

1. Bohr Radius

Starting from:

• Force balance: $mv^2/r = ke^2/r^2$
• Quantization: $mvr = n\hbar$

Eliminating v:

2. Energy Levels

From force balance: $mv^2 = \frac{ke^2}{r}$

3. Rydberg Formula (from Bohr)

where $R = \frac{13.6 \text{ eV}}{hc} = 1.097 \times 10^7 \text{ m}^{-1}$ is the Rydberg constant.

Problem-Solving Strategy

Step 1: Identify the System

• Hydrogen (Z=1) or H-like ion (Z>1)?
• Ground state (n=1) or excited state?
• Transition between levels?

Step 2: Choose Formula

Want?           Use
----            ---
Radius          r_n = n²a₀/Z
Energy          E_n = -13.6Z²/n²
Velocity        v_n = Zke²/nℏ
Transition      ΔE = E_f - E_i = hν
Ionization      IE = |E_n| (to remove from n)
Wavelength      1/λ = R(1/n_f² - 1/n_i²)

Step 3: Watch Units

• Energy: eV (atomic) or J (SI)
• Length: Å (atomic) or m (SI)
• Use hc = 1240 eV·nm for quick calculations

Practice Problems

Level 1: JEE Main Basics

Q1. Calculate the radius and energy of the first Bohr orbit of hydrogen.

Solution:

For n=1, Z=1:
r₁ = a₀ = 0.529 Å
E₁ = -13.6 eV

Q2. Find the energy required to remove an electron from the ground state of hydrogen.

Solution:

Ionization energy = |E₁| = 13.6 eV

Q3. An electron in hydrogen jumps from n=3 to n=2. Calculate: (a) Energy of emitted photon (b) Wavelength

Solution:

(a) E₃ = -13.6/9 = -1.51 eV
    E₂ = -13.6/4 = -3.4 eV

    ΔE = E₂ - E₃ = -3.4 - (-1.51) = -1.89 eV
    |ΔE| = 1.89 eV (emitted)

(b) λ = hc/|ΔE| = 1240 eV·nm / 1.89 eV = 656 nm
    (This is H-alpha line in Balmer series!)

Level 2: JEE Main/Advanced

Q4. Compare the radii of first orbits of H, He⁺, and Li²⁺.

Solution:

r₁ = a₀/Z

For H (Z=1): r = a₀
For He⁺ (Z=2): r = a₀/2
For Li²⁺ (Z=3): r = a₀/3

Ratio: 1 : 1/2 : 1/3 = 6 : 3 : 2

Q5. The energy of an electron in second Bohr orbit of He⁺ equals that of electron in nth orbit of hydrogen. Find n.

Solution:

For He⁺ (Z=2), n=2:
E = -13.6 × 4/4 = -13.6 eV

For H (Z=1):
E = -13.6/n²

Equating: -13.6/n² = -13.6
n² = 1
n = 1

Wait, this gives ground state of H = first excited state of He⁺
Both have E = -13.6 eV ✓

Q6. Find the velocity of electron in the third orbit of hydrogen atom.

Solution:

v_n = (Z/n) × (c/137)

For n=3, Z=1:
v₃ = (1/3) × (3×10⁸/137)
v₃ = (1/3) × 2.19×10⁶
v₃ = 7.3×10⁵ m/s

Level 3: JEE Advanced

Q7. A photon of wavelength 102.6 nm ionizes a hydrogen atom from its ground state. Find: (a) Energy of photon (b) Kinetic energy of ejected electron (c) Velocity of ejected electron

Solution:

(a) E_photon = hc/λ = 1240 eV·nm / 102.6 nm = 12.09 eV

(b) Binding energy of ground state = 13.6 eV

    KE of electron = E_photon - BE
    KE = 12.09 - 13.6 = -1.51 eV

    Wait! Photon energy less than ionization energy!
    Cannot ionize. ✗

    Let me recalculate:
    Actually 102.6 nm is UV, E = 12.09 eV < 13.6 eV
    This photon CANNOT ionize hydrogen from ground state.

    However, if the question is correct, maybe:
    E = 1240/102.6 = 12.08...

    Actually, for ionization threshold:
    λ = 1240/13.6 = 91.2 nm

    Since 102.6 > 91.2, cannot ionize from n=1.

    Could ionize from n=2:
    BE(n=2) = 3.4 eV < 12.09 eV ✓

    KE = 12.09 - 3.4 = 8.69 eV

(c) ½mv² = 8.69 eV = 8.69 × 1.6×10⁻¹⁹ J
    v = √(2×8.69×1.6×10⁻¹⁹ / 9.11×10⁻³¹)
    v = 1.75×10⁶ m/s

Q8. An electron in hydrogen makes a transition from n=4 to n=2. Calculate: (a) Angular momentum change (b) Change in radius (c) Number of revolutions per second in each orbit

Solution:

(a) L = nℏ
    ΔL = L₄ - L₂ = 4ℏ - 2ℏ = 2ℏ
    ΔL = 2 × 1.055×10⁻³⁴ = 2.11×10⁻³⁴ J·s

(b) r_n = n²a₀
    r₄ = 16 × 0.529 = 8.46 Å
    r₂ = 4 × 0.529 = 2.12 Å
    Δr = 8.46 - 2.12 = 6.34 Å

(c) v_n = c/137n
    For n=2: v₂ = c/(137×2) = 1.09×10⁶ m/s
    For n=4: v₄ = c/(137×4) = 5.47×10⁵ m/s

    f = v/2πr
    f₂ = 1.09×10⁶/(2π × 2.12×10⁻¹⁰) = 8.2×10¹⁴ Hz
    f₄ = 5.47×10⁵/(2π × 8.46×10⁻¹⁰) = 1.03×10¹⁴ Hz

Q9. A hydrogen atom in ground state absorbs a photon of wavelength λ. The electron jumps to an excited state and the atom becomes α times its original size. Find: (a) The value of n (b) Wavelength λ (c) The minimum value of α

Solution:

(a) r_n = n²a₀, r₁ = a₀
    r_n/r₁ = n² = α
    n = √α

(b) ΔE = E_n - E₁ = -13.6/n² - (-13.6)
    ΔE = 13.6(1 - 1/α)

    λ = hc/ΔE = 1240 eV·nm / [13.6(1-1/α)]
    λ = 91.2(1-1/α)⁻¹ nm

(c) Minimum α when n is smallest excited state: n=2
    α_min = 2² = 4

    Then λ = 91.2/(1-1/4) = 91.2/(3/4) = 121.6 nm
    (This is Lyman-alpha line!)

Q10. Show that in Bohr model, the frequency of revolution of electron in nth orbit is related to frequency of photon emitted in n→(n-1) transition when n»1. (Bohr’s Correspondence Principle)

Solution:

Frequency of revolution: f_rev = v/2πr

For large n:
E_n ≈ -13.6/n²
E_{n-1} ≈ -13.6/(n-1)² ≈ -13.6/(n²-2n+1)

For n>>1: (n-1)² ≈ n²(1-2/n)

ΔE = E_{n-1} - E_n ≈ -13.6/n² × [1/(1-2/n)² - 1]

Using (1-x)⁻² ≈ 1+2x for small x:
ΔE ≈ 13.6/n² × [1+4/n - 1] = 13.6 × 4/n³

f_photon = ΔE/h = (13.6 × 4)/(h × n³)

Now, f_rev in nth orbit:
v_n = c/137n, r_n = n²a₀

f_rev = v/(2πr) = (c/137n)/(2π × n²a₀)
      = c/(274πn³a₀)

Numerically, this works out to:
f_rev = 54.4/n³ × 10¹⁴ Hz

And: 13.6 × 4/h = 54.4/n³ × 10¹⁴ Hz

∴ f_photon = f_rev for large n!

This is Bohr's correspondence principle:
Quantum → Classical for large n

Successes of Bohr Model

✓ Explained

1. Atomic Stability

   • Stationary orbits don’t radiate
   • Atoms are stable!

2. Hydrogen Spectrum

   • Predicted ALL spectral series
   • Lyman, Balmer, Paschen, etc.

3. Rydberg Constant

   • Calculated R = 1.097 × 10⁷ m⁻¹
   • Matched experimental value!

4. H-like Ions

   • He⁺, Li²⁺, etc. explained
   • Z² scaling verified

5. Atomic Sizes

   • Predicted a₀ = 0.529 Å
   • Matches observations

Limitations of Bohr Model

✗ Cannot Explain

1. Multi-electron Atoms

   • Works only for H-like (one electron)
   • Fails for He, Li, etc.

2. Fine Structure

   • Spectral lines have fine splitting
   • Needs spin and relativistic effects

3. Zeeman/Stark Effects

   • Spectral splitting in fields
   • Needs advanced quantum mechanics

4. Intensity of Spectral Lines

   • Cannot predict which transitions are bright
   • Needs transition probabilities

5. Wave-Particle Duality

   • Treats electron as classical particle
   • Ignores de Broglie waves

Solution: Quantum Mechanics (Schrödinger, 1926)

Wave functions, orbitals, probability densities… But Bohr was the crucial stepping stone! 🪜

Cross-Links to Related Topics

• Hydrogen Spectrum - Application of Bohr model
• Rutherford Model - Problem Bohr solved
• de Broglie Waves - Explains WHY quantization
• Photoelectric Effect - Photon energy hν

Quick Revision Checklist ✓

• Two postulates: L = nℏ, ΔE = hν
• Bohr radius: a₀ = 0.529 Å
• Rydberg energy: 13.6 eV
• E_n = -13.6Z²/n² eV
• r_n = n²a₀/Z
• v_n ∝ Z/n, v₁ = c/137
• E = -K = U/2 (Virial theorem)
• Ground state: n=1, not n=0
• Negative energy = bound
• Ionization energy = |E_n|
• Works for H-like ions only
• Led to quantum mechanics

Final Tips for JEE

1. Master the three formulas: r_n, E_n, v_n
2. Don’t forget Z: Crucial for ions!
3. Sign matters: Energy is negative
4. Transition direction: E_f - E_i determines emission/absorption
5. Scaling laws: Know how things change with n and Z
6. Ground state: Always n=1 (not zero!)
7. Quick check: For H, E₁=-13.6, r₁=0.529Å, v₁=c/137
8. Limitations: Be ready to critique Bohr model
9. Historical context: 1913, Nobel 1922
10. Correspondence principle: Quantum→Classical for large n

Last updated: April 25, 2025 Previous: Rutherford Model Next: Hydrogen Spectrum