Physics Optics and Modern - How to Solve: Bohr’s Model & Hydrogen Spectrum (IIT JEE Guide)

How to Solve: Bohr’s Model & Hydrogen Spectrum (IIT JEE Guide)

Hook: Mastering Bohr’s model unlocks 5-7 marks in IIT JEE (Main + Advanced) on energy levels, ionization, and wavelength calculations—enough to push you into the top 10%. It’s also the foundation for quantum mechanics, lasers, and even MRI machines.

WHAT YOU NEED TO KNOW FIRST

1. Electromagnetic spectrum basics – Wavelength (λ), frequency (ν), and energy (E) relationship: E = hν = hc/λ.
2. Coulomb’s law – Force between charged particles: F = k(q₁q₂)/r².
3. Circular motion – Centripetal force: F = mv²/r.

KEY TERMS & FORMULAS

Key Terms

• Ground state (n=1): Lowest energy level of an electron.
• Excited state (n>1): Higher energy levels.
• Ionization energy: Energy needed to remove an electron from n=1 to n=∞.
• Rydberg constant (R): R = 1.097 × 10⁷ m⁻¹ (given on exam sheet).
• Bohr radius (a₀): a₀ = 0.529 Å (given on exam sheet).

Formulas

1. Energy of nth level (Eₙ):
   Eₙ = – (13.6 eV) / n²
2. MEMORISE THIS (13.6 eV is the ionization energy of hydrogen).

3. n = principal quantum number (1, 2, 3…).

4. Wavelength of emitted/absorbed photon (Rydberg formula):
   1/λ = R (1/n₁² – 1/n₂²)

5. MEMORISE THIS (R is given, but the form is critical).

6. n₁ = lower energy level, n₂ = higher energy level.

7. Velocity of electron in nth orbit (vₙ):
   vₙ = (2.18 × 10⁶ m/s) / n

8. MEMORISE THIS (derived from Bohr’s quantization condition).

9. Radius of nth orbit (rₙ):
   rₙ = a₀ × n²

10. a₀ = Bohr radius (given).

11. Energy difference between levels (ΔE):
    ΔE = Eₙ₂ – Eₙ₁ = 13.6 (1/n₁² – 1/n₂²) eV

12. MEMORISE THIS (use for photon energy calculations).

STEP-BY-STEP METHOD

Step 1: Identify the transition

• If the question gives two energy levels (n₁, n₂), note them.
• If it gives wavelength (λ), use 1/λ = R (1/n₁² – 1/n₂²) to find n₁, n₂.
• If it asks for ionization energy, set n₂ = ∞ (since ionization = removing electron to infinity).

Step 2: Calculate energy difference (ΔE)

• Use ΔE = 13.6 (1/n₁² – 1/n₂²) eV.
• If n₂ = ∞, ΔE = 13.6 / n₁² (ionization energy from level n₁).

Step 3: Relate ΔE to wavelength (if needed)

• ΔE = hc/λ → λ = hc/ΔE.
• Convert ΔE to joules (1 eV = 1.6 × 10⁻¹⁹ J) before using hc = 1240 eV·nm.

Step 4: Check units

• Energy: eV or joules? Convert if needed.
• Wavelength: nm or meters? 1 nm = 10⁻⁹ m.
• Rydberg constant (R): Given in m⁻¹, so λ will be in meters.

Step 5: Solve for the unknown

• Plug numbers into the correct formula.
• Cross-check with given data (e.g., if λ is given, use Rydberg formula first).

WORKED EXAMPLES

Example 1 – Basic: Energy of n=3 level

Question: What is the energy of an electron in the n=3 level of hydrogen? Steps: 1. Use Eₙ = –13.6 / n². 2. n = 3 → E₃ = –13.6 / 9 = –1.51 eV. Answer: –1.51 eV. What we did and why: Direct formula application. Negative sign means bound state.

Example 2 – Medium: Wavelength of emitted photon (n=3 → n=2)

Question: Find the wavelength of light emitted when an electron jumps from n=3 to n=2. Steps: 1. ΔE = 13.6 (1/2² – 1/3²) = 13.6 (1/4 – 1/9) = 13.6 × (5/36) = 1.89 eV. 2. Convert to joules: 1.89 eV × 1.6 × 10⁻¹⁹ = 3.02 × 10⁻¹⁹ J. 3. λ = hc/ΔE = (6.63 × 10⁻³⁴ × 3 × 10⁸) / (3.02 × 10⁻¹⁹) = 6.58 × 10⁻⁷ m = 658 nm. Answer: 658 nm (red light). What we did and why: Used energy difference to find wavelength via ΔE = hc/λ.

Example 3 – Exam-Style: Ionization energy from n=2

Question: What is the minimum energy required to ionize a hydrogen atom from the n=2 state? Steps: 1. Ionization = n₂ = ∞ → ΔE = 13.6 (1/2² – 1/∞²) = 13.6 / 4 = 3.4 eV. 2. Convert to joules: 3.4 × 1.6 × 10⁻¹⁹ = 5.44 × 10⁻¹⁹ J. Answer: 3.4 eV or 5.44 × 10⁻¹⁹ J. What we did and why: Recognized ionization as n₂ = ∞ and used the energy formula.

COMMON MISTAKES

1. MISTAKE: Forgetting the negative sign in Eₙ.
   WHY IT HAPPENS: Students treat energy as always positive.
   CORRECT APPROACH: Negative sign means bound state (electron is trapped).

2. MISTAKE: Mixing up n₁ and n₂ in Rydberg formula.
   WHY IT HAPPENS: Not noting which level is higher.
   CORRECT APPROACH: n₂ > n₁ (emission: higher → lower; absorption: lower → higher).

3. MISTAKE: Using eV in λ = hc/ΔE without converting to joules.
   WHY IT HAPPENS: Forgetting hc is in joule-seconds.
   CORRECT APPROACH: Convert ΔE to joules or use hc = 1240 eV·nm.

4. MISTAKE: Assuming R is in nm⁻¹ instead of m⁻¹.
   WHY IT HAPPENS: Confusing units.
   CORRECT APPROACH: R = 1.097 × 10⁷ m⁻¹ → λ in meters.

5. MISTAKE: Calculating ionization energy as 13.6 eV for all levels.
   WHY IT HAPPENS: Not adjusting for n₁.
   CORRECT APPROACH: Ionization from n=1 is 13.6 eV; from n=2 it’s 3.4 eV.

EXAM TRAPS

1. TRAP: Giving wavelength in nm but asking for energy in joules.
   HOW TO SPOT IT: Units in question don’t match formula.
   HOW TO AVOID IT: Convert λ to meters or use hc = 1240 eV·nm.

2. TRAP: Asking for "minimum wavelength" (ionization) but not specifying n.
   HOW TO SPOT IT: "Minimum wavelength" implies n=1 → ∞.
   HOW TO AVOID IT: Assume n=1 unless stated otherwise.

3. TRAP: Using E = hν but giving λ instead of ν.
   HOW TO SPOT IT: Question gives λ but asks for E via hν.
   HOW TO AVOID IT: Convert λ to ν using ν = c/λ first.

1-MINUTE RECAP (Night Before Exam)

"Listen up—this is 5-7 marks in your pocket. Bohr’s model is all about energy levels and wavelengths. Memorize these two formulas: 1. Eₙ = –13.6 / n² (energy of level n). 2. 1/λ = R (1/n₁² – 1/n₂²) (wavelength of emitted/absorbed light).

For ionization, set n₂ = ∞. For wavelength, use ΔE = hc/λ. Always check units—R is in m⁻¹, so λ must be in meters. If they give λ in nm, convert it. If they ask for energy in joules, convert eV to joules. And remember: emission is higher n to lower n; absorption is the reverse.

That’s it. Now go crush it."