JEE Physics • Official PYQs

Modern Physics & Semiconductors

Practice JEE Main Photoelectric effect, Bohr Model, Nuclear Physics & Semiconductor Diodes PYQs online. Includes de Broglie shortcuts & 4-step solutions.

📝 15 Solved PYQs
📅 Exam Years: 2019–2024
📊 Weightage: 10%–12% (3–4 Questions per shift — Easy & High Yield)
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🎯 High-Yield Syllabus Breakdown

These high-frequency concepts account for over 90% of all questions asked in Modern Physics & Semiconductors:

⚡ Essential Formulas & Speed Shortcuts

Formulas and time-saving shortcuts to solve Modern Physics questions in under 45 seconds:

Speed Trick #1:

Electron de Broglie Wavelength: λ = 12.27 / √V Å = 1.227 / √V nm.

Speed Trick #2:

First excitation energy of Hydrogen: ΔE = E₂ - E₁ = -3.4 - (-13.6) = 10.2 eV.

Speed Trick #3:

Zener diode operates in reverse breakdown to regulate DC output voltage.

📖 Sample Solved Exam Questions

Preview authentic questions from this chapter with full step-by-step solutions:

Question 1 of 3 JEE Main 2024 (Online) 27th Jan Shift-1
The de Broglie wavelength of an electron accelerated from rest through a potential difference of V volts is approximately:
(A) 1.227 / √V nm
(B) 12.27 / √V nm
(C) 0.286 / √V nm
(D) 0.101 / √V nm
💡 View Step-by-Step Solution & Answer

Correct Answer: 1.227 / √V nm

  1. Step 1: de Broglie wavelength is λ = h / p = h / √(2 m e V).
  2. Step 2: Substitute Planck's constant h = 6.626 × 10⁻³⁴ J·s, m_e = 9.11 × 10⁻³¹ kg, e = 1.6 × 10⁻¹⁹ C.
  3. Step 3: λ = (6.626 × 10⁻³⁴) / √(2 × 9.11 × 10⁻³¹ × 1.6 × 10⁻¹⁹ × V).
  4. Step 4: Simplifying yields λ = (12.27 / √V) Å = 1.227 / √V nm.

⚡ Speed Tip: Standard electron formula: λ = 12.27 / √V Å = 1.227 / √V nm. For V = 100 V: λ = 0.1227 nm.

Question 2 of 3 JEE Main 2024 (Online) 29th Jan Shift-1
The stopping potential for photoelectrons emitted from a surface when illuminated by light of frequency ν is V₀. Einstein's photoelectric equation is:
(A) e V₀ = h ν - Φ₀
(B) e V₀ = h ν + Φ₀
(C) V₀ = h ν - Φ₀
(D) e V₀ = Φ₀ - h ν
💡 View Step-by-Step Solution & Answer

Correct Answer: e V₀ = h ν - Φ₀

  1. Step 1: Einstein's photoelectric equation is K_max = h ν - Φ₀, where Φ₀ is the work function.
  2. Step 2: Stopping potential V₀ is the potential required to stop the most energetic photoelectron.
  3. Step 3: Maximum kinetic energy is related to stopping potential by K_max = e V₀.
  4. Step 4: Equating gives e V₀ = h ν - Φ₀.

⚡ Speed Tip: Photoelectric equation: e V₀ = h ν - h ν₀ = h ν - Φ₀.

Question 3 of 3 JEE Main 2023 (Online) 24th Jan Shift-2
The ratio of radii of the first and second Bohr orbits of hydrogen atom (r₁ : r₂) is:
(A) 1 : 4
(B) 1 : 2
(C) 1 : 8
(D) 1 : 16
💡 View Step-by-Step Solution & Answer

Correct Answer: 1 : 4

  1. Step 1: Bohr's formula for radius of n-th orbit is r_n = 0.529 (n² / Z) Å.
  2. Step 2: For hydrogen atom, atomic number Z = 1, so r_n ∝ n².
  3. Step 3: For n = 1: r₁ ∝ 1² = 1. For n = 2: r₂ ∝ 2² = 4.
  4. Step 4: Ratio of radii: r₁ : r₂ = 1 : 4.

⚡ Speed Tip: Bohr radius: r ∝ n² / Z. Doubling principal quantum number quadruples orbital radius.

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❓ Frequently Asked Questions (FAQs)

Why is Modern Physics considered the highest scoring section in JEE Main?

Modern Physics questions are direct formula-based, have predictable patterns (photoelectric, Bohr model, de Broglie), and carry 12 to 16 marks per shift.

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