CBSE Class 12 Physics: Dual Nature of Radiation and Atoms — Important Questions with Answers 2026

Short Answer Questions (2-3 Marks)

• Q: Explain the photoelectric effect. State Einstein's photoelectric equation.
  Ans: Photoelectric effect: When light of sufficient frequency falls on a metal surface, electrons are emitted. Observations: (1) Emission is instantaneous. (2) Below threshold frequency (ν₀), no emission regardless of intensity. (3) KE of photoelectrons depends on frequency, not intensity. (4) Number of electrons depends on intensity. Einstein's equation: E = hν = φ + ½mv²_max. hν = hν₀ + eV₀ (stopping potential). Work function φ = hν₀.
• Q: State de Broglie hypothesis. Calculate the wavelength of an electron accelerated through 100 V.
  Ans: De Broglie hypothesis: Every moving particle has an associated wave. Wavelength λ = h/p = h/(mv). For electron: KE = eV = ½mv². mv = √(2meV). λ = h/√(2meV). Given V = 100 V: λ = 6.63×10⁻³⁴/√(2 × 9.1×10⁻³¹ × 1.6×10⁻¹⁹ × 100) = 6.63×10⁻³⁴/5.4×10⁻²⁴ = 1.23 × 10⁻¹⁰ m = 1.23 Å.
• Q: State Bohr's postulates for the hydrogen atom. Derive the expression for energy of nth orbit.
  Ans: Postulates: (1) Electrons revolve in fixed circular orbits (stationary states) without radiating energy. (2) Angular momentum is quantised: L = mvr = nh/(2π). (3) Energy is emitted/absorbed only when electron jumps between orbits: E = hν = E₂ - E₁. Derivation: From Coulomb force = centripetal force: ke²/r² = mv²/r → mv² = ke²/r. Using L = mvr = nh/2π → v = nh/(2πmr). Substituting: r_n = n²h²/(4π²mke²). E_n = -mk²e⁴/(2n²ℏ²) = -13.6/n² eV.

Exam Tips for This Chapter

• Revise all definitions and laws from Dual Nature of Radiation and Atoms — commonly asked as 1-2 mark questions
• Practice diagrams related to Dual Nature of Radiation and Atoms — neat labelled diagrams carry 2-3 marks
• For numericals, always show formula → substitution → answer with correct units
• Previous year analysis shows Dual Nature of Radiation and Atoms carries significant marks in the board exam