Dual Nature of Radiation and Matter — Exam Practice

Answer Each photon carries a fixed energy hν, and photoemission is a strictly one-photon-one-electron process. If hν < φ₀ (the work function), no single photon carries enough energy to eject an electron — increasing intensity only increases the number of photons, not the energy each one carries.

Answer De Broglie proposed that every moving particle has an associated wavelength λ = h/p, extending wave-particle duality (already established for light) to matter.

An electron accelerated through potential V gains kinetic energy eV = ½mv², so p = mv = √(2meV).

Substituting: λ = h/√(2meV), which simplifies numerically to λ ≈ 1.227/√V nm (V in volts).

Answer (a) KE_max = hν − φ₀. This explains the threshold frequency (below which hν < φ₀, so KE_max would be negative — impossible, hence no emission), the frequency-dependence (not intensity-dependence) of KE_max, and instantaneous emission (each photon acts in a single event, no accumulation time needed).

(b) λ₀ = hc/φ₀ = (6.63×10⁻³⁴ × 3×10⁸) / (4.2 × 1.6×10⁻¹⁹) ≈ 2.96×10⁻⁷ m ≈ 296 nm.
At λ = 200 nm: E = hc/λ = (6.63×10⁻³⁴×3×10⁸)/(200×10⁻⁹) ≈ 9.95×10⁻¹⁹ J ≈ 6.22 eV.
KE_max = E − φ₀ = 6.22 − 4.2 = 2.02 eV.

Solution More intense light means more photons arriving per second, ejecting more electrons per second (higher current) — but each electron's individual maximum kinetic energy stays fixed by frequency alone.

Solution eV₀ = KE_max = hν − φ₀ — depends only on frequency (for a given metal), completely independent of intensity, echoing the same reasoning as Q4.

Solution λ = h/p — larger momentum means a smaller de Broglie wavelength, which is exactly why macroscopic (high-momentum) objects have wavelengths far too small to ever observe.

Solution Electrons diffracted off a nickel crystal exactly as de Broglie's wavelength predicted — direct evidence that matter genuinely has wave properties, not just theoretical speculation.

Solution λ ≈ 1.227/√V nm = 1.227/√100 = 1.227/10 ≈ 0.123 nm.

Solution A photon has zero rest mass and always travels at exactly c — it can never be brought to rest. Its momentum is p = E/c = h/λ, entirely consistent with relativistic energy-momentum relations for a massless particle.

Solution E = hc/λ = (6.63×10⁻³⁴ × 3×10⁸) / (400×10⁻⁹) ≈ 4.97×10⁻¹⁹ J ≈ 3.11 eV.

eV₀ = E − φ₀ = 3.11 − 2 = 1.11 eV → V₀ ≈ 1.11 V.