The threshold wavelength for a metal is \( 600 \, \text{nm} \). What is its work function in eV? (Take \( h c = 1240 \,
**De Broglie wavelength calculations** for electron moving at 2×10⁶ m/s λ= h/(m v)=6.63×10⁻³⁴/(9.11×10⁻³¹×2×10⁶)=3.64×10⁻¹⁰ m=0.364 nm, for mass 5×10⁻³⁰ kg v=2×10⁵ m/s λ=6.63×10⁻³⁴/(1×10⁻²⁴)=6.63×10⁻¹⁰ m, showing λ decreases with increasing m v. Work function Φ₀ = (h c/λ₀) . Φ₀ = (1240/600) ≈ 2.07 eV . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ = h/√(2 m e V) and f_max = e V/h, result 2.07 eV follows, reflecting photoelectric and de Broglie relations.
Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves
