A light beam emits \( 2.5 \times 10^{15} \) photons per second, each of energy \( 4.0 \times 10^{-19} \, \text{J} \). Wh
**De Broglie wavelength** λ = h/p = h/(m v) = h/√(2 m e V) for electron accelerated through V, for V=100 V λ= h/√(2 m e×100)=1.227/√V nm=0.1227 nm, for 50 V 0.173 nm, for particle mass 3.0×10⁻³⁰ kg v=10⁶ m/s λ=6.63×10⁻³⁴/(3×10⁻³⁰×10⁶)=2.21×10⁻¹⁰ m=0.221 nm, inversely proportional to momentum, property inversely proportional to de Broglie wavelength is momentum. Power P = N × E . P = 2.5 × 10¹⁵ × 4.0 × 10⁻¹⁹ = 1.0 × 10⁻³ W = 1.0 mW . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀,
Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves
