Practice question

**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. The stopping potential is the minimum negative voltage that stops the most energetic photoelectrons, directly related to their maximum kinetic energy ( e V₀ = Kₘₐₓ ). Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ =