Skip to content

#eV

2 public questions tagged with this topic.

The work function of a metal is \( 1.5 \, \text{eV} \). Light of wavelength \( 400 \, \text{nm} \) is incident on it. 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. E = (h c/λ) = (1240/400) = 3.1 eV . Kₘₐₓ = E - Φ₀ = 3.1 - 1.5 = 1.6 eV . Applying E = h f = h c/λ, p = h/λ, K_max = h f - Φ, Φ = h f₀ = h c/λ₀, λ =

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves

The threshold wavelength of a metal is \( 620 \, \text{nm} \). What is the work function in eV? (Take \( h c = 1240 \, \

**Matter waves** wave-like properties of moving particles, macroscopic objects not exhibit measurable wave-like because λ = h/(m v) extremely small for large m, e.g., 1 kg at 1 m/s λ=6.6×10⁻³⁴ m undetectable, for electron 9.11×10⁻³¹ kg at 2×10⁶ m/s λ=0.36 nm measurable via diffraction, de Broglie hypothesis λ = h/p proposed by de Broglie 1924, Davisson-Germer experiment confirmed. Φ₀ = (h c/λ₀) = (1240/620) = 2.0 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 2.0

Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves