The de Broglie wavelength of a particle of mass \( 3.0 \times 10^{-30} \, \text{kg} \) moving at \( 1.0 \times 10^6 \, \
**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. p = m v = 3.0 × 10⁻³⁰ × 1.0 × 10⁶ = 3.0 × 10⁻²⁴ kg m/s . λ = (h/p) = (6.63 × 10⁻³⁴/3.0 × 10⁻²⁴) = 2.21 × 10⁻¹⁰ m = 0.221 nm . Applying E = h f = h c/λ, p = h/λ, K_max
Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves
