The minimum wavelength of X-rays from a \( 25 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \
**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 = e V = 1.6 × 10⁻¹⁹ × 25 × 10³ = 4.0 × 10⁻¹⁵ J . λₘiₙ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/4.0 × 10⁻¹⁵) = 4.9725 × 10⁻¹¹ m ≈ 0.0497 nm . Applying E = h f = h c/λ,
Ref: NCERT > Physics Book > Dual Nature of Matter > De Broglie Wavelength and Matter Waves