The minimum wavelength of X-rays from a \( 15 \, \text{kV} \) tube is: (Take \( h = 6.63 \times 10^{-34} \, \text{J s} \
**X-rays from tube** maximum frequency f_max = e V/h, minimum wavelength λ_min = h c/(e V), Duane-Hunt law, for 10 kV f_max=1.6×10⁻¹⁹×10⁴/6.63×10⁻³⁴=2.41×10¹⁸ Hz, λ_min= c/f_max=0.124 nm, for 35 kV λ_min=1240/(35000) eV·nm? Actually 35 keV photon λ=1240/35000=0.0354 nm, for 15 kV 0.0827 nm, for 25 kV 0.0496 nm, for 30 kV 0.0413 nm. E = e V = 1.6 × 10⁻¹⁹ × 15 × 10³ = 2.4 × 10⁻¹⁵ J . λₘiₙ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/2.4 × 10⁻¹⁵) ≈ 8.2875 × 10⁻¹¹ m = 0.082875 nm . Applying E = h f = h c/λ, p = h/λ, K_max
Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays