A light source emits \( 3.0 \times 10^{16} \) photons per second with a power of \( 9.0 \, \text{mW} \). What is the wav
**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 = (P/N) = (9.0 × 10⁻³/3.0 × 10¹⁶) = 3.0 × 10⁻¹⁹ J . λ = (h c/E) = (6.63 × 10⁻³⁴ × 3 × 10⁸/3.0 × 10⁻¹⁹) = 6.63 × 10⁻⁷ m = 663 nm . Applying E = h f = h c/λ, p = h/λ, K_max = h f
Ref: NCERT > Physics Book > Dual Nature of Matter > Photon Energy, Momentum, Power and X-rays