19 Worked examples (Lesson 3)
Example 3.1 — Cut-off wavelength. A Coolidge tube operates at 30 kV. Find the shortest wavelength and the highest frequency of the X-rays. eV = hc/λmin → λmin = 1240/30 000 = 0.0413 nm. νmax = c/λmin = 3 × 10⁸/4.13 × 10⁻¹¹ = 7.3 × 10¹⁸ Hz. (Check: ν = eV/h = 1.6 × 10⁻¹⁹ × 3 × 10⁴/6.63 × 10⁻³⁴ = 7.2 × 10¹⁸ Hz.)
Example 3.2 — Speed of the electrons. At 25 kV, kinetic energy at the target = eV = 4 × 10⁻¹⁵ J, speed v = √(2KE/m) = 9.4 × 10⁷ m/s (about 31 % of c, so the result is approximate).
Example 3.3 — Characteristic line. An inner-shell transition in a target atom releases 8.0 keV. λ = 1240/8000 = 0.155 nm. If the voltage is raised from 40 kV to 60 kV, this wavelength is unchanged.