An X-ray tube operates at 40 kV . Assume that during each collision, the electron converts 70 % of its energy into a photon. Determine the lowest three wavelengths emitted from the tube. Disregard the energy imparted to the atom with which the electron collides.

Step-by-step solution

Initial energy of the electron, E₀ = 40 keV . In the first collision, the electron converts 70 % of its energy into a photon. The energy of the first photon is: E₁ = 0.70 40 keV = 28 keV The remaining kinetic energy of the electron is: E₁' = 40 keV - 28 keV = 12 keV In the second collision, the electron again converts 70 % of its remaining energy into a photon. The energy of the second photon is: E₂ = 0.70 12 keV = 8.4 keV The remaining kinetic energy of the electron is: E₂' = 12 keV - 8.4 keV = 3.6 keV In the thir