An X-ray tube operates at 40 kV. Suppose the electron converts 70% of its energy into a photon at each collision. Find the lowest three wavelengths emitted from the tube. Neglect the energy imparted to the atom with which the electron collides.

Answer

`lambda(hc)/(20xx10^(3)xx(0.7xx0.3)e)=295.6 "pm"`
`lambda_(3)=(hc)/(20xx10^(3)xx0.7xx(0.3)^(2)e)=985.6"pm"`