Question

An X-ray tube operates at 40 kV. Suppose the electrons convert 70% its energy into a photon in each collision. Find the lowest 3 wavelength emitted.

Answer 1

${\lambda }_1=\frac{1240\times {10}^{-9}}{.7\left(40\right)\times {10}^3}=44.3pm,$
${\lambda }_2=\frac{1240\times {10}^{-9}}{0.21\left(40\right)\times {10}^3}=148pm,$
${\lambda }_3=\frac{1240\times {10}^{-9}}{.063\times 40\times {10}^3}=493pm$