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.

Explanation

Given `v=40KV=40xx10^3 `
` Energy =40xx10^3eV `
` Energy Utilised=70/100xx40xx10^3 `
` =28xx10^3eV `
` lambda=(hc)/E=(1242-eVnm)/(28xx10^3eV) `
` =44.35xx10^3nm `
` 44.35p m`
For other wavelength
` E=70%(Left over energy ) `
` =70/100xx(40-28)10^3 `
` =70/100xx12xx10^3 `
` 84xx10^2 `
` lambda=(hc)/E=1242/(8.4xx10^3) `
` =147.86xx10^3nm `
` 147.86p m =148p m `
for third wavelength ,
` E=70/100(12-8.4)xx10^3 `
` =7xx3.6 xx10^2 =25.2xx10^2 `
` lambda =(hc)/E=1242/(25.2xx10^2) ~
` =49.2857xx10^-2 `ltbr` =493p m`