(a) An X-ray tube produces a continuous spectrum of radiation with its shorts wavelength end ate`0.45 Å`. What is the maximum energy of the photon in the radiation? (b) From your answer to (a) , guess what order of accelerating voltage (for electrons) is required in such a tube?

Explanation

(a) `lambda_(min)=0.45Å =0.45xx10^(-10)m, h=6.6xx10^(-34)Js, c=3xx10^(8) ms^(-1)` We know, maximum energy of X-ray photon is
`E_(max)=hv_(max)=(hc)/(lambda_(min))=(6.63xx10^(-34)xx3xx10^(8))/(0.45xx10^(-10)) J=(6.63xx10^(-34)xx3xx10^(8))/(0.45xx10^(-10)xx1.6xx10^(-19)) eV`
`=27.6xx10^(3)eV=27.6keV`
(b) In x-ray tube, accelerating voltage provides the energy to the electrons which on striking the anticathode, produce X-rays. For getting X-ray photons of 27.6 keV, it is required that the incident electrons must posses kinetic energy of atleast 27.6keV. Therefore, accelerating voltage of the order of 30 keV should be applied across the X-ray tube to get the required X-ray photons.