(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) Given, minimum wavelength,
`lamda_(min)=0.45Å`
`=0.45xx10^(-10)m`
Maximum energy of photon is
`E = hve = (hc)/lamda`
`=(6.625xx10^(-34)xx3xx10^8)/(0.45xx10^(-10)xx1.6xx10^(-19))`
`=27.6xx10^3eV`
`=27.6xx10^3xx1.6xx10^(-19)J`
`=4.42xx10^(-15)J`
(b) The accelerating potential provides the energy to the electron which strikes the target to produce X - rays. Since the wavelength is shortest , the energy emitted is maximum which must be equal to that energy provided by electric field.
`:.` Order of accelerating voltage
E = 27.6 kV
`~~30 kV`