Answer :

Given:


The wavelength of radiation produced by tube, λ = 0.45 Å


i.e. 0.45 × 10-10 m


(a) The maximum energy of the photon is given by,


…(1)


Where,


E = Energy of photon


h = Planck’s constant = 6.6 × 10-34 Js


c = 3 × 108 ms-1


λ = wavelength of photon


by putting the values in equation (1), we get,



E = 44 × 10-16 J or 27.6 × 103 eV


The maximum energy of the photon in x-ray is 44 × 10-16J.


(b) For a photon to have 27 KeV of energy, the accelerating potential must be of the order of 30 KeV.


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