(a) An X-ray tube produces a continuous spectrum of radiation with its short wavelength end at 0.45 Å. What is the maximum energy of a photon in the radiation?

(b) From your answer to (a), guess what order of accelerating voltage (for electrons) is required in such a tube?

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.