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Question
(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?
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Solution
(a) Wavelength produced by an X-ray tube, `lambda` = 0.45 Å = 0.45 × 10−10 m
Planck’s constant, h = 6.626 × 10−34 Js
Speed of light, c = 3 × 108 m/s
The maximum energy of a photon is given as:
`"E" = "hc"/lambda`
= `(6.626 xx 10^(-34) xx 3 xx 10^8)/(0.45 xx 10^(-10) xx 1.6 xx 10^(19))`
= 27.6 × 103 eV
= 27.6 keV
Therefore, the maximum energy of an X-ray photon is 27.6 keV.
(b) Accelerating voltage provides energy to the electrons for producing X-rays. To get an X-ray of 27.6 keV, the incident electrons must possess at least 27.6 keV of kinetic electric energy. Hence, an accelerating voltage of the order of 30 keV is required for producing X-rays.
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