The potential difference between the target and cathode of an X-ray tube is 50 kV and the current in the tube is 20 mA. Only 1% of the total energy is emitted as X-radiation.
(a) What is the maximum frequency of the emitted radiation?
(b) At what rate must heat be removed from the target in order to keep it at a steady temperature?
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Answer:
(a) hfmax = eV
Thus,
fmax
= eV/h
= (1.6 * 10-19) * (50 * 103) / (6.63 * 10-34)
= 1.2 * 1019 Hz
(b) Rate of heat removed
= electrical power supplied - rate of emitted X-ray energy
= 99% * electrical power supplied
= 0.99 VI
= 0.99 * 50000 * 20 * 10-3
= 900 Js-1
