(a) An X-ray tube produces a continuous spectrum of radiation withits short wavelength end at 0.45 Å. What is the maximum energyof 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 ?
(a)
It is given that the wavelength produced by an X-ray tube, λ = 0 ⋅ 45 A o or 0 ⋅ 45 × 10 − 10 m .
The formula of maximum energy of photon is,
E = h c λ
Substitute the values in above expression.
E = 6 ⋅ 626 × 10 − 34 × 3 × 10 8 0 ⋅ 45 × 10 − 10 × 1 ⋅ 6 × 10 19 = 27 ⋅ 6 × 10 3 eV = 27 ⋅ 6 keV
Hence, the maximum energy of a photon is 27 ⋅ 6 keV .
(b)
The maximum energy for an X-ray photon is 27 ⋅ 6 keV . Thus, for producing X-rays the incident electrons must possess at least 27 ⋅ 6 keV of kinetic energy.
Hence, the order of 30 keV of an accelerating voltage is required to produce X-rays.
(a)
It is given that the wavelength produced by an X-ray tube,
The formula of maximum energy of photon is,
Substitute the values in above expression.
Hence, the maximum energy of a photon is
(b)
The maximum energy for an X-ray photon is
Hence, the order of
