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Question
The wavelength of Kα X-ray of tungsten is 21.3 pm. It takes 11.3 keV to knock out an electron from the L shell of a tungsten atom. What should be the minimum accelerating voltage across an X-ray tube having tungsten target which allows production of Kα X-ray?
(Use Planck constant h = 6.63 × 10-34 Js= 4.14 × 10-15 eVs, speed of light c = 3 × 108 m/s.)
Solution
Given:-
Wavelength of X-ray of tungsten,
`lambda` = 21.3 pm
Energy required to take out electron from the L shell of a tungsten atom, EL = 11.3 keV
Voltage required to take out electron from the L shell of a tungsten atom, VL = 11.3 kV
Let EK and EL be the energies of K and L, respectively.
`E_K - E_L = (hc)/lambda`
Here,
h = Planck's constant
c = Speed of light
`E_K - E_L = (1242 "eV" - "nm")/(21.3 xx 10^-12)`
`E_K - E_L = (1242 xx 10^-9 "eV")/(21.3 xx 10^-12)`
`E_K - E_L = 58.309 "keV"`
`E_L = 11.3 "keV"`
`therefore E_K = 69.609 "keV"`
Thus, the accelerating voltage across an X-ray tube that allows the production of Kα X-ray is given by
VK = 69.609 kV
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