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Question
The distance between the cathode (filament) and the target in an X-ray tube is 1.5 m. If the cutoff wavelength is 30 pm, find the electric field between the cathode and the target.
(Use Planck constant h = 6.63 × 10-34 Js= 4.14 × 10-15 eVs, speed of light c = 3 × 108 m/s.)
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Solution
Given:
Distance between the filament and the target in the X-ray tube, d = 1.5 m
Cut off wavelength, `lambda = 30 "pm"`
Energy (E) is given by
`E = (hc)/lambda`
Here,
h = Planck's constant
c = Speed of light
`lambda` = Wavelength of light
Thus, we have
`E = (1242 "eV" - "nm")/(30 xx 10^-3)`
⇒ `E = (1242 xx 10^-9)/(30 xx 10^-12)`
⇒ `E = 41.4 xx 10^3 "eV"`
Now ,
`"Electric field" = V/d = (41.4 xx 10^3)/1.5`
= `27.6 xx 10^3 "V/m"`
= `27.6 "kV/m"`
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