Department of Pre-University Education, Karnataka course PUC Karnataka Science Class 12
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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. - Physics

ConceptElectromagnetic Spectrum

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.)

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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Solution 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. Concept: Electromagnetic Spectrum.
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