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 wavelength decrease?

Question

Aditi Chauhan · Answer

d = 1.5 m, λ = 30 pm = 30 * 10^-3 nm E = hc/λ = 1242/30* 10^3 = 41.4 * 10^eV Electric field = V/d = 41.4* 10^3/1.5 = 27.6 * 10^3 V/m = 27.6 KV/m.

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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 wavelength decrease?

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 wavelength decrease?

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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 wavelength decrease?

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 wavelength decrease?

1 Answers

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Other Related Questions on Modern Physics

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