What is the minimum potential difference between the filament and the target of an x-ray tube if the tube is to produce x-rays with a wavelength of 0.140 nm ?

Respuesta :

Potential difference is V = 8.87KV

what is x -ray?

Atomic electrons and unbound electrons that are decelerating close to atoms are what produce X-rays (i.e., Bremsstrahlung). Electrons are accelerated through an electrical voltage potential and stopped in a target by radiation-producing equipment to create X-rays.

There are various  types of x-ray:

  • plain radiography, or plain x-ray
  • computed tomography, known as CT scanning
  • fluoroscopy — which produces moving images of an organ
  • mammography — an x-ray of the breasts
  • angiography — an x-ray of the blood vessels

λ = 0.140nm  =  1.4 × 10⁻¹⁰m

h=6.63×10⁻³⁴ J/s

C=3×10⁸m/s

e=1.6×10⁻¹⁹C

λmax = hc/eV or V( potential difference) = hc/eλ

V = (6.63×10⁻³⁴ J/s) × (3×10⁸m/s) / (1.6×10⁻¹⁹C) × ( 1.4 × 10⁻¹⁰m)

V = 8.8794 × 10³V

V = 8.87KV

to learn more about X- ray go to - https://brainly.com/question/14306992

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