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What is the accelerating voltage of an x-ray tube that produces x-rays with a shortest wavelength of 0.0103 nm?
Question by OpenStax is licensed under CC BY 4.0.
$120 \textrm{ kV}$
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OpenStax College Physics Solution, Chapter 29, Problem 29 (Problems & Exercises) (2:15)

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Video Transcript
This is College Physics Answers with Shaun Dychko. An x-ray tube is producing x-rays with a wavelength of 0.0103 nanometers and these photons have a certain energy, which we'll figure out from this equation from [29.12]. And then after we know that energy, we'll say that this energy is equal to the kinetic energy of the electrons that are accelerated across the potential difference in the x-ray tube. So these electrons that are accelerated can produce photons with an energy equal to the kinetic energy of those electrons. And that kinetic energy will be the charge of an electron multiplied by the potentail difference through which it is accelerated. So we are going to substitute in place of f, an expression in terms of λ. So we have the wave equation says that the speed of light equals its frequency times λ and we'll divide both sides by λ to find that f equals c over λ and then we'll substitute that in place of f here. And this energy is also going to be the kinetic energy of the electron and so that's the charge times the voltage and so qV can be substituted in place of E and that equals hc over λ. And we'll solve for V by dividing both sides by q. So the voltage is Planck's constant times speed of light divided by the elementary charge of an electron times the wavelength of x-rays produced. So that's 1240 electron volt nanometers times 1.602 times 10 to the minus 19 joules per electron volt to make these electron volts cancel giving us joule nanometers. If you don't like the numbers being used here, you could instead substitute 6.626 times 10 to the minus 34 joule seconds for Planck's constant and then you know, 3 times 10 to the 8 meters per second for speed of light, if you prefer, and you'd get the same answer. Divide by the elementary charge times the 0.0103 nanometers I can write nanometers instead of times 10 to the minus 9 meters because the numerator has units of nanometers and so these nanometers are gonna cancel. This works out to 120 kilovolts.