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Chapter 27: Problem 16

What is the minimum potential difference applied to an x-ray tube if \(x\) -rays of wavelength \(0.250 \mathrm{nm}\) are produced?

Short Answer

Expert verified
Question: Calculate the minimum potential difference that needs to be applied to an x-ray tube to produce x-rays of wavelength 0.250 nm. Answer: The minimum potential difference to be applied is 4969.5 V.

Step by step solution

01

Convert the wavelength to meters

Before we proceed with the solution, let's convert the given wavelength from nm to meters. \(\lambda = 0.250 \mathrm{nm} = 0.250 \times 10^{-9} \mathrm{m}\)
02

Calculate the frequency of the x-rays

Now we'll use the relationship between frequency and wavelength to find the frequency of the x-rays. \(f = \dfrac{c}{\lambda}\) \(f = \dfrac{3.0 \times 10^8 \mathrm{m/s}}{0.250 \times 10^{-9} \mathrm{m}}\) \(f = 1.2 \times 10^{18} \mathrm{Hz}\)
03

Calculate the energy of the photon in Joules

Next, we'll use Planck's equation to calculate the energy of the photon. \(E = h \cdot f\) \(E = (6.626 \times 10^{-34} \mathrm{Js}) \cdot (1.2 \times 10^{18} \mathrm{Hz})\) \(E = 7.9512 \times 10^{-16} \mathrm{J}\)
04

Convert the energy to electron volts

Now convert the energy from Joules to electron volts using the conversion factor. \(E_\mathrm{photon} = \dfrac{E}{1.6 \times 10^{-19} \mathrm{J/eV}}\) \(E_\mathrm{photon} = \dfrac{7.9512 \times 10^{-16} \mathrm{J}}{1.6 \times 10^{-19} \mathrm{J/eV}}\) \(E_\mathrm{photon} = 4969.5\,\mathrm{eV}\)
05

Determine the minimum potential difference

Finally, calculate the minimum potential difference using the energy of the photon in electron volts. \(V_\mathrm{min} = \dfrac{E_\mathrm{photon}}{e}\) \(V_\mathrm{min} = \dfrac{4969.5\,\mathrm{eV}}{1\,\mathrm{eV}}\) \(V_\mathrm{min} = 4969.5\,\mathrm{V}\) Therefore, the minimum potential difference applied to the x-ray tube to produce x-rays of wavelength 0.250 nm is \(\boxed{4969.5\,\mathrm{V}}\).

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Most popular questions from this chapter

A \(100-\) W lightbulb radiates visible light at a rate of about $10 \mathrm{W} ;$ the rest of the EM radiation is mostly infrared. Assume that the lightbulb radiates uniformly in all directions. Under ideal conditions, the eye can see the lightbulb if at least 20 visible photons per second enter a dark-adapted eye with a pupil diameter of \(7 \mathrm{mm}.\) (a) Estimate how far from the source the lightbulb can be seen under these rather extreme conditions. Assume an average wavelength of 600 nm. (b) Why do we not normally see lightbulbs at anywhere near this distance?
A 640 -nm laser emits a 1 -s pulse in a beam with a diameter of $1.5 \mathrm{mm} .\( The rms electric field of the pulse is \)120 \mathrm{V} / \mathrm{m} .$ How many photons are emitted per second? [Hint: Review Section \(22.6 .]\)
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Photons with a wavelength of 400 nm are incident on an unknown metal, and electrons are ejected from the metal. However, when photons with a wavelength of \(700 \mathrm{nm}\) are incident on the metal, no electrons are ejected. (a) Could this metal be cesium with a work function of \(1.8 \mathrm{eV} ?\) (b) Could this metal be tungsten with a work function of 4.6 eV? (c) Calculate the maximum kinetic energy of the ejected electrons for each possible metal when 200 -nm photons are incident on it.
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