What is the minimum potential difference?

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SUMMARY

The minimum potential difference required to produce x-rays with a wavelength of 0.250 nm is calculated using the equations E=hf and f=c/wavelength. The frequency is determined to be 1.2e18 Hz. However, the initial calculation of energy resulted in 7.95e-16 kV, which is incorrect as it represents the energy of the x-rays rather than the voltage. The correct approach involves equating the energy of the x-rays to the work done accelerating the electrons, expressed in electron volts (eV).

PREREQUISITES
  • Understanding of Planck's constant (h)
  • Familiarity with the relationship between energy, frequency, and wavelength
  • Knowledge of electron volts (eV) as a unit of energy
  • Basic principles of x-ray tube operation
NEXT STEPS
  • Study the derivation of energy-wavelength relationships in x-ray production
  • Learn how to convert energy in joules to electron volts (eV)
  • Research the operation principles of x-ray tubes and their voltage requirements
  • Explore the implications of potential difference on x-ray intensity and quality
USEFUL FOR

Physics students, electrical engineers, radiology technicians, and anyone involved in x-ray technology and applications.

sweetipie2216
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What is the minimum potential difference applied to an x-ray tube if x-rays of wavelength 0.250 nm are produced?



E=hf
f=c/wavelength

f=3e8/.250e-9
=1.2e18

then i applied it to the first equation using Plancks constant as h and my final answer came out to be 7.95e-16 KV.

the answer is wrong. can you please help me out? thanks.
 
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half way there,

the work done accelerating the electrons is equal to [tex]eV[/tex], but you can also find out the energy of the xrays, as you've shown, these energies should be equal.

you worked out the energy of the xrays, not the voltage.
 

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