K-alpha xrays, potential difference

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SUMMARY

The discussion centers on calculating the minimum potential difference required to observe K-alpha X-rays from copper using the equation E_n = -Z² R_∞ / n². The user calculated a value of 11,441 eV, which correctly rounds to 11 kV, matching the answer provided in Krane's Modern Physics. The calculation is confirmed as accurate, indicating that the user applied the correct methodology for estimating the potential difference needed for K-alpha X-ray emission.

PREREQUISITES
  • Understanding of X-ray production and electron interactions
  • Familiarity with the Rydberg formula for energy levels
  • Knowledge of atomic structure, specifically the role of electrons in energy levels
  • Basic concepts of potential difference in physics
NEXT STEPS
  • Study the principles of X-ray tube operation and electron acceleration
  • Learn more about the Rydberg constant and its applications in spectroscopy
  • Explore the physics of K-alpha X-ray emission and its significance in material analysis
  • Investigate the effects of varying potential differences on X-ray production
USEFUL FOR

Physics students, educators, and professionals in materials science or medical imaging who are interested in the principles of X-ray production and the calculations involved in determining potential differences for K-alpha emissions.

mateomy
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Just want someone to check my work. I suspect I didn't do it right.The question is from Krane's Modern Physics Chapter 8, prob. 9In an X-ray tube electrons strike a target after being accelerated through a potential difference V. Estimate the minimum value of V required to observe the K_\alpha X rays of copper.

Well, I know that in order to have the emission of a K-alpha x-ray an electron from 1s would need to be "knocked" from its place. And that can only be achieved by an incoming particle (in this case another electron) of the same energy level. So what I did was utilize this equation:

<br /> E_n = \frac{-Z^2 R_\infty}{n^2}<br />

I got a value of 11,441 eV, using Z=29, the Rydberg constant, and n=1. The back of the book says 11 kV. I know initially it says "estimate.." so I'm wondering if I got it right or if it was coincidentally close to the back-of-the-book answer. And if wrong, what would be a better line of attack.

Thank you.
 
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11,441 eV, rounded to keV, is just 11 keV, and this corresponds to 11 kV.
And your formula looks good.
 
Awesome. Thanks!
 

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