• Support PF! Buy your school textbooks, materials and every day products Here!

K-alpha xrays, potential difference

  • Thread starter mateomy
  • Start date
  • #1
307
0
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. 9


In 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 [itex]K_\alpha[/itex] 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:

[tex]
E_n = \frac{-Z^2 R_\infty}{n^2}
[/tex]

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.
 

Answers and Replies

  • #2
33,808
9,517
11,441 eV, rounded to keV, is just 11 keV, and this corresponds to 11 kV.
And your formula looks good.
 
  • #3
307
0
Awesome. Thanks!
 

Related Threads for: K-alpha xrays, potential difference

  • Last Post
Replies
2
Views
3K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
17
Views
3K
  • Last Post
Replies
1
Views
952
Replies
2
Views
12K
  • Last Post
2
Replies
29
Views
5K
Replies
2
Views
754
Top