Absorption of X-rays

  • Thread starter N8
  • Start date
  • #1
N8
6
0

Homework Statement



If the ratio [tex]I(K\alpha)/I(K\beta)[/tex] before filtering is 7.5:1 for a copper target, then compute the thickness of a nickel filter, that would increase this ratio of 500:1.

Mass Absorption Coefficients (cm[tex]^{2}[/tex]/g)

Cu: [tex]\rho[/tex] = 8.93 [tex]g/cm^{3}[/tex]

[tex]K\alpha[/tex] (0.1542nm)
[tex]\mu/\rho[/tex] = 51.54

[tex]K\beta[/tex] (0.1392nm)
[tex]\mu/\rho[/tex] = 38.74

Ni: [tex]\rho[/tex] = 8.91 [tex]g/cm^{3}[/tex]

[tex]K\alpha[/tex] (0.1542nm)
tex]\mu/\rho[/tex] = 48.83

[tex]K\beta[/tex] (0.1392nm)
[tex]\mu / \rho[/tex] = 282.8


Homework Equations



[tex]I_{x} = I_{0} e^{-\mu x}[/tex]

The Attempt at a Solution



The answer is 20 microns.

Trying to align these equations up and solve has not yielded the correct answer. Also, the effect of filter on the mass absorption coeffient can be modified by weight fractions [tex](\mu / \rho ) = x(\mu / \rho)_{1} + (1-x)(\mu / \rho)_{2} [/tex]where x is the weight fraction of one element.

But I am not sure how to incorporate this into my calculations.


Also, unfortunately, my book Structure of Materials (Graff) does not provide any worked out problems.

Any help would be much appreciated.
 

Answers and Replies

  • #2
N8
6
0
nevermind I was ably to solve this with the relevant equation and calculations for the linear absorption coeffecients of Kalpha and Kbeta for Nickel under copper radiation
 

Related Threads on Absorption of X-rays

  • Last Post
Replies
0
Views
865
Replies
1
Views
3K
Replies
3
Views
7K
  • Last Post
Replies
1
Views
4K
Replies
1
Views
11K
Replies
3
Views
994
  • Last Post
Replies
3
Views
681
Replies
1
Views
3K
Top