Compute Nickel Filter Thickness to Increase X-Ray Ratio

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SUMMARY

The discussion focuses on calculating the thickness of a nickel filter required to increase the X-ray intensity ratio I(Kα)/I(Kβ) from 7.5:1 to 500:1 for a copper target. The mass absorption coefficients for copper and nickel are provided, with specific values for Kα and Kβ wavelengths. The solution determined that a thickness of 20 microns of nickel is necessary to achieve the desired ratio. The participant successfully utilized the equation I_{x} = I_{0} e^{-\mu x} and adjusted for weight fractions to arrive at the correct answer.

PREREQUISITES
  • Understanding of X-ray absorption and mass absorption coefficients
  • Familiarity with the equation I_{x} = I_{0} e^{-\mu x}
  • Knowledge of weight fractions in material mixtures
  • Basic principles of X-ray spectroscopy
NEXT STEPS
  • Study the impact of different filter materials on X-ray intensity ratios
  • Learn about the calculation of linear absorption coefficients for various elements
  • Explore advanced topics in X-ray spectroscopy and its applications
  • Investigate the effects of varying thicknesses of filters on X-ray transmission
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Students and professionals in materials science, physicists working with X-ray spectroscopy, and anyone involved in optimizing X-ray filtering techniques.

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Homework Statement



If the ratio I(K\alpha)/I(K\beta) 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^{2}/g)

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

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

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

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

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

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


Homework Equations



I_{x} = I_{0} e^{-\mu x}

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 (\mu / \rho ) = x(\mu / \rho)_{1} + (1-x)(\mu / \rho)_{2}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.
 
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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
 

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