Radiation penetration

  • Thread starter Millano
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Homework Statement



Calculate minimum thickness of concrete wall of radiation room (2X2X2) such that radiographer doesnt exceed dose limit. Use info below.


X ray reading surface of patient (50 cm from source) = exposure= 4.29 X 10[tex]^{}-4[/tex]

effective energy of x ray beam = 40 kev

40% of x rays are transmitted

does limit =7.5 mGy per anum

linear attenuation coeficent = 1.16 cm

scattered radiation increases does by 5%

wair/e = 33.97
[tex]\mu[/tex]/[tex]\rho[/tex]=6.83x 10[tex]^{}-5[/tex]

50 x rays taken per day


Homework Equations



intensity =1/d^2

I (X) = I0e µx

exposure X = dq/dm = [tex]\psi[/tex].([tex]\mu[/tex]/[tex]\rho[/tex]).e/wair


The Attempt at a Solution



well, You have to find the intensity(no of photons)/radiation of the x ray beams after passing through the patient. Using this you can calculate the thickness needed using the linear attenutation coefficent.

What I dont get is how the expousre and effective energy relate to the intensity of the beam after passing through the patient and how this changes as the distance increases
 
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Answers and Replies

  • #2
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no one ?
 

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