How Thick Should a Concrete Wall Be to Safely Shield X-Rays in a Radiology Room?

AI Thread Summary
To determine the minimum thickness of a concrete wall for a radiology room, calculations must consider the exposure level of X-rays, which is 4.29 x 10^-4 at 50 cm from the source, and the effective energy of the X-ray beam at 40 keV. Given that 40% of X-rays are transmitted and the dose limit is 7.5 mGy per annum, the linear attenuation coefficient of 1.16 cm is crucial for determining the wall thickness. The intensity of the X-ray beam decreases with distance, and scattered radiation contributes an additional 5% to the dose. Understanding the relationship between exposure, effective energy, and intensity after passing through the patient is essential for accurate calculations.
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Homework Statement



Calculate minimum thickness of concrete wall of radiation room (2X2X2) such that radiographer doesn't exceed dose limit. Use info below.

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

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
\mu/\rho=6.83x 10^{}-5

50 x rays taken per day

Homework Equations



intensity =1/d^2

I (X) = I0e µx

exposure X = dq/dm = \psi.(\mu/\rho).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 don't 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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