Medical Physics (a) If a material has a half-value layer of...

Medical Physics (a) If a material has a half-value layer of 3.0 mm, what is its linear attenuation coefficient? (b) By how much is the intensity of a 100 keV x-ray beam reduced if it travels through 0.1mm of lead? Assume the Half Value Layer (HVL) of lead is 0.12 mm. (c) What thickness of aluminium (HVL = 1.59 cm) is required to give the same protection from 100keV x-rays as 0.1mm lead (HVL = 0.12mm)? (d) If a person is 1m away from an x-ray, by how much is the x-ray intensity decreased if they move to 3m away from the source?

Transcript

00:01 Okay, so question a asks, what is the linear attenuation coefficient of a material with half -value layer of 3mm.

00:11 So, the linear -attenuation coefficient is related to half -bil by this formula.

00:15 L -n -of -2 over hbl is the linear -tenvation coefficient.

00:18 So, l -n -of -2 over 3 -millimeter, this equals to 0 .23 millimeter raised to negative 1.

00:28 So this is the linear -atimation coefficient of a material that value of 3 -millimeter.

00:33 Now, question we asked, what is the reduced in intensity of 100 kilo -electrovolvex rays in lead with thickness of 0 .1 millimeter? and the half -value layer of lead in a hundred -culelelelelele of x -ray is 0 .12 millimeter.

00:53 So we released a formula for intensity of radiation.

00:56 So the current density is equal to the original intensity multiplied by e -rays to the negative of the product of the linear automation coefficient and the thickness.

01:06 Then, equation, so if was this, from part 8, then solving, we have e -raised to negative nl2 over 0 .12 millimeter multiplied by 0 .1 millimeter.

01:29 Then this is equals to 0 .56 of the original intensity.

01:36 So the present intensity, the current sense is that 3 .506.

01:41 I mean 0 .56 of the original, so 56 % of the original.

01:49 Therefore, the reduced in intensity is 44 % since only 56 % remains, so 100 minus 56.

02:02 Now, question she asked, what would be the thickness, the aluminum that will provide the same protection as lead with the thickness of 0 .1 millimeter, given that the half -value layer of lead, is, is 0 .12 millimeter and the half value layer of aluminum is 1 .59 centimeters.

02:30 So to have the same section and then the ratio of the current intensity with the original intensity if it pressed to aluminum is equals to the ratio if it passed through lead.

02:52 We have this...

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