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a 50 gbq point isotropic source emitting 1-mev photons with a frequency of 0.75 photons per transformation is laying on top of a laminate shield consisting of 5 cm of lead and 10 cm of iron. the lead layer is in contact with the source. find the maximum total flux density in (cm-2 s-1) on the opposite side of the shield. use on the berger form of the buildup factor.

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Final answer:

To find the maximum total flux density on the opposite side of the shield, we need to calculate the buildup factor using the Berger form and then use it in a formula to calculate the flux density. The maximum total flux density is approximately 1.918 GBq/cm².

Step-by-step explanation:

To calculate the maximum total flux density on the opposite side of the shield, we need to first calculate the buildup factor using the Berger form. The buildup factor is given by the formula:

B = ln(1 + Σexb),

where ln is the natural logarithm, Σ represents the summation, and exb is the exposure per transformation. In this case, the exposure per transformation is 1 MeV x 0.75 photons per transformation = 0.75 MeV. Therefore, the buildup factor is:

B = ln(1 + 50 x 0.75) ≈ 3.806.

Next, we can calculate the maximum total flux density using the formula:

Flux Density = (Activity of source / Area of shield) x B,

where the activity of the source is given as 50 GBq, and the shield consists of 5 cm of lead and 10 cm of iron. The area of shield can be calculated as:

Area of shield = π x (radius of shield)^2,

where the radius of the shield is equal to the thickness of the lead layer. Therefore, the area of shield is given by:

Area of shield = π x (5 cm)^2 = 78.54 cm².

Finally, substituting the values into the formula:

Flux Density = (50 GBq / 78.54 cm²) x 3.806 ≈ 1.918 GBq/cm².

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