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Introduction to Health Physics: Fourth Edition - Ruang Baca FMIPA UB

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528 CHAPTER 10<br />

The HVL of lead for 2.75-MeV gammas is 1.43 cm. To account for the 1.37-MeV<br />

gamma, let us add one-half an HVL, or 0.72 cm, <strong>to</strong> the thickness calculated above<br />

for the 2.75-MeV gamma. We will then calculate the attenuation of the 2.75-MeV<br />

gamma with the new trial thickness, 10.16 + 0.72 = 10.88 cm.<br />

−0.485 × 10.88<br />

I = 1380 × e<br />

I = 7.1mR/h.<br />

With this thickness of lead, the 1.37-MeV gamma, whose attenuation coefficient is<br />

0.629 cm −1 (Table 5-2) will be attenuated <strong>to</strong><br />

−0.629 × 10.88<br />

I = 690 × e<br />

I = 0.7mR/h,<br />

and the <strong>to</strong>tal exposure rate at a distance of 1 m will be the sum of the exposure rates<br />

due <strong>to</strong> the two quantum energies, or 7.8 mR/h.<br />

The calculation above is based on good geometry. Let us now account for buildup<br />

in the shield. Considering at this time only the high-energy pho<strong>to</strong>n, let us add<br />

another half of one HVL <strong>to</strong> the shield thickness, which gives us 11.6 cm. Since<br />

one relaxation length of lead for 2.75-MeV gammas is 2.1 cm, the shield thickness<br />

corresponds <strong>to</strong> 5.5 relaxation lengths. From Figure 10-9, we find the buildup fac<strong>to</strong>r<br />

<strong>to</strong> be 3.1. The attenuation of the shield, therefore, according <strong>to</strong> Eq. (10.17), is<br />

I = 1380 × 3.1 × e −0.485×11.6<br />

I = 15.4 mR/h.<br />

This exposure rate is <strong>to</strong>o high; the shield thickness must therefore be increased. If we<br />

add another HVL <strong>to</strong> the shield <strong>to</strong> give us 13 cm, or 6.3 relaxation lengths, and using<br />

the corresponding dose buildup fac<strong>to</strong>r of 3.4, we find the exposure rate <strong>to</strong> be 8.5<br />

mR/h for the high-energy pho<strong>to</strong>n. For the lower-energy pho<strong>to</strong>n, whose relaxation<br />

length in lead is 1.6 cm, this shield thickness corresponds <strong>to</strong> 8.2 relaxation lengths,<br />

and the dose buildup fac<strong>to</strong>r is 3.6. With these values in Eq. (10.17), the exposure<br />

rate due <strong>to</strong> the 1.37-MeV pho<strong>to</strong>ns is calculated as 0.7 mR/h. The <strong>to</strong>tal exposure<br />

rate at a distance of 1 m from the shielded 24 Na source is thus 9.2 mR/h. Since this<br />

rate may be considered, for most practical purposes, <strong>to</strong> be equivalent <strong>to</strong> the design<br />

value of 10 mR/h, the required shield thickness is 13 cm. Since the shield may be<br />

interposed anywhere between the source and the point where the desired attenuated<br />

dose rate is located, and since the volume of lead, for a given wall thickness in<br />

a spherical shield, increases rapidly with increasing outer radius according <strong>to</strong> the<br />

expression<br />

Volume = 4<br />

3 π(r 3 0 − r 3 1 ),<br />

the inner radius of the shield is kept as small as possible, consistent with the space<br />

requirements set by the physical dimensions of the source. The outside radius then<br />

is equal <strong>to</strong> the sum of the inside radius and the shield thickness.

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