Page 55 - ARNM-3-2

Advances in Radiotherapy
& Nuclear Medicine Shielding exaggeration in medical linac bunkers

Table 3. The expected equivalent dose received by personnel behind the primary barriers B and B’
Barrier Dose limit Dose constraint Expected annual dose % of annual dose constraint
B 1 mSv/year 1 mSv/year 40 patient/day: 71.7 µSv 100×71.1/1,000=7.17
60 patients/day: 134.2 µSv 100×134.2/1,000=13.42
B’ 20 mSv/year 6 mSv/year 40 patient/day: 71.45 µSv 100×71.1/6,000=1.191
60 patients/day: 134 µSv 100×134.2/6,000=2.23

(i) using a 40 × 40 cm field size instead of the average To reduce this value to the weekly dose constraints for
20 × 20 cm and (ii) assigning a use factor of 1 for secondary the public, the transmission factor becomes B = 4.28 × 10 .
−4
barriers adjacent to primary barriers. Therefore, the first Accordingly, the number of TVLs needed to achieve this
assumption was retained, while the second was revised to value is 3.28; considering that the TVL leakage for 6 MV
reflect practical clinical conditions better, as previously is 26.1 cm, the secondary barrier should be 0.86 m thick.
indicated. A simple calculation shows that the total patient scatter
radiation from 6 and 10 MV at the existing slant thickness
3.3.1. Case 1: Working load of 40 patients/day (1.14 m) equals 4.75 µSv/week.
The weekly unshielded leakage dose at 10 MV (20 patients) Accordingly, the total weekly dose from leakage and
is presented in Equation XV. scatter radiation at point C equals 10.7 µSv, less than the
855 × 0.001/5.12 = 32.6 mSv/week (XV) weekly dose constraints, 20 µSv/week. Hence, the existing
2
The transmission factor was deemed as B = 6.13 × 10 thickness is adequate for the workload of 40 patients/day.
−4
to reduce this value to the weekly dose constraints for the 3.3.2. Case 2: Working load of 60 patients/day
public (20 µSv/week). Accordingly, the number of TVLs
needed to achieve this value is 3.21; considering that the The weekly unshielded leakage dose at 10 MV (40 patients)
TVL leakage for 10 MV is 30.5 cm, the secondary barrier is presented in Equation XIX.
should be 0.98 m thick. 1,710 × 0.001/5.12 = 65.2 mSv/week (XIX)
2
The weekly unshielded leakage dose at 6 MV To reduce this value to the weekly dose constraints for
(20 patients) is presented in Equation XVI. the public, the transmission factor becomes B = 3.06 × 10 .
−4
855 × 0.001/5.12 = 32.6 mSv/week (XVI) Accordingly, the number of TVLs needed to achieve this
2
value is 3.51; considering that the TVL leakage for 10 MV
Similarly, the transmission factor becomes B = 6.13 × is 30.5 cm, the secondary barrier should be 1.07 m thick.
10 to reduce the obtained value. The number of TVLs
−4
needed to achieve this value is 3.21; considering that the The weekly unshielded dose at 6 MV (20 patients) is
TVL leakage for 6 MV is 27.9 cm, the secondary barrier presented in Equation XX.
should be 0.90 m thick. A simple calculation shows that the 855 × 0.001/5.12 = 32.6 mSv/week (XX)
2
total leakage radiation from 6 and 10 MV at the existing
−4
slant thickness (1.14 m) equals 5.95 µSv/week. The transmission factor becomes B = 6.14 × 10 to
reduce this value to the weekly dose constraints for the
The weekly unshielded patient scatter dose at 10 MV public. Hence, the number of TVLs needed to achieve
(20 patients) is presented in Equation XVII. this value is 3.21. Given that the TVL leakage for 6 MV
W × T × U × a × F /400 × d =360 × 1 × 0.25 × 3.18 × 10 −3 is 27.9 cm, the secondary barrier should be 0.90 m
2
2
× 1,600/400 × 5.12 = 43.7 mSv/week (XVII) thick. A simple calculation shows that the total leakage
2
radiation from 6 and 10 MV at the existing slant thickness
To reduce this value to the weekly dose constraints for
−4
the public, the transmission factor becomes B = 4.57 × 10 . is 14.57 µSv/week.
Accordingly, the number of TVLs needed to achieve this value The weekly unshielded patient scatter dose at 10 MV
is 3.34; taking into consideration that the TVL leakage for (40 patients) is presented in Equation XXI.
10 MV is 27.5 cm, the secondary barrier should be 0.92 m thick. 720 × 1 × 0.25 × 3.18×10 × 1,600/400 × 5.12 = 87.3 mSv/
2
−3
The weekly unshielded patient scatter dose at 6 MV week (XXI)
(20 patients) is presented in Equation XVIII. To reduce this value to the weekly dose constraints for
−4
W × T × U × a × F /400 × d =360 × 1 × 0.25 × 2.77 × 10 -3 the public, the transmission factor becomes B = 2.3 × 10 .
2
2
× 1,600/400 × 5.12 = 38 mSv/week (XVIII) Accordingly, the number of TVLs needed to achieve this
2
Volume 3 Issue 2 (2025) 47 doi: 10.36922/ARNM025070007