If the original gamma intensity is 211 mr/hr for 99mTc, the attenuator (Pb) is 2.5 mm in length, and attenuation coefficient is 23 what is the new gamma intensity? This calculation is similar to the decay formula
Let's take a look at the different ways you can use the attenuation formula
- While this is an approach used to determining the attenuation of gamma radiation after shielding, in nuclear medicine we can also apply the concept of half-value layer (HVL). This concept ties into the shielding formula noted above, however, it is "more basic" in its approach.
- To have an attenuator that has a value of 1 HVL means that it reduces the intensity of radiation by 1/2. Similar to the value x in the formula above where the density of the shield determines the HVL. HVL is also determined by the amount of energy being emitted by the incoming gamma. In addition, the greater the Eγ the need more shielding and/or greater density is required Below are some HVL values
HVL - Attenuator in gm/cm3 |
Pb - 11.34 |
H2O - 1.0 |
NaI - 3.67 |
140 keV |
0.0256 cm |
4.53 cm |
0.278 cm |
364 keV |
0.219 cm |
6.245 cm |
1.51 cm |
511 keV |
0.4021 cm |
7.22 cm |
2.12 cm |
- Therefore, to reduce an incoming gamma by 50% with an Eγ of 140 keV you would need 0.256 mm of lead. How much NaI would you need to reduce a positron gamma to 12.5%?
- Let's first look at HVLs (the easy way). The original rate of exposure for 99mTc is 734.5 mr/hr. If 1.24 mm of Pb is used as a shielding device. What is the new rate of exposure?
- See PDF file - contains the different formulas for attenuation and decay
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