Question

1-A shielded vial of I131 gives off a reading of 30 mR/hr at 1 meter. If the vial is placed in a drawer lined with 5mm lead, what will the exposure rate be at 1-meter?

(The HVL of lead for I131= 0.21mm)

2-A survey meter reading taken outside the shield surrounding a radionuclide storage area gives a reading of 12.8 mR/hr. Only 99Tcm is being stored. The HVL for lead is 0.27mm. How much lead must be added to bring the reading down to a background reading of 0.02mR/hr?

3-A shielded dose of Tl201 registers as 25mR/hr on an ionization survey meter held at 4 inches from the surface. If the HVL for lead is 0.2mm, what will be the reading at 4 inches after the shielded dose is placed in a lead pig constructed of 1.35 mm thick lead?

4-An unshielded dose of 99mTc MAA gives off a reading of 35mR/hr. You put this dose in a syringe shield which has a thickness of 1.8mm. What would be the exposure when you use this shield while injecting the dose? The HVL of Tc99m =0.27mm.

5-The HVL of lead is 0.27 mm for Tc99m. A dose generates 5300 mR/hr before being shielded.

a-What will be the exposure rate after the dose is placed in a shield made with 0.90 mm of lead?

A thicker shield is used for the same dose above (5300 mR/hr). The shield is made with 3.8 mm of lead.

b-What will be the exposure rate after using this shielding?

6-Someone spills a dose of 99mTc-MIBI on the treadmill and after removing all of the “removable contamination” there is still an exposure of 45mR/hr coming from the treadmill. You need to cover this so that the stress lab can still be used, so this exposure NEEDS to be 0.3mR/hr. How much lead will you need to accomplish this?

7-A radioactive source is giving off 1.5 mr/hr at 1 meter after shielding. Three half value layers are used for shielding. What was the original reading of this source before shielding?

8-A radioactive source is measuring 2 mr/hr at a distance of 1 meter. This source has been shielded with 4 half value layers of lead. What would the reading of the source be if you removed all of the lead?

Answer 1

(1) We will first calculate the number of half value layers, this can be calculated by the equation,

After every pass from one half value layer the intensity reduces by (1/2) Therefore intensity of the radiation after passing through n layers is, (I0 is initial intensity).

Substituting the values we get, The exposure rate at 1 meter is,

(Since the reading is taken at the same distance there will be no change in intensity due to change in distance)

(2) The given reading is 12.8 mR/hr, and we have to bring down the reading to 0.02 mR/hr. Let n be the number of half value layers. Then we can write,

Here x is the thickness of layer and HVL is the value for half value layer, Substituting the values we get,

Therefore the thickness of the lead should be 2.517 mm to bring down the reading.

(3) We will use the same method as used in part (1)

The number of half value layers are,

The intensity of radiation coming out will be given by,

Substituting the values we get,

(Since the reading is taken at the same distance there will be no change in intensity due to change in distance)

(4) This question can also be solved by the same method as above,

The number of half value layers is,

Therefore the intensity of the radiation coming out of this layer is,

Substituting the values we get,